output
stringlengths 52
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stringlengths 296
182k
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|---|---|
#include <bits/stdc++.h>
using namespace std;
void qmax(int &x, int y) {
if (x < y) x = y;
}
void qmin(long long &x, long long y) {
if (x > y) x = y;
}
inline long long read() {
char s;
long long k = 0, base = 1;
while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
;
if (s == EOF) exit(0);
if (s == '-') base = -1, s = getchar();
while (isdigit(s)) {
k = k * 10 + (s ^ '0');
s = getchar();
}
return k * base;
}
inline void write(int x) {
static char cnt, num[15];
cnt = 0;
if (!x) {
putchar('0');
return;
}
for (; x; x /= 10) num[++cnt] = x % 10;
for (; cnt; putchar(num[cnt--] + 48))
;
}
int n, m, q;
char b[1010], a[1010];
int dp[1010][1010];
int Next[1010];
int main() {
n = read(), m = read(), q = read();
scanf("%s", b + 1);
scanf("%s", a + 1);
int p;
Next[1] = 0;
p = 0;
for (int i = 2; i <= m; i++) {
while (p && a[p + 1] != a[i]) p = Next[p];
if (a[p + 1] == a[i]) {
p++;
Next[i] = p;
}
}
p = 0;
for (int i = 1; i <= n; i++) {
while ((p == m) || (p && a[p + 1] != b[i])) p = Next[p];
if (a[p + 1] == b[i]) {
p++;
}
if (p == m) {
dp[i - m + 1][i] = 1;
}
}
for (int l = m; l < n; l++)
for (int i = 1, j = i + l; j <= n; i++, j++) {
dp[i][j] = dp[i][j - 1];
if (j - m + 1 > 0) dp[i][j] = dp[i][j - 1] + dp[j - m + 1][j];
}
int X, Y;
while (q--) {
X = read();
Y = read();
printf("%d\n", dp[X][Y]);
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void qmax(int &x, int y) {
if (x < y) x = y;
}
void qmin(long long &x, long long y) {
if (x > y) x = y;
}
inline long long read() {
char s;
long long k = 0, base = 1;
while ((s = getchar()) != '-' && s != EOF && !(isdigit(s)))
;
if (s == EOF) exit(0);
if (s == '-') base = -1, s = getchar();
while (isdigit(s)) {
k = k * 10 + (s ^ '0');
s = getchar();
}
return k * base;
}
inline void write(int x) {
static char cnt, num[15];
cnt = 0;
if (!x) {
putchar('0');
return;
}
for (; x; x /= 10) num[++cnt] = x % 10;
for (; cnt; putchar(num[cnt--] + 48))
;
}
int n, m, q;
char b[1010], a[1010];
int dp[1010][1010];
int Next[1010];
int main() {
n = read(), m = read(), q = read();
scanf("%s", b + 1);
scanf("%s", a + 1);
int p;
Next[1] = 0;
p = 0;
for (int i = 2; i <= m; i++) {
while (p && a[p + 1] != a[i]) p = Next[p];
if (a[p + 1] == a[i]) {
p++;
Next[i] = p;
}
}
p = 0;
for (int i = 1; i <= n; i++) {
while ((p == m) || (p && a[p + 1] != b[i])) p = Next[p];
if (a[p + 1] == b[i]) {
p++;
}
if (p == m) {
dp[i - m + 1][i] = 1;
}
}
for (int l = m; l < n; l++)
for (int i = 1, j = i + l; j <= n; i++, j++) {
dp[i][j] = dp[i][j - 1];
if (j - m + 1 > 0) dp[i][j] = dp[i][j - 1] + dp[j - m + 1][j];
}
int X, Y;
while (q--) {
X = read();
Y = read();
printf("%d\n", dp[X][Y]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int N, M, Q;
string S, T;
int ok[1010];
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> N >> M >> Q >> S >> T;
for (i = 0; i < (N); i++) {
ok[i + 1] = ok[i];
if (i + M <= N) {
x = 1;
for (j = 0; j < (M); j++)
if (S[i + j] != T[j]) x = 0;
ok[i + 1] += x;
}
}
while (Q--) {
cin >> x >> y;
y -= M - 1;
if (y < x) {
cout << 0 << endl;
} else {
cout << ok[y] - ok[x - 1] << endl;
}
}
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int N, M, Q;
string S, T;
int ok[1010];
void solve() {
int i, j, k, l, r, x, y;
string s;
cin >> N >> M >> Q >> S >> T;
for (i = 0; i < (N); i++) {
ok[i + 1] = ok[i];
if (i + M <= N) {
x = 1;
for (j = 0; j < (M); j++)
if (S[i + j] != T[j]) x = 0;
ok[i + 1] += x;
}
}
while (Q--) {
cin >> x >> y;
y -= M - 1;
if (y < x) {
cout << 0 << endl;
} else {
cout << ok[y] - ok[x - 1] << endl;
}
}
}
int main(int argc, char** argv) {
string s;
int i;
if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
cout.tie(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
int n, m, q, l, r;
cin >> n >> m >> q;
cin >> s >> t;
int arr[10000] = {0};
for (int i = 0; i <= n - m; ++i) {
arr[i + 1] = arr[i] + (s.substr(i, m) == t);
}
while (q--) {
cin >> l >> r;
if (r - l + 1 < m)
cout << "0\n";
else
cout << arr[r - m + 1] - arr[l - 1] << endl;
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
string s, t;
int n, m, q, l, r;
cin >> n >> m >> q;
cin >> s >> t;
int arr[10000] = {0};
for (int i = 0; i <= n - m; ++i) {
arr[i + 1] = arr[i] + (s.substr(i, m) == t);
}
while (q--) {
cin >> l >> r;
if (r - l + 1 < m)
cout << "0\n";
else
cout << arr[r - m + 1] - arr[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, pre[2005], q;
char s[2005], t[2005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s, t);
for (int i = 0; i < n - m + 1; i++) {
int flag = 1;
for (int j = 0; j < m; j++)
if (s[i + j] != t[j]) {
flag = 0;
break;
}
if (flag) pre[i + 1]++;
}
for (int i = 1; i <= n; i++) pre[i] += pre[i - 1];
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", pre[max(r - m + 1, l - 1)] - pre[l - 1]);
}
return 0;
}
| ### Prompt
Your task is to create a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, pre[2005], q;
char s[2005], t[2005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s, t);
for (int i = 0; i < n - m + 1; i++) {
int flag = 1;
for (int j = 0; j < m; j++)
if (s[i + j] != t[j]) {
flag = 0;
break;
}
if (flag) pre[i + 1]++;
}
for (int i = 1; i <= n; i++) pre[i] += pre[i - 1];
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", pre[max(r - m + 1, l - 1)] - pre[l - 1]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, Q, cnt = 0;
int a[1010], L[1010];
char s[1010], t[1010];
int main() {
scanf("%d%d%d", &n, &m, &Q);
scanf("%s%s", s, t);
for (int i = 0; i < n; i++) {
if (s[i] == t[0]) {
bool bo = 0;
for (int j = 1; j < m; j++) {
if (s[i + j] != t[j]) {
bo = 1;
break;
}
}
if (!bo) {
a[cnt] = i + 1;
L[i + 1] = cnt;
cnt++;
}
}
}
int x, y;
while (Q--) {
scanf("%d%d", &x, &y);
int sum = 0;
for (int i = 0; i < cnt; i++) {
if (a[i] >= x && a[i] + m - 1 <= y) sum++;
if (a[i] > y) break;
}
printf("%d\n", sum);
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, Q, cnt = 0;
int a[1010], L[1010];
char s[1010], t[1010];
int main() {
scanf("%d%d%d", &n, &m, &Q);
scanf("%s%s", s, t);
for (int i = 0; i < n; i++) {
if (s[i] == t[0]) {
bool bo = 0;
for (int j = 1; j < m; j++) {
if (s[i + j] != t[j]) {
bo = 1;
break;
}
}
if (!bo) {
a[cnt] = i + 1;
L[i + 1] = cnt;
cnt++;
}
}
}
int x, y;
while (Q--) {
scanf("%d%d", &x, &y);
int sum = 0;
for (int i = 0; i < cnt; i++) {
if (a[i] >= x && a[i] + m - 1 <= y) sum++;
if (a[i] > y) break;
}
printf("%d\n", sum);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
map<long long, long long> d;
long long cont = 0;
for (long long i = (0); i < (long long)(n); i++) {
if (s.substr(i, m) == t) {
d[i] = ++cont;
}
}
while (q--) {
long long l, r;
cin >> l >> r;
l--;
r--;
auto it = d.lower_bound(l);
auto jt = d.upper_bound(r - m + 1);
if (jt == d.begin()) {
puts("0");
continue;
}
jt--;
long long x = jt->second - it->second + 1;
if (x < 0) x = 0;
printf("%lld\n", x);
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
long long n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
map<long long, long long> d;
long long cont = 0;
for (long long i = (0); i < (long long)(n); i++) {
if (s.substr(i, m) == t) {
d[i] = ++cont;
}
}
while (q--) {
long long l, r;
cin >> l >> r;
l--;
r--;
auto it = d.lower_bound(l);
auto jt = d.upper_bound(r - m + 1);
if (jt == d.begin()) {
puts("0");
continue;
}
jt--;
long long x = jt->second - it->second + 1;
if (x < 0) x = 0;
printf("%lld\n", x);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int fx[] = {-1, 1, 0, 0};
int fy[] = {0, 0, -1, 1};
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int kx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ky[] = {2, -2, 2, -2, 1, -1, 1, -1};
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return a.first < b.first;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
s = '?' + s;
int ara[1010], sum[1010];
memset(ara, 0, sizeof(ara)), memset(sum, 0, sizeof(sum));
for (int i = 1; i <= n - m + 1; i++) {
ara[i] = 1;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
ara[i] = 0;
break;
}
}
sum[i] = sum[i - 1] + ara[i];
}
for (int i = 1; i <= q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0\n";
} else {
cout << sum[r - m + 1] - sum[l - 1] << "\n";
}
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int fx[] = {-1, 1, 0, 0};
int fy[] = {0, 0, -1, 1};
int dx[] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy[] = {0, 1, 1, 1, 0, -1, -1, -1};
int kx[] = {1, 1, -1, -1, 2, 2, -2, -2};
int ky[] = {2, -2, 2, -2, 1, -1, 1, -1};
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
return a.first < b.first;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
s = '?' + s;
int ara[1010], sum[1010];
memset(ara, 0, sizeof(ara)), memset(sum, 0, sizeof(sum));
for (int i = 1; i <= n - m + 1; i++) {
ara[i] = 1;
for (int j = 0; j < t.size(); j++) {
if (s[i + j] != t[j]) {
ara[i] = 0;
break;
}
}
sum[i] = sum[i - 1] + ara[i];
}
for (int i = 1; i <= q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0\n";
} else {
cout << sum[r - m + 1] - sum[l - 1] << "\n";
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int main() {
int n, m, q;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
int mark[1000];
memset(mark, 0, sizeof(mark));
for (int i = 0; i < n; i++) {
if (a[i] == b[0]) {
int flag = 1, j;
for (j = 1; j < m && i + j < n; j++) {
if (a[i + j] != b[j]) {
flag = 0;
break;
}
}
if (flag && j == m) mark[i] = 1;
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int cnt = 0;
for (int i = l - 1; i <= r - 1; i++) {
if (mark[i] == 1 && i + m - 1 <= r - 1) cnt++;
}
cout << cnt << endl;
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int main() {
int n, m, q;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
int mark[1000];
memset(mark, 0, sizeof(mark));
for (int i = 0; i < n; i++) {
if (a[i] == b[0]) {
int flag = 1, j;
for (j = 1; j < m && i + j < n; j++) {
if (a[i + j] != b[j]) {
flag = 0;
break;
}
}
if (flag && j == m) mark[i] = 1;
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int cnt = 0;
for (int i = l - 1; i <= r - 1; i++) {
if (mark[i] == 1 && i + m - 1 <= r - 1) cnt++;
}
cout << cnt << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 5;
char a[maxn], b[maxn];
int sum[maxn];
int n, m, q;
void solve() {
for (int i = 0; i < n; i++) {
int flag = 1;
for (int j = 0; j < m; j++)
if (a[i + j] != b[j]) {
flag = 0;
break;
}
if (flag) sum[i + 1]++;
sum[i + 1] += sum[i];
}
}
int main() {
scanf("%d %d %d", &n, &m, &q);
scanf(" %s %s", a, b);
solve();
while (q--) {
int l, r;
scanf("%d %d", &l, &r);
if (r - m + 1 >= 0 && r - m + 1 >= l - 1)
printf("%d\n", sum[r - m + 1] - sum[l - 1]);
else
puts("0");
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e3 + 5;
char a[maxn], b[maxn];
int sum[maxn];
int n, m, q;
void solve() {
for (int i = 0; i < n; i++) {
int flag = 1;
for (int j = 0; j < m; j++)
if (a[i + j] != b[j]) {
flag = 0;
break;
}
if (flag) sum[i + 1]++;
sum[i + 1] += sum[i];
}
}
int main() {
scanf("%d %d %d", &n, &m, &q);
scanf(" %s %s", a, b);
solve();
while (q--) {
int l, r;
scanf("%d %d", &l, &r);
if (r - m + 1 >= 0 && r - m + 1 >= l - 1)
printf("%d\n", sum[r - m + 1] - sum[l - 1]);
else
puts("0");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int ans[1005];
int main() {
while (scanf("%d%d%d", &n, &m, &q) != EOF) {
string s1, s2;
cin >> s1 >> s2;
for (int i = 0; i + m <= n; i++) {
if (s1.substr(i, m) == s2) ans[i + 1]++;
}
for (int i = 2; i <= n; i++) ans[i] += ans[i - 1];
while (q--) {
int li, ri;
scanf("%d%d", &li, &ri);
ri = ri - m + 1;
printf("%d\n", max(0, ans[max(0, ri)] - ans[li - 1]));
}
}
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int ans[1005];
int main() {
while (scanf("%d%d%d", &n, &m, &q) != EOF) {
string s1, s2;
cin >> s1 >> s2;
for (int i = 0; i + m <= n; i++) {
if (s1.substr(i, m) == s2) ans[i + 1]++;
}
for (int i = 2; i <= n; i++) ans[i] += ans[i - 1];
while (q--) {
int li, ri;
scanf("%d%d", &li, &ri);
ri = ri - m + 1;
printf("%d\n", max(0, ans[max(0, ri)] - ans[li - 1]));
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[1002], t[1002];
int v[1002];
int main() {
int n, m, q, l, r, sol, i, ok, j;
cin >> n >> m >> q;
cin.get();
cin.getline(s, 1001);
cin.getline(t, 1001);
for (i = 0; i < n - m + 1; i++) {
ok = 1;
for (j = i; j <= i + m - 1; j++)
if (t[j - i] != s[j]) {
ok = 0;
break;
}
if (ok == 1) v[i + m - 1] += 1;
}
for (i = 1; i < n; i++) v[i] += v[i - 1];
for (i = 1; i <= q; i++) {
cin >> l >> r;
l--;
r--;
l += m - 1;
if (l > r)
cout << 0 << '\n';
else {
sol = v[r] - v[l - 1];
cout << sol << '\n';
}
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[1002], t[1002];
int v[1002];
int main() {
int n, m, q, l, r, sol, i, ok, j;
cin >> n >> m >> q;
cin.get();
cin.getline(s, 1001);
cin.getline(t, 1001);
for (i = 0; i < n - m + 1; i++) {
ok = 1;
for (j = i; j <= i + m - 1; j++)
if (t[j - i] != s[j]) {
ok = 0;
break;
}
if (ok == 1) v[i + m - 1] += 1;
}
for (i = 1; i < n; i++) v[i] += v[i - 1];
for (i = 1; i <= q; i++) {
cin >> l >> r;
l--;
r--;
l += m - 1;
if (l > r)
cout << 0 << '\n';
else {
sol = v[r] - v[l - 1];
cout << sol << '\n';
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
vector<int> Zalgo(const string& s) {
int n = (int)s.size();
vector<int> res(n);
res[0] = n;
int i = 1, j = 0;
while (i < n) {
while (i + j < n && s[j] == s[i + j]) ++j;
res[i] = j;
if (j == 0) {
++i;
continue;
}
int k = 1;
while (i + k < n && k + res[k] < j) res[i + k] = res[k], ++k;
i += k, j -= k;
}
return res;
}
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
string u = t + s;
auto res = Zalgo(u);
vector<int> cnt(n + 1, 0);
for (int i = 0; i < n; i++) {
cnt[i + 1] = cnt[i] + (res[m + i] >= m ? 1 : 0);
}
for (int _ = 0; _ < q; _++) {
int l, r;
cin >> l >> r;
cout << cnt[max(r - m + 1, l - 1)] - cnt[l - 1] << endl;
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
solve();
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
const long long INF = 1LL << 60;
const long long mod = 1e9 + 7;
const long double eps = 1e-8;
const long double pi = acos(-1.0);
template <class T>
inline bool chmax(T& a, T b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
vector<int> Zalgo(const string& s) {
int n = (int)s.size();
vector<int> res(n);
res[0] = n;
int i = 1, j = 0;
while (i < n) {
while (i + j < n && s[j] == s[i + j]) ++j;
res[i] = j;
if (j == 0) {
++i;
continue;
}
int k = 1;
while (i + k < n && k + res[k] < j) res[i + k] = res[k], ++k;
i += k, j -= k;
}
return res;
}
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
string u = t + s;
auto res = Zalgo(u);
vector<int> cnt(n + 1, 0);
for (int i = 0; i < n; i++) {
cnt[i + 1] = cnt[i] + (res[m + i] >= m ? 1 : 0);
}
for (int _ = 0; _ < q; _++) {
int l, r;
cin >> l >> r;
cout << cnt[max(r - m + 1, l - 1)] - cnt[l - 1] << endl;
}
}
signed main() {
ios::sync_with_stdio(false);
cin.tie(0);
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void z_function(string &s, vector<int> &z) {
int n = s.length(), l = -1, r = -1;
z.clear();
z.resize(n, 0);
z[0] = n;
for (int i = 0; i < n; i++) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
if (z[i] + i - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
}
int n, m, q, l, r;
string s, t;
vector<int> z, pr;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
n = s.length();
m = t.length();
s = t + '#' + s;
z_function(s, z);
pr.resize(n, 0);
if (z[m + 1] == m) pr[0] = 1;
for (int i = 1; i < n; i++) {
pr[i] = pr[i - 1];
if (z[m + 1 + i] == m) pr[i]++;
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
l--;
r--;
if (r - m + 1 < 0) {
cout << 0 << endl;
continue;
}
if (l == 0)
cout << pr[r - m + 1];
else
cout << max(pr[r - m + 1] - pr[l - 1], 0);
cout << endl;
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void z_function(string &s, vector<int> &z) {
int n = s.length(), l = -1, r = -1;
z.clear();
z.resize(n, 0);
z[0] = n;
for (int i = 0; i < n; i++) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (z[i] + i < n && s[z[i]] == s[z[i] + i]) z[i]++;
if (z[i] + i - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
}
int n, m, q, l, r;
string s, t;
vector<int> z, pr;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
n = s.length();
m = t.length();
s = t + '#' + s;
z_function(s, z);
pr.resize(n, 0);
if (z[m + 1] == m) pr[0] = 1;
for (int i = 1; i < n; i++) {
pr[i] = pr[i - 1];
if (z[m + 1 + i] == m) pr[i]++;
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
l--;
r--;
if (r - m + 1 < 0) {
cout << 0 << endl;
continue;
}
if (l == 0)
cout << pr[r - m + 1];
else
cout << max(pr[r - m + 1] - pr[l - 1], 0);
cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e3;
int sum[MX + 2];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m, q, l, r;
string str, str2;
cin >> n >> m >> q;
cin >> str >> str2;
for (long long i = 0; i < n - m + 1; i++) {
sum[i] = (str.substr(i, m) == str2);
}
while (q-- > 0) {
cin >> l >> r;
l--, r--;
r = r - m + 1;
int ans = 0;
for (long long i = l; i <= r; i++) ans += sum[i];
cout << ans << '\n';
}
}
| ### Prompt
In Cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MX = 1e3;
int sum[MX + 2];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
long long n, m, q, l, r;
string str, str2;
cin >> n >> m >> q;
cin >> str >> str2;
for (long long i = 0; i < n - m + 1; i++) {
sum[i] = (str.substr(i, m) == str2);
}
while (q-- > 0) {
cin >> l >> r;
l--, r--;
r = r - m + 1;
int ans = 0;
for (long long i = l; i <= r; i++) ans += sum[i];
cout << ans << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> vet;
int p[201010];
void kmp(string &s, int sz) {
p[0] = 0;
int idx = 0, n = (int)s.size();
for (int i = 1; i < n; i++) {
while (s[idx] != s[i] && idx > 0) idx = p[idx - 1];
if (s[idx] == s[i]) idx++;
p[i] = idx;
if (p[i] == sz) {
vet.push_back(i - sz - sz + 1);
}
}
}
int solve(int l, int r) {
if (l > r) return 0;
int a = lower_bound(vet.begin(), vet.end(), l) - vet.begin();
int b = upper_bound(vet.begin(), vet.end(), r) - vet.begin();
return b - a;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, m, q, l, r;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
string aux = t + "$" + s;
kmp(aux, (int)t.size());
sort(vet.begin(), vet.end());
while (q--) {
cin >> l >> r;
cout << solve(l, r - (int)t.size() + 1) << "\n";
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> vet;
int p[201010];
void kmp(string &s, int sz) {
p[0] = 0;
int idx = 0, n = (int)s.size();
for (int i = 1; i < n; i++) {
while (s[idx] != s[i] && idx > 0) idx = p[idx - 1];
if (s[idx] == s[i]) idx++;
p[i] = idx;
if (p[i] == sz) {
vet.push_back(i - sz - sz + 1);
}
}
}
int solve(int l, int r) {
if (l > r) return 0;
int a = lower_bound(vet.begin(), vet.end(), l) - vet.begin();
int b = upper_bound(vet.begin(), vet.end(), r) - vet.begin();
return b - a;
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int n, m, q, l, r;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
string aux = t + "$" + s;
kmp(aux, (int)t.size());
sort(vet.begin(), vet.end());
while (q--) {
cin >> l >> r;
cout << solve(l, r - (int)t.size() + 1) << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e3 + 5;
int sum[N];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int n, m, q, l, r;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i + m - 1 < n; i++) {
bool f = true;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
f = false;
break;
}
}
if (f) sum[i + 1]++;
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
while (q--) {
cin >> l >> r;
r = r - m + 1;
if (r >= l)
cout << sum[r] - sum[l - 1] << endl;
else
cout << 0 << endl;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
using namespace std;
const int N = 1e3 + 5;
int sum[N];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
int n, m, q, l, r;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i + m - 1 < n; i++) {
bool f = true;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
f = false;
break;
}
}
if (f) sum[i + 1]++;
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
while (q--) {
cin >> l >> r;
r = r - m + 1;
if (r >= l)
cout << sum[r] - sum[l - 1] << endl;
else
cout << 0 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0), EPS = 1e-9;
const int INF = 1e9, NINF = -1e9, MOD = 1e9 + 7;
void GO() { cout << fixed << setprecision(10); }
int main() {
GO();
int n, m, q, r, l, c, a[1001] = {};
string second, t;
cin >> n >> m >> q >> second >> t;
for (int i = 0; i <= n - m; i++)
if (second.substr(i, m) == t) a[i] = 1;
while (q--) {
c = 0;
cin >> r >> l;
for (int i = r - 1; i <= l - m; i++)
if (a[i]) c++;
cout << c << "\n";
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const double PI = 2 * acos(0.0), EPS = 1e-9;
const int INF = 1e9, NINF = -1e9, MOD = 1e9 + 7;
void GO() { cout << fixed << setprecision(10); }
int main() {
GO();
int n, m, q, r, l, c, a[1001] = {};
string second, t;
cin >> n >> m >> q >> second >> t;
for (int i = 0; i <= n - m; i++)
if (second.substr(i, m) == t) a[i] = 1;
while (q--) {
c = 0;
cin >> r >> l;
for (int i = r - 1; i <= l - m; i++)
if (a[i]) c++;
cout << c << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1111;
string s, t;
int sum[N];
int n, m, q;
signed main() {
ios_base::sync_with_stdio(false);
cin >> n >> m >> q >> s >> t;
for (int i = (1); i <= (n); i++) {
bool found = true;
for (int j = (0); j <= (m - 1); j++) {
int ii = i - 1 + j;
if (ii < n && s[ii] != t[j]) {
found = false;
break;
}
}
sum[i] = sum[i - 1] + found;
}
while (q--) {
int l, r;
cin >> l >> r;
int ans = 0;
if (r - m + 1 >= 1) {
ans = max(0, sum[r - m + 1] - sum[l - 1]);
}
cout << ans << "\n";
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1111;
string s, t;
int sum[N];
int n, m, q;
signed main() {
ios_base::sync_with_stdio(false);
cin >> n >> m >> q >> s >> t;
for (int i = (1); i <= (n); i++) {
bool found = true;
for (int j = (0); j <= (m - 1); j++) {
int ii = i - 1 + j;
if (ii < n && s[ii] != t[j]) {
found = false;
break;
}
}
sum[i] = sum[i - 1] + found;
}
while (q--) {
int l, r;
cin >> l >> r;
int ans = 0;
if (r - m + 1 >= 1) {
ans = max(0, sum[r - m + 1] - sum[l - 1]);
}
cout << ans << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 1010;
char a[maxn], b[maxn];
int n, m, q;
vector<int> pos;
void solve() {
while (q--) {
int l, r, tr;
scanf("%d%d", &l, &r);
tr = r - m + 1;
if (l > tr)
puts("0");
else {
printf("%d\n",
upper_bound(pos.begin(), pos.end(), tr) - pos.begin() -
(lower_bound(pos.begin(), pos.end(), l) - pos.begin()));
}
}
}
int main() {
cin >> n >> m >> q;
cin >> a + 1;
cin >> b + 1;
for (int i = 1; i <= n; ++i) {
bool ok = true;
for (int j = 0; j < m; ++j) {
if (i + j > n || (a[i + j] != b[j + 1])) {
ok = false;
break;
}
}
if (ok) pos.push_back(i);
}
solve();
return 0;
}
| ### Prompt
Your task is to create a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const int maxn = 1010;
char a[maxn], b[maxn];
int n, m, q;
vector<int> pos;
void solve() {
while (q--) {
int l, r, tr;
scanf("%d%d", &l, &r);
tr = r - m + 1;
if (l > tr)
puts("0");
else {
printf("%d\n",
upper_bound(pos.begin(), pos.end(), tr) - pos.begin() -
(lower_bound(pos.begin(), pos.end(), l) - pos.begin()));
}
}
}
int main() {
cin >> n >> m >> q;
cin >> a + 1;
cin >> b + 1;
for (int i = 1; i <= n; ++i) {
bool ok = true;
for (int j = 0; j < m; ++j) {
if (i + j > n || (a[i + j] != b[j + 1])) {
ok = false;
break;
}
}
if (ok) pos.push_back(i);
}
solve();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, dp[10004];
string s, t;
int main() {
cin >> n >> m >> q >> s >> t;
for (int i = 0, ok = true; i <= n - m; ++i, ok = true) {
for (int j = 0; j < m; ++j) {
if (s[i + j] != t[j]) ok = false;
}
dp[i + 1] = dp[i] + ok;
}
for (int i = 0, l, r; i < q; ++i) {
cin >> l >> r;
cout << (l > r - m + 1 ? 0 : dp[r - m + 1] - dp[l - 1]) << endl;
}
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, dp[10004];
string s, t;
int main() {
cin >> n >> m >> q >> s >> t;
for (int i = 0, ok = true; i <= n - m; ++i, ok = true) {
for (int j = 0; j < m; ++j) {
if (s[i + j] != t[j]) ok = false;
}
dp[i + 1] = dp[i] + ok;
}
for (int i = 0, l, r; i < q; ++i) {
cin >> l >> r;
cout << (l > r - m + 1 ? 0 : dp[r - m + 1] - dp[l - 1]) << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
int occ[n];
int sta[n];
int sum[n];
for (int o = 0; o < n; o++) {
occ[o] = 0;
sum[o] = 0;
sta[o] = 0;
}
int v = 0;
for (int e = 0; e < n - m + 1; e++) {
int fl = 1;
for (int l = 0; l < m; l++) {
if (s1[e + l] != s2[l]) {
fl = 0;
break;
}
}
if (fl == 1) occ[e + m - 1] = 1;
}
for (int g = 0; g < n; g++) {
v = v + occ[g];
sum[g] = v;
}
for (int a = 1; a <= q; a++) {
int p, q;
cin >> p >> q;
if (q - p < m - 1)
cout << "0" << endl;
else if (p != 1)
cout << sum[q - 1] - sum[p - 2 + m - 1] << endl;
else
cout << sum[q - 1] << endl;
}
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
int occ[n];
int sta[n];
int sum[n];
for (int o = 0; o < n; o++) {
occ[o] = 0;
sum[o] = 0;
sta[o] = 0;
}
int v = 0;
for (int e = 0; e < n - m + 1; e++) {
int fl = 1;
for (int l = 0; l < m; l++) {
if (s1[e + l] != s2[l]) {
fl = 0;
break;
}
}
if (fl == 1) occ[e + m - 1] = 1;
}
for (int g = 0; g < n; g++) {
v = v + occ[g];
sum[g] = v;
}
for (int a = 1; a <= q; a++) {
int p, q;
cin >> p >> q;
if (q - p < m - 1)
cout << "0" << endl;
else if (p != 1)
cout << sum[q - 1] - sum[p - 2 + m - 1] << endl;
else
cout << sum[q - 1] << endl;
}
}
``` |
#include <bits/stdc++.h>
int n, m, q, a0, a1;
char s[1003], t[1003];
int pp[1003];
int main() {
scanf("%d %d %d", &n, &m, &q);
scanf("%s", s);
scanf("%s", t);
for (int i = 0; i <= n - m; ++i) {
int j = 0;
for (; j < m && s[i + j] == t[j]; ++j)
;
if (j == m) pp[i] = 1;
}
for (int i = 1; i <= n - m; ++i) pp[i] += pp[i - 1];
for (int i = 0; i < q; ++i) {
scanf("%d %d", &a0, &a1);
a0--;
a1--;
if (a0 + m - 1 <= a1) {
if (a0 > 0)
printf("%d\n", pp[a1 - m + 1] - pp[a0 - 1]);
else
printf("%d\n", pp[a1 - m + 1]);
} else
printf("0\n");
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int n, m, q, a0, a1;
char s[1003], t[1003];
int pp[1003];
int main() {
scanf("%d %d %d", &n, &m, &q);
scanf("%s", s);
scanf("%s", t);
for (int i = 0; i <= n - m; ++i) {
int j = 0;
for (; j < m && s[i + j] == t[j]; ++j)
;
if (j == m) pp[i] = 1;
}
for (int i = 1; i <= n - m; ++i) pp[i] += pp[i - 1];
for (int i = 0; i < q; ++i) {
scanf("%d %d", &a0, &a1);
a0--;
a1--;
if (a0 + m - 1 <= a1) {
if (a0 > 0)
printf("%d\n", pp[a1 - m + 1] - pp[a0 - 1]);
else
printf("%d\n", pp[a1 - m + 1]);
} else
printf("0\n");
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int z[1000005];
string s;
void Z() {
int n = s.size(), l = 0, r = 0, i;
for (i = 0; i < n; i++) {
if (i > r) {
l = r = i;
while (r < n && s[r] == s[r - l]) {
r++;
}
z[i] = r - l;
r--;
} else {
if (i + z[i - l] <= r) {
z[i] = z[i - l];
} else {
l = i;
while (r < n && s[r] == s[r - l]) {
r++;
}
z[i] = r - l;
r--;
}
}
}
}
int main() {
int n, m, q;
scanf("%d %d %d", &n, &m, &q);
string d, t;
cin >> d >> t;
s = t + '$' + d;
Z();
vector<int> a;
for (int i = 0; i < t.size() + d.size() + 1; i++) {
if (z[i] >= t.size()) {
a.push_back(i - t.size() - 1);
}
}
int i, l, r, u, v, mi;
for (i = 0; i < q; i++) {
scanf("%d %d", &u, &v);
u--;
v--;
l = 0, r = a.size() - 1;
if (r == -1) {
printf("0\n");
continue;
}
while (l < r) {
mi = (l + r) >> 1;
if (a[mi] >= u) {
r = mi;
} else {
l = mi + 1;
}
}
if ((a[r] < u) || (a[r] > v) || (a[r] + t.size() - 1 > v)) {
printf("0\n");
continue;
}
int idxst = r;
l = 0, r = a.size() - 1;
while (l < r) {
mi = (l + r + 1) >> 1;
if (a[mi] + t.size() - 1 <= v) {
l = mi;
} else {
r = mi - 1;
}
}
if ((a[l] < u) || (a[l] > v)) {
printf("0\n");
continue;
}
int ans = l - idxst + 1;
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Generate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int z[1000005];
string s;
void Z() {
int n = s.size(), l = 0, r = 0, i;
for (i = 0; i < n; i++) {
if (i > r) {
l = r = i;
while (r < n && s[r] == s[r - l]) {
r++;
}
z[i] = r - l;
r--;
} else {
if (i + z[i - l] <= r) {
z[i] = z[i - l];
} else {
l = i;
while (r < n && s[r] == s[r - l]) {
r++;
}
z[i] = r - l;
r--;
}
}
}
}
int main() {
int n, m, q;
scanf("%d %d %d", &n, &m, &q);
string d, t;
cin >> d >> t;
s = t + '$' + d;
Z();
vector<int> a;
for (int i = 0; i < t.size() + d.size() + 1; i++) {
if (z[i] >= t.size()) {
a.push_back(i - t.size() - 1);
}
}
int i, l, r, u, v, mi;
for (i = 0; i < q; i++) {
scanf("%d %d", &u, &v);
u--;
v--;
l = 0, r = a.size() - 1;
if (r == -1) {
printf("0\n");
continue;
}
while (l < r) {
mi = (l + r) >> 1;
if (a[mi] >= u) {
r = mi;
} else {
l = mi + 1;
}
}
if ((a[r] < u) || (a[r] > v) || (a[r] + t.size() - 1 > v)) {
printf("0\n");
continue;
}
int idxst = r;
l = 0, r = a.size() - 1;
while (l < r) {
mi = (l + r + 1) >> 1;
if (a[mi] + t.size() - 1 <= v) {
l = mi;
} else {
r = mi - 1;
}
}
if ((a[l] < u) || (a[l] > v)) {
printf("0\n");
continue;
}
int ans = l - idxst + 1;
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
vector<int> l, r;
int main() {
string s, t;
int n, m, q, x, y;
cin >> n >> m >> q >> s >> t;
for (int i = 0; i < n; i++) {
int flag = 1;
if (s[i] == t[0]) {
for (int j = i, k = 0; j < n && k < m; j++, k++)
if (s[j] != t[k]) {
flag = 0;
break;
}
if (flag) {
l.push_back(i + 1);
r.push_back(i + m);
}
}
}
while (q--) {
cin >> x >> y;
int a, cnt = 0;
a = lower_bound(l.begin(), l.end(), x) - l.begin();
for (int i = a; i < l.size(); i++) {
if (l[i] >= x && r[i] <= y) cnt++;
if (r[i] > y) break;
}
cout << cnt << endl;
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
vector<int> l, r;
int main() {
string s, t;
int n, m, q, x, y;
cin >> n >> m >> q >> s >> t;
for (int i = 0; i < n; i++) {
int flag = 1;
if (s[i] == t[0]) {
for (int j = i, k = 0; j < n && k < m; j++, k++)
if (s[j] != t[k]) {
flag = 0;
break;
}
if (flag) {
l.push_back(i + 1);
r.push_back(i + m);
}
}
}
while (q--) {
cin >> x >> y;
int a, cnt = 0;
a = lower_bound(l.begin(), l.end(), x) - l.begin();
for (int i = a; i < l.size(); i++) {
if (l[i] >= x && r[i] <= y) cnt++;
if (r[i] > y) break;
}
cout << cnt << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int64_t MOD = 1e9 + 7;
const int64_t N = 2e6;
bool isQuery = false;
int64_t power(int64_t x, int64_t n) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return power(x, n / 2) * power(x, n / 2);
else
return x * power(x, n / 2) * power(x, n / 2);
}
int64_t powerMod(int64_t x, int64_t n, int64_t mod) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return (powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
else
return (x * powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
}
void solve() {
int64_t n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int64_t> dp(n, 0), pref(n, -1), suff(n, -1), id;
int64_t ans = 0, p = -1;
for (int64_t i = 0, j; i < n; i++) {
for (j = 0; j < m && i + j < n; j++) {
if (s[i + j] != t[j]) break;
}
if (j == m)
pref[i + m - 1] = i + m - 1, id.push_back(i + m - 1), dp[i + m - 1] = 1;
}
for (int64_t i = 1; i < n; i++)
if (pref[i] == -1) pref[i] = pref[i - 1];
for (int64_t i = n - 1, j; i >= 0; i--) {
for (j = m - 1; j >= 0 && i - j >= 0; j--) {
if (s[i - (m - 1 - j)] != t[j]) break;
}
if (j == -1) suff[i - m + 1] = i;
}
for (int64_t i = n - 2; i >= 0; i--)
if (suff[i] == -1) suff[i] = suff[i + 1];
for (int64_t i = 1; i < n; i++) dp[i] += dp[i - 1];
while (q--) {
int64_t l, r;
cin >> l >> r;
l--, r--;
if (pref[r] == -1 || suff[l] == -1 | pref[r] < suff[l])
cout << 0;
else if (suff[l] > 0)
cout << dp[pref[r]] - dp[suff[l] - 1];
else
cout << dp[pref[r]];
cout << endl;
}
}
signed main() {
int64_t t = 1;
if (isQuery) cin >> t;
while (t--) {
solve();
cout << endl;
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int64_t MOD = 1e9 + 7;
const int64_t N = 2e6;
bool isQuery = false;
int64_t power(int64_t x, int64_t n) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return power(x, n / 2) * power(x, n / 2);
else
return x * power(x, n / 2) * power(x, n / 2);
}
int64_t powerMod(int64_t x, int64_t n, int64_t mod) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return (powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
else
return (x * powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
}
void solve() {
int64_t n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int64_t> dp(n, 0), pref(n, -1), suff(n, -1), id;
int64_t ans = 0, p = -1;
for (int64_t i = 0, j; i < n; i++) {
for (j = 0; j < m && i + j < n; j++) {
if (s[i + j] != t[j]) break;
}
if (j == m)
pref[i + m - 1] = i + m - 1, id.push_back(i + m - 1), dp[i + m - 1] = 1;
}
for (int64_t i = 1; i < n; i++)
if (pref[i] == -1) pref[i] = pref[i - 1];
for (int64_t i = n - 1, j; i >= 0; i--) {
for (j = m - 1; j >= 0 && i - j >= 0; j--) {
if (s[i - (m - 1 - j)] != t[j]) break;
}
if (j == -1) suff[i - m + 1] = i;
}
for (int64_t i = n - 2; i >= 0; i--)
if (suff[i] == -1) suff[i] = suff[i + 1];
for (int64_t i = 1; i < n; i++) dp[i] += dp[i - 1];
while (q--) {
int64_t l, r;
cin >> l >> r;
l--, r--;
if (pref[r] == -1 || suff[l] == -1 | pref[r] < suff[l])
cout << 0;
else if (suff[l] > 0)
cout << dp[pref[r]] - dp[suff[l] - 1];
else
cout << dp[pref[r]];
cout << endl;
}
}
signed main() {
int64_t t = 1;
if (isQuery) cin >> t;
while (t--) {
solve();
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, i, q, j, k, l, count, c, p = 0, a[1001] = {0};
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (i = 0; i <= n - m; i++) {
if (s.substr(i, m) == t) a[i] = {1};
}
for (i = 1; i <= q; i++) {
long long d, e;
c = 0;
cin >> d >> e;
for (j = d - 1; j <= e - m; j++) {
if (a[j]) c++;
}
cout << c << "\n";
}
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, i, q, j, k, l, count, c, p = 0, a[1001] = {0};
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (i = 0; i <= n - m; i++) {
if (s.substr(i, m) == t) a[i] = {1};
}
for (i = 1; i <= q; i++) {
long long d, e;
c = 0;
cin >> d >> e;
for (j = d - 1; j <= e - m; j++) {
if (a[j]) c++;
}
cout << c << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s;
string t;
int d[1001];
bool check(string s) {
for (int i = 0; i < s.length(); i++) {
if (s[i] != t[i]) return false;
}
return true;
}
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
string tmp = "";
for (int j = i; j < i + m; j++) {
tmp += s[j];
}
if (check(tmp)) {
d[i + m]++;
}
}
for (int i = 1; i <= n; i++) {
d[i] += d[i - 1];
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int len = r - l + 1;
if (len < t.length()) {
cout << 0 << "\n";
continue;
}
cout << d[r] - d[l + m - 2] << "\n";
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s;
string t;
int d[1001];
bool check(string s) {
for (int i = 0; i < s.length(); i++) {
if (s[i] != t[i]) return false;
}
return true;
}
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
string tmp = "";
for (int j = i; j < i + m; j++) {
tmp += s[j];
}
if (check(tmp)) {
d[i + m]++;
}
}
for (int i = 1; i <= n; i++) {
d[i] += d[i - 1];
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
int len = r - l + 1;
if (len < t.length()) {
cout << 0 << "\n";
continue;
}
cout << d[r] - d[l + m - 2] << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int a[1001];
int main() {
int i, n, m, q, j, l, r;
string str1, str2;
cin >> n >> m >> q;
cin >> str1;
cin >> str2;
for (i = 0; i <= n - m; i++) {
for (j = 0; j < m; j++) {
if (str1[i + j] != str2[j]) {
break;
}
}
if (j == m) a[i + 1] = 1;
}
for (i = 2; i <= n; i++) {
a[i] = a[i] + a[i - 1];
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
if (l == 1) {
cout << a[r - m + 1] << endl;
;
} else {
cout << a[r - m + 1] - a[l - 1] << endl;
}
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int a[1001];
int main() {
int i, n, m, q, j, l, r;
string str1, str2;
cin >> n >> m >> q;
cin >> str1;
cin >> str2;
for (i = 0; i <= n - m; i++) {
for (j = 0; j < m; j++) {
if (str1[i + j] != str2[j]) {
break;
}
}
if (j == m) a[i + 1] = 1;
}
for (i = 2; i <= n; i++) {
a[i] = a[i] + a[i - 1];
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
if (l == 1) {
cout << a[r - m + 1] << endl;
;
} else {
cout << a[r - m + 1] - a[l - 1] << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10005;
int Next[maxn];
char str[maxn], th[maxn];
int change(char s[]) {
int i = 0, j = 1;
Next[i] = 0;
while (s[j]) {
if (s[i] == s[j]) {
i++;
Next[j] = i;
j++;
} else {
if (i > 0) {
i = Next[i - 1];
} else {
Next[j] = i;
j++;
}
}
}
for (int i = j - 1; i > 0; i--) {
Next[i] = Next[i - 1];
}
Next[0] = -1;
return j;
}
int num[maxn] = {0};
int main() {
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
scanf("%s%s", str, th);
int l = change(th);
int i = 0, j = 0, cnt = 0;
while (str[i]) {
if (str[i] == th[j]) {
if (j == l - 1) {
num[i - m + 1]++;
cnt++;
j = Next[j];
} else {
i++;
j++;
}
} else {
if (j >= 0) {
j = Next[j];
} else {
i++;
j = 0;
}
}
}
while (k--) {
int l, r;
scanf("%d%d", &l, &r);
l--;
r--;
int sum = 0;
for (int i = l; i <= r; i++) {
if ((r - i + 1) >= m) sum += num[i];
}
printf("%d\n", sum);
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 10005;
int Next[maxn];
char str[maxn], th[maxn];
int change(char s[]) {
int i = 0, j = 1;
Next[i] = 0;
while (s[j]) {
if (s[i] == s[j]) {
i++;
Next[j] = i;
j++;
} else {
if (i > 0) {
i = Next[i - 1];
} else {
Next[j] = i;
j++;
}
}
}
for (int i = j - 1; i > 0; i--) {
Next[i] = Next[i - 1];
}
Next[0] = -1;
return j;
}
int num[maxn] = {0};
int main() {
int n, m, k;
scanf("%d%d%d", &n, &m, &k);
scanf("%s%s", str, th);
int l = change(th);
int i = 0, j = 0, cnt = 0;
while (str[i]) {
if (str[i] == th[j]) {
if (j == l - 1) {
num[i - m + 1]++;
cnt++;
j = Next[j];
} else {
i++;
j++;
}
} else {
if (j >= 0) {
j = Next[j];
} else {
i++;
j = 0;
}
}
}
while (k--) {
int l, r;
scanf("%d%d", &l, &r);
l--;
r--;
int sum = 0;
for (int i = l; i <= r; i++) {
if ((r - i + 1) >= m) sum += num[i];
}
printf("%d\n", sum);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
vector<int> v;
v.clear();
if (n < m) {
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
cout << "0" << endl;
}
return 0;
}
for (int i = 0; i <= s.size() - t.size(); i++) {
int j;
int val = i;
for (j = 0; j < t.size(); j++) {
if (t[j] == s[val++])
continue;
else
break;
}
if (j == t.size()) v.push_back(i);
}
if (v.size() == 0) {
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
cout << "0" << endl;
}
return 0;
}
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
a = a - 1;
b = b - 1;
int count = 0;
for (int p = 0; p < v.size(); p++) {
if (v[p] >= a && v[p] + m - 1 <= b) count++;
}
cout << count << endl;
}
return 0;
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
vector<int> v;
v.clear();
if (n < m) {
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
cout << "0" << endl;
}
return 0;
}
for (int i = 0; i <= s.size() - t.size(); i++) {
int j;
int val = i;
for (j = 0; j < t.size(); j++) {
if (t[j] == s[val++])
continue;
else
break;
}
if (j == t.size()) v.push_back(i);
}
if (v.size() == 0) {
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
cout << "0" << endl;
}
return 0;
}
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
a = a - 1;
b = b - 1;
int count = 0;
for (int p = 0; p < v.size(); p++) {
if (v[p] >= a && v[p] + m - 1 <= b) count++;
}
cout << count << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, sum[1001];
string s1, s2;
int main() {
cin >> n >> m >> q >> s1 >> s2;
for (int i = 1; i <= n; i += 1) {
sum[i] += sum[i - 1];
if (i > n - m + 1) {
continue;
}
bool f = true;
for (int j = i; j < i + m; j += 1) {
if (s1[j - 1] != s2[j - i]) {
f = false;
break;
}
}
if (f) {
sum[i + m - 1] += 1;
}
}
for (int i = 0; i < q; i += 1) {
int a, b;
cin >> a >> b;
if (a + m - 2 > b) {
cout << 0 << endl;
continue;
}
cout << sum[b] - sum[a + m - 2] << endl;
}
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, sum[1001];
string s1, s2;
int main() {
cin >> n >> m >> q >> s1 >> s2;
for (int i = 1; i <= n; i += 1) {
sum[i] += sum[i - 1];
if (i > n - m + 1) {
continue;
}
bool f = true;
for (int j = i; j < i + m; j += 1) {
if (s1[j - 1] != s2[j - i]) {
f = false;
break;
}
}
if (f) {
sum[i + m - 1] += 1;
}
}
for (int i = 0; i < q; i += 1) {
int a, b;
cin >> a >> b;
if (a + m - 2 > b) {
cout << 0 << endl;
continue;
}
cout << sum[b] - sum[a + m - 2] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
char s[1005], t[1005];
int sum[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf(" %s %s", s + 1, t + 1);
for (int i = 1; i <= n; i++) {
int good = 1;
for (int j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) good = 0;
}
sum[i] = sum[i - 1] + good;
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
r = max(l - 1, r - m + 1);
printf("%d\n", sum[r] - sum[l - 1]);
}
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
char s[1005], t[1005];
int sum[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf(" %s %s", s + 1, t + 1);
for (int i = 1; i <= n; i++) {
int good = 1;
for (int j = 1; j <= m; j++) {
if (s[i + j - 1] != t[j]) good = 0;
}
sum[i] = sum[i - 1] + good;
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
r = max(l - 1, r - m + 1);
printf("%d\n", sum[r] - sum[l - 1]);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int dp[2000];
char str[2000], tr[2000];
int main() {
int i, j, k, l, m, n, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str + 1);
scanf("%s", tr + 1);
for (int i = 1; str[i]; i++) {
for (j = 1, k = i; tr[j] && str[k]; j++, k++) {
if (tr[j] != str[k]) break;
}
dp[i] += dp[i - 1];
if (j <= m) continue;
dp[i]++;
}
while (q--) {
scanf("%d%d", &l, &k);
if ((k - l + 1) < m) {
printf("0\n");
continue;
}
int v = k - m + 1;
int ans = dp[v] - dp[l - 1];
cout << ans << endl;
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[2000];
char str[2000], tr[2000];
int main() {
int i, j, k, l, m, n, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str + 1);
scanf("%s", tr + 1);
for (int i = 1; str[i]; i++) {
for (j = 1, k = i; tr[j] && str[k]; j++, k++) {
if (tr[j] != str[k]) break;
}
dp[i] += dp[i - 1];
if (j <= m) continue;
dp[i]++;
}
while (q--) {
scanf("%d%d", &l, &k);
if ((k - l + 1) < m) {
printf("0\n");
continue;
}
int v = k - m + 1;
int ans = dp[v] - dp[l - 1];
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1001;
int a[maxn], cnt = 0;
char s1[maxn], s2[maxn];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
cin >> s1 >> s2;
for (int i = 0; i < n; i++) {
if (s1[i] == s2[0]) {
bool fa = 1;
int now = i;
for (int j = 1; j < m; j++) {
now++;
if (s2[j] != s1[now]) {
fa = 0;
break;
}
}
if (fa == 1) {
a[cnt++] = i + 1;
}
}
}
while (q--) {
int u, v;
scanf("%d%d", &u, &v);
int sum = 0;
for (int i = 0; i < cnt; i++) {
if (u <= a[i] && v >= a[i] + m - 1) sum++;
}
printf("%d\n", sum);
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1001;
int a[maxn], cnt = 0;
char s1[maxn], s2[maxn];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
cin >> s1 >> s2;
for (int i = 0; i < n; i++) {
if (s1[i] == s2[0]) {
bool fa = 1;
int now = i;
for (int j = 1; j < m; j++) {
now++;
if (s2[j] != s1[now]) {
fa = 0;
break;
}
}
if (fa == 1) {
a[cnt++] = i + 1;
}
}
}
while (q--) {
int u, v;
scanf("%d%d", &u, &v);
int sum = 0;
for (int i = 0; i < cnt; i++) {
if (u <= a[i] && v >= a[i] + m - 1) sum++;
}
printf("%d\n", sum);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long z[3000019], arr[3000019], pre[3000019];
void fnc(string s) {
memset(z, 0, sizeof(z));
long long l = 0, r = 0;
long long n = s.size();
for (long long i = 1; i <= n; i++) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (i + z[i] <= n && s[z[i]] == s[i + z[i]]) z[i]++;
if (i + z[i] - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
z[0] = n;
return;
}
signed main() {
std::ios::sync_with_stdio(false);
long long n, m, q, i, l, r;
string s, t, str;
cin >> n >> m >> q;
cin >> s >> t;
str = t + "$" + s;
fnc(str);
for (i = m + 1; i <= n + m; i++) arr[i - m] = (z[i] == m);
for (i = 1; i <= n; i++) pre[i] = arr[i] + pre[i - 1];
while (q--) {
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
cout << pre[r - m + 1] - pre[l - 1] << endl;
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1000000007;
long long z[3000019], arr[3000019], pre[3000019];
void fnc(string s) {
memset(z, 0, sizeof(z));
long long l = 0, r = 0;
long long n = s.size();
for (long long i = 1; i <= n; i++) {
if (i <= r) z[i] = min(r - i + 1, z[i - l]);
while (i + z[i] <= n && s[z[i]] == s[i + z[i]]) z[i]++;
if (i + z[i] - 1 > r) {
l = i;
r = i + z[i] - 1;
}
}
z[0] = n;
return;
}
signed main() {
std::ios::sync_with_stdio(false);
long long n, m, q, i, l, r;
string s, t, str;
cin >> n >> m >> q;
cin >> s >> t;
str = t + "$" + s;
fnc(str);
for (i = m + 1; i <= n + m; i++) arr[i - m] = (z[i] == m);
for (i = 1; i <= n; i++) pre[i] = arr[i] + pre[i - 1];
while (q--) {
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
cout << pre[r - m + 1] - pre[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
int64_t n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
vector<pair<int, int>> ins;
for (int i = 0; i < s.size(); ++i) {
if (i + t.size() - 1 < s.size() && s.substr(i, t.size()) == t) {
ins.push_back(make_pair(i, i + t.size() - 1));
}
}
for (int i = 0; i < q; ++i) {
int64_t l, r, cnt = 0;
cin >> l >> r;
--l;
--r;
for (auto now : ins) {
if (now.first >= l && now.second <= r) {
++cnt;
}
}
cout << cnt << '\n';
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, const char* argv[]) {
int64_t n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
vector<pair<int, int>> ins;
for (int i = 0; i < s.size(); ++i) {
if (i + t.size() - 1 < s.size() && s.substr(i, t.size()) == t) {
ins.push_back(make_pair(i, i + t.size() - 1));
}
}
for (int i = 0; i < q; ++i) {
int64_t l, r, cnt = 0;
cin >> l >> r;
--l;
--r;
for (auto now : ins) {
if (now.first >= l && now.second <= r) {
++cnt;
}
}
cout << cnt << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> I;
for (int i = 0; i < n - m + 1; i++) {
if (t == s.substr(i, m)) I.push_back(i);
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
int ans = 0;
for (int j = 0; j < I.size(); j++) {
if (I[j] < l)
continue;
else if (I[j] + m - 1 > r)
break;
else
ans++;
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> I;
for (int i = 0; i < n - m + 1; i++) {
if (t == s.substr(i, m)) I.push_back(i);
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
int ans = 0;
for (int j = 0; j < I.size(); j++) {
if (I[j] < l)
continue;
else if (I[j] + m - 1 > r)
break;
else
ans++;
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> v;
for (int i = 0; i < int((s).size()); i++) {
int j;
for (j = 0; j < int((t).size()) && (i + j) < int((s).size()); j++) {
if (s[j + i] != t[j]) break;
}
if (j == t.size()) v.push_back(i);
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
r = r - (int)t.size() + 1;
if (l > r) {
cout << 0 << endl;
continue;
}
int ind1 = lower_bound((v).begin(), (v).end(), l) - v.begin();
int ind2 = upper_bound((v).begin(), (v).end(), r) - v.begin();
cout << (ind2 - ind1) << endl;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int kickstart = 0;
int test = 1;
for (int i = 0; i < test; i++) {
if (kickstart)
cout << "Case #" << i + 1 << ": ", solve(), cout << endl;
else
solve();
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> v;
for (int i = 0; i < int((s).size()); i++) {
int j;
for (j = 0; j < int((t).size()) && (i + j) < int((s).size()); j++) {
if (s[j + i] != t[j]) break;
}
if (j == t.size()) v.push_back(i);
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--;
r--;
r = r - (int)t.size() + 1;
if (l > r) {
cout << 0 << endl;
continue;
}
int ind1 = lower_bound((v).begin(), (v).end(), l) - v.begin();
int ind2 = upper_bound((v).begin(), (v).end(), r) - v.begin();
cout << (ind2 - ind1) << endl;
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int kickstart = 0;
int test = 1;
for (int i = 0; i < test; i++) {
if (kickstart)
cout << "Case #" << i + 1 << ": ", solve(), cout << endl;
else
solve();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, dp[1010][1010];
char s[1010], t[1010];
bool used[1010];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t);
for (int i = 1; i + m - 1 <= n; i++) {
bool f = true;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
f = false;
break;
}
}
used[i] = f;
}
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
dp[i][j] = max(dp[i][j], dp[i][j - 1]);
if (j - m + 1 >= i && used[j - m + 1]) {
dp[i][j]++;
}
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", dp[l][r]);
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, dp[1010][1010];
char s[1010], t[1010];
bool used[1010];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t);
for (int i = 1; i + m - 1 <= n; i++) {
bool f = true;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
f = false;
break;
}
}
used[i] = f;
}
for (int i = 1; i <= n; i++) {
for (int j = i; j <= n; j++) {
dp[i][j] = max(dp[i][j], dp[i][j - 1]);
if (j - m + 1 >= i && used[j - m + 1]) {
dp[i][j]++;
}
}
}
while (q--) {
int l, r;
scanf("%d%d", &l, &r);
printf("%d\n", dp[l][r]);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<bool> occ(n);
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) {
occ[i] = 1;
}
}
vector<int> prefixOcc(n + 1);
prefixOcc[0] = 0;
for (int i = 1; i <= n; i++) {
prefixOcc[i] = prefixOcc[i - 1] + occ[i - 1];
}
while (q--) {
int left, right;
cin >> left >> right;
left--;
right--;
if (right - left + 1 < m or right - m + 1 < 0) {
cout << '0' << endl;
;
continue;
}
int prefLeft = prefixOcc[left];
int prefRight = prefixOcc[right - m + 2];
cout << prefRight - prefLeft << endl;
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<bool> occ(n);
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) {
occ[i] = 1;
}
}
vector<int> prefixOcc(n + 1);
prefixOcc[0] = 0;
for (int i = 1; i <= n; i++) {
prefixOcc[i] = prefixOcc[i - 1] + occ[i - 1];
}
while (q--) {
int left, right;
cin >> left >> right;
left--;
right--;
if (right - left + 1 < m or right - m + 1 < 0) {
cout << '0' << endl;
;
continue;
}
int prefLeft = prefixOcc[left];
int prefRight = prefixOcc[right - m + 2];
cout << prefRight - prefLeft << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
long long m, n, q, dp[1010][1010];
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
bool b = false;
for (int j = i; j < i + m; j++) {
if (s[j] != t[j - i]) b = true;
}
if (b == false) {
dp[i][i + m] = 1;
}
}
for (int i = 0; i <= n; i++) {
for (int j = 0; j < n; j++) {
if (i >= m) {
dp[j][j + i] = dp[j][j + i] + dp[j][j + i - 1] + dp[j + 1][j + i] -
dp[j + 1][j + i - 1];
}
}
}
for (int i = 0, a, b; i < q; i++) {
cin >> a >> b;
cout << dp[a - 1][b] << endl;
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
long long m, n, q, dp[1010][1010];
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
bool b = false;
for (int j = i; j < i + m; j++) {
if (s[j] != t[j - i]) b = true;
}
if (b == false) {
dp[i][i + m] = 1;
}
}
for (int i = 0; i <= n; i++) {
for (int j = 0; j < n; j++) {
if (i >= m) {
dp[j][j + i] = dp[j][j + i] + dp[j][j + i - 1] + dp[j + 1][j + i] -
dp[j + 1][j + i - 1];
}
}
}
for (int i = 0, a, b; i < q; i++) {
cin >> a >> b;
cout << dp[a - 1][b] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
vector<long long> build(string s) {
int n = s.size();
vector<long long> kmp(n, 0);
for (int i = 1; i < n; i++) {
int j = kmp[i - 1];
while (j > 0 && s[i] != s[j]) {
j = kmp[j - 1];
}
if (s[i] == s[j]) j++;
kmp[i] = j;
}
return kmp;
}
void icchhipadey() {
long long n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
while (q--) {
long long l, r;
cin >> l >> r;
string s = s1.substr(l - 1, r - l + 1);
long long ind = 0;
long long cnt = 0;
string temp = s2 + "$" + s;
vector<long long> kmp = build(temp);
for (auto x : kmp) {
if (x == m) {
cnt++;
}
}
cout << cnt << endl;
}
}
int main() { icchhipadey(); }
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long MOD = 1e9 + 7;
vector<long long> build(string s) {
int n = s.size();
vector<long long> kmp(n, 0);
for (int i = 1; i < n; i++) {
int j = kmp[i - 1];
while (j > 0 && s[i] != s[j]) {
j = kmp[j - 1];
}
if (s[i] == s[j]) j++;
kmp[i] = j;
}
return kmp;
}
void icchhipadey() {
long long n, m, q;
cin >> n >> m >> q;
string s1, s2;
cin >> s1 >> s2;
while (q--) {
long long l, r;
cin >> l >> r;
string s = s1.substr(l - 1, r - l + 1);
long long ind = 0;
long long cnt = 0;
string temp = s2 + "$" + s;
vector<long long> kmp = build(temp);
for (auto x : kmp) {
if (x == m) {
cnt++;
}
}
cout << cnt << endl;
}
}
int main() { icchhipadey(); }
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 20;
int sum[maxn];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (int i = 0; i < n; i++)
sum[i + 1] = sum[i] + (i + m <= n && s.substr(i, m) == t);
while (q--) {
int l, r;
cin >> l >> r;
r -= m;
l--;
if (r < l)
cout << 0 << endl;
else
cout << sum[r + 1] - sum[l] << endl;
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 20;
int sum[maxn];
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
for (int i = 0; i < n; i++)
sum[i + 1] = sum[i] + (i + m <= n && s.substr(i, m) == t);
while (q--) {
int l, r;
cin >> l >> r;
r -= m;
l--;
if (r < l)
cout << 0 << endl;
else
cout << sum[r + 1] - sum[l] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string t;
cin >> t;
string str1 = s;
size_t pos1;
vector<long long> arr;
pos1 = str1.find(t);
while (pos1 != string::npos) {
arr.push_back(pos1);
pos1 = str1.find(t, pos1 + 1);
}
while (q--) {
long long l, r;
cin >> l >> r;
long long count = 0;
for (long long i = 0; i < arr.size(); i++) {
if (arr[i] >= (l - 1) && (r - 1) >= (arr[i] + m - 1)) {
count++;
}
}
cout << count << endl;
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string t;
cin >> t;
string str1 = s;
size_t pos1;
vector<long long> arr;
pos1 = str1.find(t);
while (pos1 != string::npos) {
arr.push_back(pos1);
pos1 = str1.find(t, pos1 + 1);
}
while (q--) {
long long l, r;
cin >> l >> r;
long long count = 0;
for (long long i = 0; i < arr.size(); i++) {
if (arr[i] >= (l - 1) && (r - 1) >= (arr[i] + m - 1)) {
count++;
}
}
cout << count << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int64_t MOD = 1e9 + 7;
const int64_t N = 2e6;
bool isQuery = false;
int64_t power(int64_t x, int64_t n) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return power(x, n / 2) * power(x, n / 2);
else
return x * power(x, n / 2) * power(x, n / 2);
}
int64_t powerMod(int64_t x, int64_t n, int64_t mod) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return (powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
else
return (x * powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
}
void solve() {
int64_t n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int64_t> dp(n, 0), pref(n, -1), suff(n, -1), id;
int64_t ans = 0, p = -1;
for (int64_t i = 0, j; i < n; i++) {
for (j = 0; j < m && i + j < n; j++) {
if (s[i + j] != t[j]) break;
}
if (j == m)
pref[i + m - 1] = i + m - 1, id.push_back(i + m - 1), dp[i + m - 1] = 1;
}
for (int64_t i = 1; i < n; i++)
if (pref[i] == -1) pref[i] = pref[i - 1];
for (int64_t i = n - 1, j; i >= 0; i--) {
for (j = m - 1; j >= 0 && i - j >= 0; j--) {
if (s[i - (m - 1 - j)] != t[j]) break;
}
if (j == -1) suff[i - m + 1] = i;
}
for (int64_t i = n - 2; i >= 0; i--)
if (suff[i] == -1) suff[i] = suff[i + 1];
for (int64_t i = 1; i < n; i++) dp[i] += dp[i - 1];
while (q--) {
int64_t l, r;
cin >> l >> r;
l--, r--;
if (pref[r] == -1 || suff[l] == -1 | pref[r] < suff[l])
cout << 0;
else if (suff[l] > 0)
cout << dp[pref[r]] - dp[suff[l] - 1];
else
cout << dp[pref[r]];
cout << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int64_t t = 1;
if (isQuery) cin >> t;
while (t--) {
solve();
cout << endl;
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int64_t MOD = 1e9 + 7;
const int64_t N = 2e6;
bool isQuery = false;
int64_t power(int64_t x, int64_t n) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return power(x, n / 2) * power(x, n / 2);
else
return x * power(x, n / 2) * power(x, n / 2);
}
int64_t powerMod(int64_t x, int64_t n, int64_t mod) {
if (n == 0)
return 1;
else if (n % 2 == 0)
return (powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
else
return (x * powerMod(x, n / 2, mod) * powerMod(x, n / 2, mod)) % mod;
}
void solve() {
int64_t n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int64_t> dp(n, 0), pref(n, -1), suff(n, -1), id;
int64_t ans = 0, p = -1;
for (int64_t i = 0, j; i < n; i++) {
for (j = 0; j < m && i + j < n; j++) {
if (s[i + j] != t[j]) break;
}
if (j == m)
pref[i + m - 1] = i + m - 1, id.push_back(i + m - 1), dp[i + m - 1] = 1;
}
for (int64_t i = 1; i < n; i++)
if (pref[i] == -1) pref[i] = pref[i - 1];
for (int64_t i = n - 1, j; i >= 0; i--) {
for (j = m - 1; j >= 0 && i - j >= 0; j--) {
if (s[i - (m - 1 - j)] != t[j]) break;
}
if (j == -1) suff[i - m + 1] = i;
}
for (int64_t i = n - 2; i >= 0; i--)
if (suff[i] == -1) suff[i] = suff[i + 1];
for (int64_t i = 1; i < n; i++) dp[i] += dp[i - 1];
while (q--) {
int64_t l, r;
cin >> l >> r;
l--, r--;
if (pref[r] == -1 || suff[l] == -1 | pref[r] < suff[l])
cout << 0;
else if (suff[l] > 0)
cout << dp[pref[r]] - dp[suff[l] - 1];
else
cout << dp[pref[r]];
cout << endl;
}
}
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int64_t t = 1;
if (isQuery) cin >> t;
while (t--) {
solve();
cout << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int st[1001];
int main() {
int n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string t;
cin >> t;
st[0] = 0;
for (int i = 1; i <= n - m + 1; i++) {
bool b = true;
for (int j = i; j < i + m; j++) {
if (s.at(j - 1) != t.at(j - i)) {
b = false;
break;
}
}
if (b) {
st[i] = st[i - 1] + 1;
} else {
st[i] = st[i - 1];
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l < m - 1) {
cout << 0 << endl;
} else {
cout << st[r - m + 1] - st[l - 1] << endl;
}
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int st[1001];
int main() {
int n, m, q;
cin >> n >> m >> q;
string s;
cin >> s;
string t;
cin >> t;
st[0] = 0;
for (int i = 1; i <= n - m + 1; i++) {
bool b = true;
for (int j = i; j < i + m; j++) {
if (s.at(j - 1) != t.at(j - i)) {
b = false;
break;
}
}
if (b) {
st[i] = st[i - 1] + 1;
} else {
st[i] = st[i - 1];
}
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l < m - 1) {
cout << 0 << endl;
} else {
cout << st[r - m + 1] - st[l - 1] << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
long long j = 0, pre = i;
while (j < m) {
if (str[i] == s[j]) {
i++;
j++;
} else
break;
}
if (j == m) ara[pre + 1]++;
i = pre;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
long long j = 0, pre = i;
while (j < m) {
if (str[i] == s[j]) {
i++;
j++;
} else
break;
}
if (j == m) ara[pre + 1]++;
i = pre;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, lct1, lct2;
string s, t;
int z[3005];
int cnt;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> m >> q;
cin >> s >> t;
s = t + s;
int len = s.length();
z[0] = 0;
for (int i = 1, l = 0, r = 0; i < len; ++i) {
if (i <= r) z[i] = min(z[i - l], r - i + 1);
while (i + z[i] < len && s[z[i]] == s[i + z[i]]) z[i]++;
if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
}
for (int i = 1; i <= q; ++i) {
cnt = 0;
cin >> lct1 >> lct2;
for (int i = lct1 + m - 1; i <= lct2 + m - 1; ++i)
if (z[i] >= m && i + m - 1 <= lct2 + m - 1) cnt++;
cout << cnt << "\n";
}
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, lct1, lct2;
string s, t;
int z[3005];
int cnt;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> m >> q;
cin >> s >> t;
s = t + s;
int len = s.length();
z[0] = 0;
for (int i = 1, l = 0, r = 0; i < len; ++i) {
if (i <= r) z[i] = min(z[i - l], r - i + 1);
while (i + z[i] < len && s[z[i]] == s[i + z[i]]) z[i]++;
if (i + z[i] - 1 > r) l = i, r = i + z[i] - 1;
}
for (int i = 1; i <= q; ++i) {
cnt = 0;
cin >> lct1 >> lct2;
for (int i = lct1 + m - 1; i <= lct2 + m - 1; ++i)
if (z[i] >= m && i + m - 1 <= lct2 + m - 1) cnt++;
cout << cnt << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
struct Node {
int d;
struct Node *left, *right;
};
int p[1000][1000] = {0};
string s, t;
void dp(int i, int j) {
if (j - i + 1 == t.size()) {
int f = 1;
for (int z = 0; z < t.size(); z++) {
if (s[i + z] != t[z]) {
f = 0;
break;
}
}
if (f == 1) p[i][j] = 1;
} else if (j - i + 1 < t.size())
return;
else if (j - i + 1 > t.size()) {
int maxm = 0, l = t.size();
for (int x = i; x <= j - l + 1; x++) {
if (p[x][x + l - 1] > 0) p[i][j]++;
}
}
return;
}
int main() {
int n, m, q;
cin >> n >> m >> q;
int a[q][2];
getchar();
cin >> s;
getchar();
cin >> t;
for (int i = 0; i <= q - 1; i++) {
cin >> a[i][0] >> a[i][1];
}
int l = t.size();
for (int si = l; si <= n; si++) {
for (int j = 0; j <= n - si; j++) dp(j, j + si - 1);
}
for (int i = 0; i <= q - 1; i++) cout << p[a[i][0] - 1][a[i][1] - 1] << "\n";
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
struct Node {
int d;
struct Node *left, *right;
};
int p[1000][1000] = {0};
string s, t;
void dp(int i, int j) {
if (j - i + 1 == t.size()) {
int f = 1;
for (int z = 0; z < t.size(); z++) {
if (s[i + z] != t[z]) {
f = 0;
break;
}
}
if (f == 1) p[i][j] = 1;
} else if (j - i + 1 < t.size())
return;
else if (j - i + 1 > t.size()) {
int maxm = 0, l = t.size();
for (int x = i; x <= j - l + 1; x++) {
if (p[x][x + l - 1] > 0) p[i][j]++;
}
}
return;
}
int main() {
int n, m, q;
cin >> n >> m >> q;
int a[q][2];
getchar();
cin >> s;
getchar();
cin >> t;
for (int i = 0; i <= q - 1; i++) {
cin >> a[i][0] >> a[i][1];
}
int l = t.size();
for (int si = l; si <= n; si++) {
for (int j = 0; j <= n - si; j++) dp(j, j + si - 1);
}
for (int i = 0; i <= q - 1; i++) cout << p[a[i][0] - 1][a[i][1] - 1] << "\n";
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
vector<int> can;
int n, m, k;
int main() {
cin >> n >> m >> k;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
if (s.substr(i, m) != t) continue;
can.push_back(i + 1);
}
while (k--) {
int l, r;
cin >> l >> r;
r = r - m + 1;
int first = lower_bound(can.begin(), can.end(), l) - can.begin();
int last = upper_bound(can.begin(), can.end(), r) - can.begin();
cout << max(0, last - first) << endl;
}
return 0;
}
| ### Prompt
Please create a solution in Cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
vector<int> can;
int n, m, k;
int main() {
cin >> n >> m >> k;
cin >> s >> t;
for (int i = 0; i <= n - m; i++) {
if (s.substr(i, m) != t) continue;
can.push_back(i + 1);
}
while (k--) {
int l, r;
cin >> l >> r;
r = r - m + 1;
int first = lower_bound(can.begin(), can.end(), l) - can.begin();
int last = upper_bound(can.begin(), can.end(), r) - can.begin();
cout << max(0, last - first) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, l, r, vf, ans, i, j, k;
const int LMAX = 1e3;
char s[LMAX + 5], t[LMAX + 5];
int st[LMAX + 5], dr[LMAX + 5], cnt = 0;
int main() {
cin >> n >> m >> q;
for (i = 1; i <= n; i++) cin >> s[i];
for (i = 1; i <= m; i++) cin >> t[i];
for (i = 1; i <= n - m + 1; i++)
if (s[i] == t[1]) {
bool ok = true;
int vf = 0;
for (j = i; j <= i + m - 1; j++)
if (s[j] != t[++vf]) {
ok = false;
break;
}
if (ok == true) {
st[++cnt] = i;
dr[cnt] = i + m - 1;
}
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
ans = 0;
for (j = 1; j <= cnt; j++)
if (st[j] >= l && dr[j] <= r) ans++;
cout << ans << '\n';
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, l, r, vf, ans, i, j, k;
const int LMAX = 1e3;
char s[LMAX + 5], t[LMAX + 5];
int st[LMAX + 5], dr[LMAX + 5], cnt = 0;
int main() {
cin >> n >> m >> q;
for (i = 1; i <= n; i++) cin >> s[i];
for (i = 1; i <= m; i++) cin >> t[i];
for (i = 1; i <= n - m + 1; i++)
if (s[i] == t[1]) {
bool ok = true;
int vf = 0;
for (j = i; j <= i + m - 1; j++)
if (s[j] != t[++vf]) {
ok = false;
break;
}
if (ok == true) {
st[++cnt] = i;
dr[cnt] = i + m - 1;
}
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
ans = 0;
for (j = 1; j <= cnt; j++)
if (st[j] >= l && dr[j] <= r) ans++;
cout << ans << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
int n, m, q, ct = 0, flag, l, r;
cin >> n >> m >> q;
vector<vector<int>> ans(n + 1, vector<int>(n + 1, 0));
cin >> s;
cin >> t;
for (int i = 0; i < n; i++) {
flag = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
flag = 0;
break;
}
}
if (flag) {
ct++;
for (int j = 0; j <= i; j++) ans[j + 1][i + m]++;
}
}
for (int i = 1; i <= n; i++) {
for (int j = i; j < n; j++) ans[i][j + 1] += ans[i][j];
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
cout << ans[l][r] << endl;
}
return 0;
}
| ### Prompt
Please create a solution in cpp to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
int n, m, q, ct = 0, flag, l, r;
cin >> n >> m >> q;
vector<vector<int>> ans(n + 1, vector<int>(n + 1, 0));
cin >> s;
cin >> t;
for (int i = 0; i < n; i++) {
flag = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
flag = 0;
break;
}
}
if (flag) {
ct++;
for (int j = 0; j <= i; j++) ans[j + 1][i + m]++;
}
}
for (int i = 1; i <= n; i++) {
for (int j = i; j < n; j++) ans[i][j + 1] += ans[i][j];
}
for (int i = 0; i < q; i++) {
cin >> l >> r;
cout << ans[l][r] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string str;
vector<int> z;
int len = 0;
void zfunc() {
len = str.size();
z.resize(len);
int ind = 0;
for (int i = 1; i < len; i++) {
if (ind + z[ind] - 1 >= i) {
z[i] = min(ind + z[ind] - i, z[i - ind]);
}
for (; i + z[i] < len && str[i + z[i]] == str[z[i]]; z[i]++)
;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
str = t + '#' + s;
zfunc();
vector<int> pref(s.size() + 1);
for (int i = 1; i <= s.size(); i++) {
pref[i] = pref[i - 1] + int(z[i + t.size()] == t.size());
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < t.size()) {
cout << 0 << '\n';
} else {
cout << pref[r - t.size() + 1] - pref[l - 1] << '\n';
}
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string str;
vector<int> z;
int len = 0;
void zfunc() {
len = str.size();
z.resize(len);
int ind = 0;
for (int i = 1; i < len; i++) {
if (ind + z[ind] - 1 >= i) {
z[i] = min(ind + z[ind] - i, z[i - ind]);
}
for (; i + z[i] < len && str[i + z[i]] == str[z[i]]; z[i]++)
;
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
str = t + '#' + s;
zfunc();
vector<int> pref(s.size() + 1);
for (int i = 1; i <= s.size(); i++) {
pref[i] = pref[i - 1] + int(z[i + t.size()] == t.size());
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
if (r - l + 1 < t.size()) {
cout << 0 << '\n';
} else {
cout << pref[r - t.size() + 1] - pref[l - 1] << '\n';
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[100005], t[100005];
int Q, m, n, l1, l2;
int sum[100005], fr[100005], ma[100005];
void init() {
int i, np;
scanf("%d%d%d", &l1, &l2, &Q);
scanf("%s", s);
scanf("%s", t);
for (i = l1; i > 0; i--) s[i] = s[i - 1];
for (i = l2; i > 0; i--) t[i] = t[i - 1];
fr[1] = 0;
for (i = 2; i <= l2; i++) {
np = fr[i - 1];
while (t[i] != t[np + 1] && np) np = fr[np];
if (t[i] == t[np + 1])
fr[i] = np + 1;
else
fr[i] = 0;
}
t[l2 + 1] = '&';
ma[0] = 0;
for (i = 1; i <= l1; i++) {
np = ma[i - 1];
while (np && s[i] != t[np + 1]) np = fr[np];
if (s[i] == t[np + 1])
ma[i] = np + 1;
else
ma[i] = 0;
}
sum[0] = 0;
for (i = 1; i <= l1; i++) sum[i] = sum[i - 1] + (ma[i] == l2);
}
int main() {
int i, l, r;
init();
for (i = 1; i <= Q; i++) {
scanf("%d%d", &l, &r);
if (r - l + 1 < l2) {
printf("0\n");
continue;
}
printf("%d\n", max(0, sum[r] - sum[l + l2 - 2]));
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[100005], t[100005];
int Q, m, n, l1, l2;
int sum[100005], fr[100005], ma[100005];
void init() {
int i, np;
scanf("%d%d%d", &l1, &l2, &Q);
scanf("%s", s);
scanf("%s", t);
for (i = l1; i > 0; i--) s[i] = s[i - 1];
for (i = l2; i > 0; i--) t[i] = t[i - 1];
fr[1] = 0;
for (i = 2; i <= l2; i++) {
np = fr[i - 1];
while (t[i] != t[np + 1] && np) np = fr[np];
if (t[i] == t[np + 1])
fr[i] = np + 1;
else
fr[i] = 0;
}
t[l2 + 1] = '&';
ma[0] = 0;
for (i = 1; i <= l1; i++) {
np = ma[i - 1];
while (np && s[i] != t[np + 1]) np = fr[np];
if (s[i] == t[np + 1])
ma[i] = np + 1;
else
ma[i] = 0;
}
sum[0] = 0;
for (i = 1; i <= l1; i++) sum[i] = sum[i - 1] + (ma[i] == l2);
}
int main() {
int i, l, r;
init();
for (i = 1; i <= Q; i++) {
scanf("%d%d", &l, &r);
if (r - l + 1 < l2) {
printf("0\n");
continue;
}
printf("%d\n", max(0, sum[r] - sum[l + l2 - 2]));
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long ans[1005];
int main() {
long long n, m, k;
string a, b;
cin >> n >> m >> k;
cin >> a >> b;
for (int i = 0; i <= n - m; i++) {
ans[i + 1] = ans[i] + (a.substr(i, m) == b);
}
while (k--) {
long long l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0" << endl;
} else {
cout << ans[r - m + 1] - ans[l - 1] << endl;
}
}
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long ans[1005];
int main() {
long long n, m, k;
string a, b;
cin >> n >> m >> k;
cin >> a >> b;
for (int i = 0; i <= n - m; i++) {
ans[i + 1] = ans[i] + (a.substr(i, m) == b);
}
while (k--) {
long long l, r;
cin >> l >> r;
if (r - l + 1 < m) {
cout << "0" << endl;
} else {
cout << ans[r - m + 1] - ans[l - 1] << endl;
}
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 + 10;
int main() {
int n, q, g;
int index;
int a[1005];
int k = 0;
string s, t;
string ss[1005];
scanf("%d%d%d", &n, &q, &g);
getchar();
cin >> s >> t;
for (int i = 0; i < n; i++) {
string str = s.substr(i, q);
if (str == t) {
a[k++] = i + 1;
}
}
while (g--) {
int sum = 0;
int l, r;
scanf("%d%d", &l, &r);
for (int i = 0; i < k; i++) {
if (a[i] >= l && a[i] + q - 1 <= r) sum++;
}
printf("%d\n", sum);
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000 + 10;
int main() {
int n, q, g;
int index;
int a[1005];
int k = 0;
string s, t;
string ss[1005];
scanf("%d%d%d", &n, &q, &g);
getchar();
cin >> s >> t;
for (int i = 0; i < n; i++) {
string str = s.substr(i, q);
if (str == t) {
a[k++] = i + 1;
}
}
while (g--) {
int sum = 0;
int l, r;
scanf("%d%d", &l, &r);
for (int i = 0; i < k; i++) {
if (a[i] >= l && a[i] + q - 1 <= r) sum++;
}
printf("%d\n", sum);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int arr[1001] = {0};
int u[200001] = {0};
int main() {
ios::sync_with_stdio(false);
int n, m, q, i, l, r;
cin >> n >> m >> q;
string s, t, ch;
cin >> s >> t;
int pos = -1, x = 0;
while (pos + 1 < s.size() && s.find(t, pos + 1) != string::npos) {
pos = s.find(t, pos + 1);
arr[pos + t.size() - 1]++;
}
for (i = 1; i < n; i++) {
arr[i] += arr[i - 1];
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < t.size())
cout << "0";
else if (l > 1) {
cout << arr[r - 1] - arr[l - 1 + t.size() - 2];
} else if (l == 1) {
cout << arr[r - 1];
}
if (i != q) cout << endl;
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int arr[1001] = {0};
int u[200001] = {0};
int main() {
ios::sync_with_stdio(false);
int n, m, q, i, l, r;
cin >> n >> m >> q;
string s, t, ch;
cin >> s >> t;
int pos = -1, x = 0;
while (pos + 1 < s.size() && s.find(t, pos + 1) != string::npos) {
pos = s.find(t, pos + 1);
arr[pos + t.size() - 1]++;
}
for (i = 1; i < n; i++) {
arr[i] += arr[i - 1];
}
for (i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < t.size())
cout << "0";
else if (l > 1) {
cout << arr[r - 1] - arr[l - 1 + t.size() - 2];
} else if (l == 1) {
cout << arr[r - 1];
}
if (i != q) cout << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
;
const double eps = 1e-8;
const int mod = 10007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f;
const unsigned long long p = 2333;
int n, m, q, l, r, ans;
unsigned long long hs[1007], ht, pw[1007];
char s[1007], t[1007];
void check(int l, int r) {
ans = 0;
for (int i = l; i <= r - m + 1; i++) {
if (hs[i + m - 1] - hs[i - 1] * pw[m] == ht) {
ans++;
}
}
}
int main() {
pw[0] = 1;
for (int i = 1; i < 1007; i++) {
pw[i] = pw[i - 1] * p;
}
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s + 1, t + 1);
hs[0] = 0, ht = 0;
for (int i = 1; i <= n; i++) {
hs[i] = hs[i - 1] * p + (s[i] - 'a' + 1);
}
for (int i = 1; i <= m; i++) {
ht = ht * p + (t[i] - 'a' + 1);
}
while (q--) {
scanf("%d%d", &l, &r);
check(l, r);
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Please formulate a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
;
const double eps = 1e-8;
const int mod = 10007;
const int maxn = 1e6 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f;
const unsigned long long p = 2333;
int n, m, q, l, r, ans;
unsigned long long hs[1007], ht, pw[1007];
char s[1007], t[1007];
void check(int l, int r) {
ans = 0;
for (int i = l; i <= r - m + 1; i++) {
if (hs[i + m - 1] - hs[i - 1] * pw[m] == ht) {
ans++;
}
}
}
int main() {
pw[0] = 1;
for (int i = 1; i < 1007; i++) {
pw[i] = pw[i - 1] * p;
}
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", s + 1, t + 1);
hs[0] = 0, ht = 0;
for (int i = 1; i <= n; i++) {
hs[i] = hs[i - 1] * p + (s[i] - 'a' + 1);
}
for (int i = 1; i <= m; i++) {
ht = ht * p + (t[i] - 'a' + 1);
}
while (q--) {
scanf("%d%d", &l, &r);
check(l, r);
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
inline int two(int n) { return 1 << n; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
template <class T>
T gcd(T a, T b) {
return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
return (a / gcd<T>(a, b) * b);
}
const int maxx = 1e4;
int arr[maxx] = {0};
string s, t;
int n, m, q;
bool ok = true;
int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; ++i) {
if (s[i] == t[0]) {
ok = true;
for (int j = 0; j < m; ++j) {
if (j + i > n - 1) break;
if (s[j + i] != t[j]) {
ok = false;
break;
}
}
if (ok) arr[i] = 1;
}
if (i > 0) arr[i] += arr[i - 1];
}
while (q--) {
int start, end;
cin >> start >> end;
start--, end--;
if (end - start + 1 < m)
cout << 0 << '\n';
else
cout << arr[end - m + 1] - (start ? arr[start - 1] : 0) << '\n';
}
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
inline int two(int n) { return 1 << n; }
inline void set_bit(int& n, int b) { n |= two(b); }
inline void unset_bit(int& n, int b) { n &= ~two(b); }
inline int last_bit(int n) { return n & (-n); }
template <class T>
T gcd(T a, T b) {
return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
return (a / gcd<T>(a, b) * b);
}
const int maxx = 1e4;
int arr[maxx] = {0};
string s, t;
int n, m, q;
bool ok = true;
int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; ++i) {
if (s[i] == t[0]) {
ok = true;
for (int j = 0; j < m; ++j) {
if (j + i > n - 1) break;
if (s[j + i] != t[j]) {
ok = false;
break;
}
}
if (ok) arr[i] = 1;
}
if (i > 0) arr[i] += arr[i - 1];
}
while (q--) {
int start, end;
cin >> start >> end;
start--, end--;
if (end - start + 1 < m)
cout << 0 << '\n';
else
cout << arr[end - m + 1] - (start ? arr[start - 1] : 0) << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
char a[1005], b[1005];
struct qujian {
int s, e;
};
qujian qq[10005];
int num = 0;
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s\n%s", a, b);
if (n < m) {
for (int i = 0; i < q; i++) {
int x, y;
scanf("%d%d", &x, &y);
printf("0\n");
}
} else {
for (int i = 0; i <= n - m; i++) {
if (strncmp((a + i), b, m) == 0) {
qq[num].s = i;
qq[num++].e = i + m - 1;
}
}
for (int i = 0; i < q; i++) {
int x, y;
scanf("%d%d", &x, &y);
if (y - x + 1 < m)
printf("0\n");
else {
x = x - 1;
y = y - 1;
int flag = 0, tnum = 0;
for (int i = 0; i < num; i++) {
if (qq[i].s >= x && qq[i].e <= y) {
tnum++;
}
}
printf("%d\n", tnum);
}
}
}
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
char a[1005], b[1005];
struct qujian {
int s, e;
};
qujian qq[10005];
int num = 0;
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s\n%s", a, b);
if (n < m) {
for (int i = 0; i < q; i++) {
int x, y;
scanf("%d%d", &x, &y);
printf("0\n");
}
} else {
for (int i = 0; i <= n - m; i++) {
if (strncmp((a + i), b, m) == 0) {
qq[num].s = i;
qq[num++].e = i + m - 1;
}
}
for (int i = 0; i < q; i++) {
int x, y;
scanf("%d%d", &x, &y);
if (y - x + 1 < m)
printf("0\n");
else {
x = x - 1;
y = y - 1;
int flag = 0, tnum = 0;
for (int i = 0; i < num; i++) {
if (qq[i].s >= x && qq[i].e <= y) {
tnum++;
}
}
printf("%d\n", tnum);
}
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
string s1, s2;
int ans[maxn];
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> s1 >> s2;
memset(ans, 0, sizeof(ans));
for (int i = 0; i + m <= n; i++) {
if (s1.substr(i, m) == s2) {
ans[i + 1]++;
}
}
for (int i = 1; i <= n; i++) {
ans[i] += ans[i - 1];
}
while (q--) {
int a, b;
cin >> a >> b;
b -= m - 1;
if (b < 0) b = 0;
cout << max(0, ans[b] - ans[a - 1]) << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
string s1, s2;
int ans[maxn];
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> s1 >> s2;
memset(ans, 0, sizeof(ans));
for (int i = 0; i + m <= n; i++) {
if (s1.substr(i, m) == s2) {
ans[i + 1]++;
}
}
for (int i = 1; i <= n; i++) {
ans[i] += ans[i - 1];
}
while (q--) {
int a, b;
cin >> a >> b;
b -= m - 1;
if (b < 0) b = 0;
cout << max(0, ans[b] - ans[a - 1]) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int const lmt = 1e5 + 4;
long long pre[lmt];
int main() {
int n, m, q;
cin >> m >> n >> q;
string s, t;
cin >> s >> t;
string tt = t + '#' + s;
int len = tt.length();
int f[len];
int j = 0;
f[0] = 0;
for (int i = 1; i < len; i++) {
j = f[i - 1];
while (j > 0 && tt[i] != tt[j]) j = f[j - 1];
if (tt[i] == tt[j]) j++;
f[i] = j;
if (j == n) {
pre[i + 1 - (n + 1)] = 1;
}
}
for (int i = 1; i <= m; i++) {
pre[i] += pre[i - 1];
}
int l, r;
while (q--) {
cin >> l >> r;
if (r - l + 1 < n)
cout << "0\n";
else
cout << max(1ll * 0, pre[r] - pre[l + n - 2]) << "\n";
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int const lmt = 1e5 + 4;
long long pre[lmt];
int main() {
int n, m, q;
cin >> m >> n >> q;
string s, t;
cin >> s >> t;
string tt = t + '#' + s;
int len = tt.length();
int f[len];
int j = 0;
f[0] = 0;
for (int i = 1; i < len; i++) {
j = f[i - 1];
while (j > 0 && tt[i] != tt[j]) j = f[j - 1];
if (tt[i] == tt[j]) j++;
f[i] = j;
if (j == n) {
pre[i + 1 - (n + 1)] = 1;
}
}
for (int i = 1; i <= m; i++) {
pre[i] += pre[i - 1];
}
int l, r;
while (q--) {
cin >> l >> r;
if (r - l + 1 < n)
cout << "0\n";
else
cout << max(1ll * 0, pre[r] - pre[l + n - 2]) << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
int n, m, q;
int v[100000] = {};
void match() {
for (int i = 0; i <= n - m; i++) {
v[i + 1] = v[i] + (s.substr(i, m) == t);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
cin >> n >> m >> q;
cin >> s;
cin >> t;
match();
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
if ((b - a + 1) < m)
cout << "0\n";
else {
cout << v[b - m + 1] - v[a - 1] << endl;
}
}
return 0;
}
| ### Prompt
Construct a CPP code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
int n, m, q;
int v[100000] = {};
void match() {
for (int i = 0; i <= n - m; i++) {
v[i + 1] = v[i] + (s.substr(i, m) == t);
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
;
cin >> n >> m >> q;
cin >> s;
cin >> t;
match();
for (int i = 0; i < q; i++) {
int a, b;
cin >> a >> b;
if ((b - a + 1) < m)
cout << "0\n";
else {
cout << v[b - m + 1] - v[a - 1] << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
char str[maxn], s[maxn];
struct node {
char ch;
int val;
node() {}
node(char _ch, int _val) : ch(_ch), val(_val) {}
};
node stk[maxn];
int top;
int f[maxn];
int tr[maxn];
int len1, len2;
void update(int x, int v) {
while (x <= len1) {
tr[x] += v;
x += (x & (-x));
}
}
int get_sum(int x) {
int ans = 0;
while (x > 0) {
ans += tr[x];
x -= (x & (-x));
}
return ans;
}
int fail[maxn];
void getFail(char* P, int len) {
fail[0] = fail[1] = 0;
for (int i = 1; i < len; i++) {
int j = fail[i];
while (j && P[i] != P[j]) j = fail[j];
fail[i + 1] = (P[i] == P[j] ? j + 1 : 0);
}
}
int main() {
int m;
scanf("%d %d %d", &len1, &len2, &m);
scanf(" %s", str);
scanf(" %s", s);
getFail(s, len2);
int j = 0;
for (int i = 0; i < len1; i++) {
while (j && s[j] != str[i]) j = fail[j];
if (s[j] == str[i]) ++j;
if (j == len2) {
int x = i - len2 + 1;
update(x + 1, 1);
}
}
for (int i = 0; i < m; i++) {
int l, r;
scanf("%d %d", &l, &r);
if (r - l + 1 < len2) {
printf("0\n");
} else {
printf("%d\n", get_sum(r - len2 + 1) - get_sum(l - 1));
}
}
return 0;
}
| ### Prompt
Construct a Cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 10;
char str[maxn], s[maxn];
struct node {
char ch;
int val;
node() {}
node(char _ch, int _val) : ch(_ch), val(_val) {}
};
node stk[maxn];
int top;
int f[maxn];
int tr[maxn];
int len1, len2;
void update(int x, int v) {
while (x <= len1) {
tr[x] += v;
x += (x & (-x));
}
}
int get_sum(int x) {
int ans = 0;
while (x > 0) {
ans += tr[x];
x -= (x & (-x));
}
return ans;
}
int fail[maxn];
void getFail(char* P, int len) {
fail[0] = fail[1] = 0;
for (int i = 1; i < len; i++) {
int j = fail[i];
while (j && P[i] != P[j]) j = fail[j];
fail[i + 1] = (P[i] == P[j] ? j + 1 : 0);
}
}
int main() {
int m;
scanf("%d %d %d", &len1, &len2, &m);
scanf(" %s", str);
scanf(" %s", s);
getFail(s, len2);
int j = 0;
for (int i = 0; i < len1; i++) {
while (j && s[j] != str[i]) j = fail[j];
if (s[j] == str[i]) ++j;
if (j == len2) {
int x = i - len2 + 1;
update(x + 1, 1);
}
}
for (int i = 0; i < m; i++) {
int l, r;
scanf("%d %d", &l, &r);
if (r - l + 1 < len2) {
printf("0\n");
} else {
printf("%d\n", get_sum(r - len2 + 1) - get_sum(l - 1));
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1005, MAXM = 1005;
string T, P, tt;
int Next[MAXM], N, M, C;
void MakeNext(int M) {
int i = 0, j = -1;
Next[i] = -1;
while (i < M) {
if (j == -1 || P[i] == P[j])
Next[++i] = ++j;
else
j = Next[j];
}
}
int KMP(int pos, int N, int M) {
int i = pos, j = 0, ans = 0;
while (i < N) {
if (T[i] == P[j] || j == -1)
i++, j++;
else
j = Next[j];
if (j == M) {
ans++;
j = Next[j - 1];
i--;
}
}
return ans;
}
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> tt >> P;
int l, r;
while (q--) {
cin >> l >> r;
T = tt.substr(l - 1, r - l + 1);
N = r - l + 1;
M = m;
MakeNext(M);
printf("%d\n", KMP(0, N, M));
}
return 0;
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1005, MAXM = 1005;
string T, P, tt;
int Next[MAXM], N, M, C;
void MakeNext(int M) {
int i = 0, j = -1;
Next[i] = -1;
while (i < M) {
if (j == -1 || P[i] == P[j])
Next[++i] = ++j;
else
j = Next[j];
}
}
int KMP(int pos, int N, int M) {
int i = pos, j = 0, ans = 0;
while (i < N) {
if (T[i] == P[j] || j == -1)
i++, j++;
else
j = Next[j];
if (j == M) {
ans++;
j = Next[j - 1];
i--;
}
}
return ans;
}
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> tt >> P;
int l, r;
while (q--) {
cin >> l >> r;
T = tt.substr(l - 1, r - l + 1);
N = r - l + 1;
M = m;
MakeNext(M);
printf("%d\n", KMP(0, N, M));
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
int ans[1000003] = {0};
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) {
ans[i] = 1;
}
}
while (q--) {
int l, r, cnt = 0;
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
} else {
l--;
r--;
for (int i = l; i <= r; i++) {
if (ans[i] == 1 && i + m - 1 <= r) cnt++;
}
cout << cnt << endl;
}
}
return 0;
}
| ### Prompt
Please create a solution in CPP to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
int ans[1000003] = {0};
for (int i = 0; i < n; i++) {
if (s.substr(i, m) == t) {
ans[i] = 1;
}
}
while (q--) {
int l, r, cnt = 0;
cin >> l >> r;
if (r - l + 1 < m) {
cout << 0 << endl;
} else {
l--;
r--;
for (int i = l; i <= r; i++) {
if (ans[i] == 1 && i + m - 1 <= r) cnt++;
}
cout << cnt << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
string a;
string b;
int ans[1000 + 7];
int main() {
int m, n, q;
cin >> m >> n >> q;
cin >> a >> b;
int tmp = 0;
for (int i = 0; i < m; i++) {
if (a[i] == b[0]) {
int cnt = 0;
for (int j = 0; j < n; j++) {
if (a[i + j] == b[j])
cnt++;
else
break;
}
if (cnt == n) tmp += 1;
}
ans[i + 1] = tmp;
}
for (int i = 0; i < q; i++) {
int x, y;
cin >> x >> y;
y = y - n + 1;
if (x > y)
cout << "0" << endl;
else
cout << ans[y] - ans[x - 1] << endl;
}
}
| ### Prompt
In cpp, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string a;
string b;
int ans[1000 + 7];
int main() {
int m, n, q;
cin >> m >> n >> q;
cin >> a >> b;
int tmp = 0;
for (int i = 0; i < m; i++) {
if (a[i] == b[0]) {
int cnt = 0;
for (int j = 0; j < n; j++) {
if (a[i + j] == b[j])
cnt++;
else
break;
}
if (cnt == n) tmp += 1;
}
ans[i + 1] = tmp;
}
for (int i = 0; i < q; i++) {
int x, y;
cin >> x >> y;
y = y - n + 1;
if (x > y)
cout << "0" << endl;
else
cout << ans[y] - ans[x - 1] << endl;
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
template <class T1, class T2>
inline void checkmin(T1 &x, T2 y) {
if (x > y) x = y;
}
template <class T1, class T2>
inline void checkmax(T1 &x, T2 y) {
if (x < y) x = y;
}
template <class T1>
inline void sort(T1 &arr) {
sort(arr.begin(), arr.end());
}
template <class T1>
inline void rsort(T1 &arr) {
sort(arr.rbegin(), arr.rend());
}
template <class T1>
inline void reverse(T1 &arr) {
reverse(arr.begin(), arr.end());
}
template <class T1>
inline void shuffle(T1 &arr) {
for (int i = -int(arr.size()); i < int(arr.size()); ++i)
swap(arr[rand() % arr.size()], arr[rand() % arr.size()]);
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cout << fixed << setprecision(20);
srand(time(NULL));
long long n, m, q;
cin >> n >> m >> q;
string second, t;
cin >> second >> t;
vector<int> cnt(second.size());
for (int i = 0; i <= int(second.size()) - int(t.size()); ++i) {
string news = "";
for (int j = i; j < i + t.size(); ++j) news += second[j];
if (news == t) ++cnt[i];
}
for (int i = 1; i < cnt.size(); ++i) cnt[i] += cnt[i - 1];
for (int i = 0; i < q; ++i) {
int u, v;
cin >> u >> v;
--u;
v -= int(t.size());
if (v < u)
cout << 0 << '\n';
else {
if (u == 0)
cout << cnt[v] << '\n';
else
cout << cnt[v] - cnt[u - 1] << '\n';
}
}
return 0;
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-loops")
using namespace std;
template <class T1, class T2>
inline void checkmin(T1 &x, T2 y) {
if (x > y) x = y;
}
template <class T1, class T2>
inline void checkmax(T1 &x, T2 y) {
if (x < y) x = y;
}
template <class T1>
inline void sort(T1 &arr) {
sort(arr.begin(), arr.end());
}
template <class T1>
inline void rsort(T1 &arr) {
sort(arr.rbegin(), arr.rend());
}
template <class T1>
inline void reverse(T1 &arr) {
reverse(arr.begin(), arr.end());
}
template <class T1>
inline void shuffle(T1 &arr) {
for (int i = -int(arr.size()); i < int(arr.size()); ++i)
swap(arr[rand() % arr.size()], arr[rand() % arr.size()]);
}
signed main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
cout << fixed << setprecision(20);
srand(time(NULL));
long long n, m, q;
cin >> n >> m >> q;
string second, t;
cin >> second >> t;
vector<int> cnt(second.size());
for (int i = 0; i <= int(second.size()) - int(t.size()); ++i) {
string news = "";
for (int j = i; j < i + t.size(); ++j) news += second[j];
if (news == t) ++cnt[i];
}
for (int i = 1; i < cnt.size(); ++i) cnt[i] += cnt[i - 1];
for (int i = 0; i < q; ++i) {
int u, v;
cin >> u >> v;
--u;
v -= int(t.size());
if (v < u)
cout << 0 << '\n';
else {
if (u == 0)
cout << cnt[v] << '\n';
else
cout << cnt[v] - cnt[u - 1] << '\n';
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 100005;
int BIT[MaxN];
void update(int pos) {
int i, j, p, q;
for (i = pos; i < MaxN; i += i & (-i)) {
BIT[i] += 1;
}
}
int query(int pos) {
int i, j, p = 0, q;
for (i = pos; i > 0; i -= i & (-i)) {
p += BIT[i];
}
return p;
}
int main() {
cin.tie(0), ios_base::sync_with_stdio(0);
int N, M, K, i, j, p, q;
bool proof;
cin >> N >> M >> K;
string a, b;
cin >> a >> b;
for (i = 0; i <= N - M; i++) {
proof = 1;
for (j = i; j < i + M; j++) {
if (a[j] != b[j - i]) {
proof = 0;
continue;
}
}
if (proof) {
update(i + 1);
}
}
for (i = 0; i < K; i++) {
cin >> p >> q;
if (q - M - p + 1 < 0)
cout << "0\n";
else
cout << query(q - M + 1) - query(p - 1) << "\n";
}
}
| ### Prompt
Your task is to create a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int MaxN = 100005;
int BIT[MaxN];
void update(int pos) {
int i, j, p, q;
for (i = pos; i < MaxN; i += i & (-i)) {
BIT[i] += 1;
}
}
int query(int pos) {
int i, j, p = 0, q;
for (i = pos; i > 0; i -= i & (-i)) {
p += BIT[i];
}
return p;
}
int main() {
cin.tie(0), ios_base::sync_with_stdio(0);
int N, M, K, i, j, p, q;
bool proof;
cin >> N >> M >> K;
string a, b;
cin >> a >> b;
for (i = 0; i <= N - M; i++) {
proof = 1;
for (j = i; j < i + M; j++) {
if (a[j] != b[j - i]) {
proof = 0;
continue;
}
}
if (proof) {
update(i + 1);
}
}
for (i = 0; i < K; i++) {
cin >> p >> q;
if (q - M - p + 1 < 0)
cout << "0\n";
else
cout << query(q - M + 1) - query(p - 1) << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s[2000], t[2000];
int f[2000];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n - m + 1; i++) {
int flag = 1;
for (int j = 1; j <= m; j++)
if (s[i + j - 1] != t[j]) {
flag = 0;
break;
}
f[i] = f[i - 1] + flag;
}
for (int i = 1; i <= q; i++) {
int l, r;
scanf("%d%d", &l, &r);
if (r - l + 1 < m)
printf("0\n");
else
printf("%d\n", f[r - m + 1] - f[l - 1]);
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s[2000], t[2000];
int f[2000];
int main() {
int n, m, q;
scanf("%d%d%d", &n, &m, &q);
scanf("%s", s + 1);
scanf("%s", t + 1);
for (int i = 1; i <= n - m + 1; i++) {
int flag = 1;
for (int j = 1; j <= m; j++)
if (s[i + j - 1] != t[j]) {
flag = 0;
break;
}
f[i] = f[i - 1] + flag;
}
for (int i = 1; i <= q; i++) {
int l, r;
scanf("%d%d", &l, &r);
if (r - l + 1 < m)
printf("0\n");
else
printf("%d\n", f[r - m + 1] - f[l - 1]);
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n, m, l, r, q, nr;
string s;
string t;
cin >> n >> m >> q;
cin >> s >> t;
int cnt[1001] = {0};
for (int i = 0; i <= n - m; i++) {
if (s.substr(i, m) == t) {
cnt[i] = 1;
}
}
l = 0;
r = 0;
for (i = 1; i <= q; i++) {
cin >> l >> r;
nr = 0;
for (j = l - 1; j <= r - m; j++) {
if (cnt[j]) nr++;
}
cout << nr << '\n';
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n, m, l, r, q, nr;
string s;
string t;
cin >> n >> m >> q;
cin >> s >> t;
int cnt[1001] = {0};
for (int i = 0; i <= n - m; i++) {
if (s.substr(i, m) == t) {
cnt[i] = 1;
}
}
l = 0;
r = 0;
for (i = 1; i <= q; i++) {
cin >> l >> r;
nr = 0;
for (j = l - 1; j <= r - m; j++) {
if (cnt[j]) nr++;
}
cout << nr << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
long long j = 0, pre = i;
while (j < m) {
if (str[i] == s[j]) {
i++;
j++;
} else
break;
}
if (j == m) {
bra[pre + 1]++;
i--;
bra[i]--;
ara[pre + 1]++;
}
i = pre;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
bra[i] += bra[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long const MOD = 1e9 + 7;
long long const N = 1e3 + 10;
long long ara[N + 1];
long long bra[N + 1];
int main() {
(ios_base::sync_with_stdio(false), cin.tie(NULL));
long long n, m, q;
cin >> n >> m >> q;
string str, s;
cin >> str >> s;
for (long long i = 0; i < n - m + 1; i++) {
long long j = 0, pre = i;
while (j < m) {
if (str[i] == s[j]) {
i++;
j++;
} else
break;
}
if (j == m) {
bra[pre + 1]++;
i--;
bra[i]--;
ara[pre + 1]++;
}
i = pre;
}
for (long long i = 1; i <= n + 10; i++) {
ara[i] += ara[i - 1];
bra[i] += bra[i - 1];
}
while (q--) {
long long a, b;
cin >> a >> b;
b = b - m + 1;
cout << max(0ll, ara[max(b, 0ll)] - ara[a - 1]) << endl;
}
}
``` |
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long MAXN = 1e3 + 10;
long long n, m, q;
string a, b;
long long pref[MAXN];
inline void precalc() {
for (long long i = 0; i < n - m + 1; ++i) {
bool flag = true;
for (long long j = i; j < i + m; ++j) {
if (a[j] != b[j - i]) {
flag = false;
break;
}
}
if (i == 0) {
pref[i] = flag;
continue;
}
pref[i] = pref[i - 1] + flag;
}
}
inline long long get(long long i) {
if (i < 0) return 0;
return pref[i];
}
inline long long get_ans(long long L, long long R) {
if (R - m + 1 < 0 || L + m - 1 > R) return 0;
return pref[R - m + 1] - get(L - 1);
}
signed main() {
ios::sync_with_stdio(0);
cin >> n >> m >> q;
cin >> a >> b;
precalc();
for (long long i = 0; i < q; ++i) {
long long L, R;
cin >> L >> R;
--L, --R;
cout << get_ans(L, R) << "\n";
}
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long MAXN = 1e3 + 10;
long long n, m, q;
string a, b;
long long pref[MAXN];
inline void precalc() {
for (long long i = 0; i < n - m + 1; ++i) {
bool flag = true;
for (long long j = i; j < i + m; ++j) {
if (a[j] != b[j - i]) {
flag = false;
break;
}
}
if (i == 0) {
pref[i] = flag;
continue;
}
pref[i] = pref[i - 1] + flag;
}
}
inline long long get(long long i) {
if (i < 0) return 0;
return pref[i];
}
inline long long get_ans(long long L, long long R) {
if (R - m + 1 < 0 || L + m - 1 > R) return 0;
return pref[R - m + 1] - get(L - 1);
}
signed main() {
ios::sync_with_stdio(0);
cin >> n >> m >> q;
cin >> a >> b;
precalc();
for (long long i = 0; i < q; ++i) {
long long L, R;
cin >> L >> R;
--L, --R;
cout << get_ans(L, R) << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
void add_self(int& a, int b) {
a += b;
if (a >= (int)1e9 + 7) a -= (int)1e9 + 7;
}
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> a(n + 1, 0);
int cnt = 0;
for (int i = 0; i <= n - m; i++) {
bool ok = true;
int k = 0;
for (int j = i; j < i + m; j++) {
if (s[j] != t[k++]) {
ok = false;
break;
}
}
if (ok) {
cnt++;
}
a[i + 1] = cnt;
}
for (int i = n - m + 1; i <= n; i++) a[i] = cnt;
while (q--) {
int l, r;
cin >> l >> r;
int t = r - m + 1;
cout << (a[t] - a[l - 1]) * (r - l + 1 >= m) << endl;
}
}
int main() {
ios::sync_with_stdio(false);
int q = 1;
for (int i = 1; i <= q; i++) {
solve();
}
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
void add_self(int& a, int b) {
a += b;
if (a >= (int)1e9 + 7) a -= (int)1e9 + 7;
}
void solve() {
int n, m, q;
cin >> n >> m >> q;
string s, t;
cin >> s >> t;
vector<int> a(n + 1, 0);
int cnt = 0;
for (int i = 0; i <= n - m; i++) {
bool ok = true;
int k = 0;
for (int j = i; j < i + m; j++) {
if (s[j] != t[k++]) {
ok = false;
break;
}
}
if (ok) {
cnt++;
}
a[i + 1] = cnt;
}
for (int i = n - m + 1; i <= n; i++) a[i] = cnt;
while (q--) {
int l, r;
cin >> l >> r;
int t = r - m + 1;
cout << (a[t] - a[l - 1]) * (r - l + 1 >= m) << endl;
}
}
int main() {
ios::sync_with_stdio(false);
int q = 1;
for (int i = 1; i <= q; i++) {
solve();
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int mxn = 1e3 + 5, seed = 131, seed2 = 31, MOD = 1e9 + 9;
int n, m, q;
string s, t;
long long p[mxn], pp[mxn], a[mxn], aa[mxn], b[mxn], bb[mxn];
inline long long strHash(int l, int r) {
return (long long)((1LL * a[r] - 1LL * a[l - 1] * p[r - l + 1] % MOD) + MOD) %
MOD;
}
int main() {
cin >> n >> m >> q >> s >> t;
p[0] = 1;
for (int i = 1; i <= max(n, m); i++)
p[i] = ((p[i - 1] * seed % MOD) + MOD) % MOD;
for (int i = 1; i <= n; i++)
a[i] = ((a[i - 1] * seed + s[i - 1]) % MOD + MOD) % MOD;
for (int i = 1; i <= m; i++)
b[i] = ((b[i - 1] * seed + t[i - 1]) % MOD + MOD) % MOD;
for (int i = 1, L, R, ans; i <= q; i++) {
cin >> L >> R;
ans = 0;
for (int j = L; j + m - 1 <= R; j++) {
if (strHash(j, j + m - 1) == b[m]) ans++;
}
cout << ans << "\n";
}
}
| ### Prompt
Your challenge is to write a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int mxn = 1e3 + 5, seed = 131, seed2 = 31, MOD = 1e9 + 9;
int n, m, q;
string s, t;
long long p[mxn], pp[mxn], a[mxn], aa[mxn], b[mxn], bb[mxn];
inline long long strHash(int l, int r) {
return (long long)((1LL * a[r] - 1LL * a[l - 1] * p[r - l + 1] % MOD) + MOD) %
MOD;
}
int main() {
cin >> n >> m >> q >> s >> t;
p[0] = 1;
for (int i = 1; i <= max(n, m); i++)
p[i] = ((p[i - 1] * seed % MOD) + MOD) % MOD;
for (int i = 1; i <= n; i++)
a[i] = ((a[i - 1] * seed + s[i - 1]) % MOD + MOD) % MOD;
for (int i = 1; i <= m; i++)
b[i] = ((b[i - 1] * seed + t[i - 1]) % MOD + MOD) % MOD;
for (int i = 1, L, R, ans; i <= q; i++) {
cin >> L >> R;
ans = 0;
for (int j = L; j + m - 1 <= R; j++) {
if (strHash(j, j + m - 1) == b[m]) ans++;
}
cout << ans << "\n";
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int match[1009];
int prefix[1009];
string s, t;
bool isMatch(int a) {
for (int i = 0; i < t.size(); i++) {
if (i + a >= s.size() || s[i + a] != t[i]) {
return false;
}
}
return true;
}
int main() {
ios_base::sync_with_stdio(0);
int n, m;
cin >> n >> m;
int q;
cin >> q;
cin >> s >> t;
for (int i = 0; i < s.size(); i++) {
match[i] = isMatch(i);
}
prefix[0] = match[0];
for (int i = 1; i < s.size(); i++) {
prefix[i] = prefix[i - 1] + match[i];
}
cout << endl;
for (int i = 0; i < q; i++) {
int p, q;
cin >> p >> q;
q -= t.size() - 2;
q--;
p--;
q--;
p--;
int ans;
if (q < 0 || q <= p) {
ans = 0;
} else if (p == -1) {
ans = prefix[q];
} else {
ans = prefix[q] - prefix[p];
}
cout << ans << endl;
}
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int match[1009];
int prefix[1009];
string s, t;
bool isMatch(int a) {
for (int i = 0; i < t.size(); i++) {
if (i + a >= s.size() || s[i + a] != t[i]) {
return false;
}
}
return true;
}
int main() {
ios_base::sync_with_stdio(0);
int n, m;
cin >> n >> m;
int q;
cin >> q;
cin >> s >> t;
for (int i = 0; i < s.size(); i++) {
match[i] = isMatch(i);
}
prefix[0] = match[0];
for (int i = 1; i < s.size(); i++) {
prefix[i] = prefix[i - 1] + match[i];
}
cout << endl;
for (int i = 0; i < q; i++) {
int p, q;
cin >> p >> q;
q -= t.size() - 2;
q--;
p--;
q--;
p--;
int ans;
if (q < 0 || q <= p) {
ans = 0;
} else if (p == -1) {
ans = prefix[q];
} else {
ans = prefix[q] - prefix[p];
}
cout << ans << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, q, t = 0;
string a, b, s = "";
cin >> n >> m >> q >> a >> b;
vector<pair<int, int> > v;
while (t + b.size() - 1 < a.size()) {
if (a.substr(t, b.size()) == b) {
v.push_back({t + 1, t + b.size()});
}
t++;
}
while (q--) {
int l, r;
cin >> l >> r;
int c = 0;
for (int i = 0; i < v.size(); i++) {
if (l <= v[i].first && r >= v[i].second) c++;
}
cout << c << '\n';
}
}
| ### Prompt
In CPP, your task is to solve the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
int n, m, q, t = 0;
string a, b, s = "";
cin >> n >> m >> q >> a >> b;
vector<pair<int, int> > v;
while (t + b.size() - 1 < a.size()) {
if (a.substr(t, b.size()) == b) {
v.push_back({t + 1, t + b.size()});
}
t++;
}
while (q--) {
int l, r;
cin >> l >> r;
int c = 0;
for (int i = 0; i < v.size(); i++) {
if (l <= v[i].first && r >= v[i].second) c++;
}
cout << c << '\n';
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
string s, t;
int ans[100009];
int lps[100009];
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> s >> t;
int sz = t.size();
int sz1 = s.size();
for (int i = 1; i <= (sz1 + 1 - sz); i++) {
int flag = 0;
string sb = s.substr(i - 1, sz);
if (sb == t) ans[i]++;
}
for (int i = 1; i <= sz1; i++) ans[i] += ans[i - 1];
while (q--) {
int x, y;
cin >> x >> y;
if (y - x + 1 < sz)
cout << 0 << endl;
else {
int mx = y + 1 - sz;
int val;
cout << ans[mx] - ans[x - 1] << endl;
}
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
string s, t;
int ans[100009];
int lps[100009];
int main() {
int n, m, q;
cin >> n >> m >> q;
cin >> s >> t;
int sz = t.size();
int sz1 = s.size();
for (int i = 1; i <= (sz1 + 1 - sz); i++) {
int flag = 0;
string sb = s.substr(i - 1, sz);
if (sb == t) ans[i]++;
}
for (int i = 1; i <= sz1; i++) ans[i] += ans[i - 1];
while (q--) {
int x, y;
cin >> x >> y;
if (y - x + 1 < sz)
cout << 0 << endl;
else {
int mx = y + 1 - sz;
int val;
cout << ans[mx] - ans[x - 1] << endl;
}
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q, i, j, l, r, a[5000];
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
if (m > n) goto label;
if (s.compare(0, m, t) == 0) {
a[0] = 1;
} else
a[0] = 0;
for (j = 1; j <= n - m + 1; j++) {
if (s.compare(j, m, t) == 0) {
a[j] = a[j - 1] + 1;
} else {
a[j] = a[j - 1];
}
}
for (j = n - m + 2; j < n; j++) {
a[j] = a[n - m + 1];
}
label:
for (i = 0; i < q; i++) {
cin >> l >> r;
l--;
r--;
if (r - l < m - 1 || m > n)
cout << l - l << endl;
else if (l == 0)
cout << a[r - m + 1] << endl;
else
cout << a[r - m + 1] - a[l - 1] << endl;
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n, m, q, i, j, l, r, a[5000];
string s, t;
cin >> n >> m >> q;
cin >> s >> t;
if (m > n) goto label;
if (s.compare(0, m, t) == 0) {
a[0] = 1;
} else
a[0] = 0;
for (j = 1; j <= n - m + 1; j++) {
if (s.compare(j, m, t) == 0) {
a[j] = a[j - 1] + 1;
} else {
a[j] = a[j - 1];
}
}
for (j = n - m + 2; j < n; j++) {
a[j] = a[n - m + 1];
}
label:
for (i = 0; i < q; i++) {
cin >> l >> r;
l--;
r--;
if (r - l < m - 1 || m > n)
cout << l - l << endl;
else if (l == 0)
cout << a[r - m + 1] << endl;
else
cout << a[r - m + 1] - a[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int max(int a, int b) { return (a > b) ? a : b; }
int min(int a, int b) { return (a < b) ? a : b; }
int main() {
int n, m, k;
cin >> n >> m >> k;
string s, t;
cin >> s;
cin >> t;
unordered_map<char, int> umap;
vector<int> skips;
for (int i = 0; i < t.size(); i++) {
if (umap.find(t[i]) == umap.end()) {
umap[t[i]] = i;
skips.push_back(i + 1);
} else {
skips.push_back(i - umap[t[i]]);
umap[t[i]] = i;
}
}
vector<int> counter(n);
int count = 0, x = 0;
for (int j = 0; j < n && m + j - 1 < n;) {
bool ans = 0;
for (int x = 0; x < m && m + j - 1 < n; x++) {
if (t[x] != s[j + x]) {
ans = 0;
if (x != 0)
j += skips[x - 1];
else
j += 1;
break;
} else {
ans = 1;
}
}
if (ans == 1) {
count++;
counter[j + m - 1] = count;
j += skips[m - 1];
}
}
for (int i = 1; i < n; i++) {
if (counter[i - 1] != 0 && counter[i] == 0) counter[i] = counter[i - 1];
}
for (int i = 0; i < k; i++) {
int l, r;
cin >> l >> r;
if (l - 1 + m - 2 >= 0)
cout << counter[r - 1] - counter[min(r - 1, l - 1 + m - 2)] << endl;
else
cout << counter[r - 1] << endl;
}
return 0;
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int max(int a, int b) { return (a > b) ? a : b; }
int min(int a, int b) { return (a < b) ? a : b; }
int main() {
int n, m, k;
cin >> n >> m >> k;
string s, t;
cin >> s;
cin >> t;
unordered_map<char, int> umap;
vector<int> skips;
for (int i = 0; i < t.size(); i++) {
if (umap.find(t[i]) == umap.end()) {
umap[t[i]] = i;
skips.push_back(i + 1);
} else {
skips.push_back(i - umap[t[i]]);
umap[t[i]] = i;
}
}
vector<int> counter(n);
int count = 0, x = 0;
for (int j = 0; j < n && m + j - 1 < n;) {
bool ans = 0;
for (int x = 0; x < m && m + j - 1 < n; x++) {
if (t[x] != s[j + x]) {
ans = 0;
if (x != 0)
j += skips[x - 1];
else
j += 1;
break;
} else {
ans = 1;
}
}
if (ans == 1) {
count++;
counter[j + m - 1] = count;
j += skips[m - 1];
}
}
for (int i = 1; i < n; i++) {
if (counter[i - 1] != 0 && counter[i] == 0) counter[i] = counter[i - 1];
}
for (int i = 0; i < k; i++) {
int l, r;
cin >> l >> r;
if (l - 1 + m - 2 >= 0)
cout << counter[r - 1] - counter[min(r - 1, l - 1 + m - 2)] << endl;
else
cout << counter[r - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long mod[] = {(long long)1e9 + 7, (long long)1e9 + 9,
(long long)1e9 + 21, (long long)1e9 + 33};
const int maxn = 2e3 + 5;
const int Nmod = 3;
string a, b;
int i, j, n, m, q;
long long Pow[Nmod][maxn];
static long long Hash[Nmod][maxn], Hash1[Nmod][maxn], Key[Nmod];
static int dp[maxn][maxn];
long long gethash(int i, int j, int type) {
return (Hash[type][j] - Hash[type][i - 1] * Pow[type][j - i + 1] +
(long long)mod[type] * (long long)mod[type]) %
mod[type];
}
void init() {
cin >> n >> m >> q;
cin >> a;
a = " " + a;
for (int i = 0; i < Nmod; ++i) {
Pow[i][0] = 1;
for (int j = 1; j <= n; ++j) Pow[i][j] = (Pow[i][j - 1] * 31) % mod[i];
}
for (int i = 0; i < Nmod; ++i) {
Hash[i][0] = 0;
for (int j = 1; j <= n; ++j)
Hash[i][j] = (Hash[i][j - 1] * 31 + a[j] - 'a') % mod[i];
}
cin >> b;
b = " " + b;
for (int i = 0; i < Nmod; ++i) {
for (int j = 1; j <= m; ++j) {
Key[i] = (Key[i] * 31 + b[j] - 'a') % mod[i];
}
}
for (int t = (1); t <= (n); ++t) {
for (int i = t; i <= n - m + 1; ++i) {
int ok = 1;
for (int j = 0; j < Nmod; ++j) {
ok = min(ok, (int)(gethash(i, i + m - 1, j) == Key[j]));
}
dp[t][i + m - 1] = dp[t][i + m - 2] + ok;
}
}
for (int i = (1); i <= (q); ++i) {
int x, y;
cin >> x >> y;
cout << dp[x][y] << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
init();
return 0;
}
| ### Prompt
Develop a solution in cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod[] = {(long long)1e9 + 7, (long long)1e9 + 9,
(long long)1e9 + 21, (long long)1e9 + 33};
const int maxn = 2e3 + 5;
const int Nmod = 3;
string a, b;
int i, j, n, m, q;
long long Pow[Nmod][maxn];
static long long Hash[Nmod][maxn], Hash1[Nmod][maxn], Key[Nmod];
static int dp[maxn][maxn];
long long gethash(int i, int j, int type) {
return (Hash[type][j] - Hash[type][i - 1] * Pow[type][j - i + 1] +
(long long)mod[type] * (long long)mod[type]) %
mod[type];
}
void init() {
cin >> n >> m >> q;
cin >> a;
a = " " + a;
for (int i = 0; i < Nmod; ++i) {
Pow[i][0] = 1;
for (int j = 1; j <= n; ++j) Pow[i][j] = (Pow[i][j - 1] * 31) % mod[i];
}
for (int i = 0; i < Nmod; ++i) {
Hash[i][0] = 0;
for (int j = 1; j <= n; ++j)
Hash[i][j] = (Hash[i][j - 1] * 31 + a[j] - 'a') % mod[i];
}
cin >> b;
b = " " + b;
for (int i = 0; i < Nmod; ++i) {
for (int j = 1; j <= m; ++j) {
Key[i] = (Key[i] * 31 + b[j] - 'a') % mod[i];
}
}
for (int t = (1); t <= (n); ++t) {
for (int i = t; i <= n - m + 1; ++i) {
int ok = 1;
for (int j = 0; j < Nmod; ++j) {
ok = min(ok, (int)(gethash(i, i + m - 1, j) == Key[j]));
}
dp[t][i + m - 1] = dp[t][i + m - 2] + ok;
}
}
for (int i = (1); i <= (q); ++i) {
int x, y;
cin >> x >> y;
cout << dp[x][y] << '\n';
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
init();
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int main() {
ios_base::sync_with_stdio(false);
int n, m, q, l, r, tlen, cnt;
char s[N], t[N];
bool match[N] = {0};
cin >> n >> m >> q >> s >> t;
tlen = strlen(t);
for (int i = 0; s[i]; ++i) {
match[i] = !strncmp(s + i, t, tlen);
}
while (q-- > 0) {
cin >> l >> r;
--l;
--r;
cnt = 0;
for (int i = l; i <= r - tlen + 1; ++i) {
cnt += match[i];
}
cout << cnt << "\n";
}
return 0;
}
| ### Prompt
Construct a cpp code solution to the problem outlined:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10;
int main() {
ios_base::sync_with_stdio(false);
int n, m, q, l, r, tlen, cnt;
char s[N], t[N];
bool match[N] = {0};
cin >> n >> m >> q >> s >> t;
tlen = strlen(t);
for (int i = 0; s[i]; ++i) {
match[i] = !strncmp(s + i, t, tlen);
}
while (q-- > 0) {
cin >> l >> r;
--l;
--r;
cnt = 0;
for (int i = l; i <= r - tlen + 1; ++i) {
cnt += match[i];
}
cout << cnt << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char a[1005], b[1005];
int sum[1005];
int main() {
ios::sync_with_stdio(false);
int n, m, t;
cin >> n >> m >> t;
cin >> a + 1 >> b + 1;
for (int i = m; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i - m + j] != b[j]) break;
if (j == m) sum[i] = 1;
}
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r] - sum[l + m - 2] << endl;
}
return 0;
}
| ### Prompt
Create a solution in cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char a[1005], b[1005];
int sum[1005];
int main() {
ios::sync_with_stdio(false);
int n, m, t;
cin >> n >> m >> t;
cin >> a + 1 >> b + 1;
for (int i = m; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (a[i - m + j] != b[j]) break;
if (j == m) sum[i] = 1;
}
}
for (int i = 1; i <= n; i++) sum[i] += sum[i - 1];
while (t--) {
int l, r;
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else
cout << sum[r] - sum[l + m - 2] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long pw[(int)1e6 + 5];
long long base = 1331;
long long Hash[(int)1e6 + 5];
void preCal() {
pw[0] = 1;
for (int i = 1; i < (int)1e6 + 5; i++) pw[i] = pw[i - 1] * base;
}
void setHash(string s) {
Hash[0] = 0;
for (int i = 1; i < s.size(); i++) Hash[i] = Hash[i - 1] * base + s[i];
}
long long getHash(int l, int r) {
return Hash[r] - (Hash[l - 1] * pw[r - l + 1]);
}
long long Hasher(string s) {
long long hashValue = 0;
for (int i = 0; i < s.size(); i++) hashValue = hashValue * base + s[i];
return hashValue;
}
int main() {
preCal();
long long n, m, q;
cin >> n >> m >> q;
string s, ss;
cin >> s >> ss;
while (q--) {
long long le, ri;
cin >> le >> ri;
string temp = "\0";
for (int i = le - 1; i < ri; i++) temp += s[i];
temp = "$" + temp;
setHash(temp);
int l1 = temp.size(), l2 = ss.size();
long long hashValue = Hasher(ss);
int c = 0;
for (int i = 1; i + l2 <= l1; i++) {
int l = i, r = i + l2 - 1;
if (getHash(l, r) == hashValue) c++;
}
cout << c << endl;
}
}
| ### Prompt
Generate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long pw[(int)1e6 + 5];
long long base = 1331;
long long Hash[(int)1e6 + 5];
void preCal() {
pw[0] = 1;
for (int i = 1; i < (int)1e6 + 5; i++) pw[i] = pw[i - 1] * base;
}
void setHash(string s) {
Hash[0] = 0;
for (int i = 1; i < s.size(); i++) Hash[i] = Hash[i - 1] * base + s[i];
}
long long getHash(int l, int r) {
return Hash[r] - (Hash[l - 1] * pw[r - l + 1]);
}
long long Hasher(string s) {
long long hashValue = 0;
for (int i = 0; i < s.size(); i++) hashValue = hashValue * base + s[i];
return hashValue;
}
int main() {
preCal();
long long n, m, q;
cin >> n >> m >> q;
string s, ss;
cin >> s >> ss;
while (q--) {
long long le, ri;
cin >> le >> ri;
string temp = "\0";
for (int i = le - 1; i < ri; i++) temp += s[i];
temp = "$" + temp;
setHash(temp);
int l1 = temp.size(), l2 = ss.size();
long long hashValue = Hasher(ss);
int c = 0;
for (int i = 1; i + l2 <= l1; i++) {
int l = i, r = i + l2 - 1;
if (getHash(l, r) == hashValue) c++;
}
cout << c << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
vector<int> ans;
for (int i = 0; i < n; i++) {
bool flag = 1;
for (int j = 0; j < m; j++) {
if (t[j] != s[j + i]) {
flag = 0;
break;
}
}
if (flag) {
ans.push_back(i + 1);
}
}
while (q--) {
int l, r;
cin >> l >> r;
r = (r - m + 1);
if (r < l) {
cout << 0 << '\n';
} else
cout << (upper_bound(ans.begin(), ans.end(), r) -
lower_bound(ans.begin(), ans.end(), l))
<< '\n';
}
return 0;
}
| ### Prompt
Create a solution in CPP for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
int n, m, q;
string s, t;
cin >> n >> m >> q >> s >> t;
vector<int> ans;
for (int i = 0; i < n; i++) {
bool flag = 1;
for (int j = 0; j < m; j++) {
if (t[j] != s[j + i]) {
flag = 0;
break;
}
}
if (flag) {
ans.push_back(i + 1);
}
}
while (q--) {
int l, r;
cin >> l >> r;
r = (r - m + 1);
if (r < l) {
cout << 0 << '\n';
} else
cout << (upper_bound(ans.begin(), ans.end(), r) -
lower_bound(ans.begin(), ans.end(), l))
<< '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char str[1011];
char ptr[1011];
int Next[1011];
int plen, slen;
int n, m, q, x, y;
inline int in() {
int f = 1, ans, ch;
while ((ch = getchar()) < '0' || ch > '9')
if ('-' == ch) {
ch = getchar(), f = -1;
break;
}
ans = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9')
ans = (ans << 3) + (ans << 1) + ch - '0';
return f * ans;
}
void getNext() {
int i = 0, j = -1;
memset(Next, 0, sizeof(Next));
Next[0] = -1;
while (i < plen) {
if (j == -1 || ptr[i] == ptr[j]) {
i++, j++;
Next[i] = j;
} else
j = Next[j];
}
}
int KMP() {
int i = x - 1, j = 0;
int ans = 0;
while (i < y) {
if (j == -1 || str[i] == ptr[j]) {
i++, j++;
} else
j = Next[j];
if (j == plen) {
j = Next[j];
ans++;
}
}
return ans;
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str);
scanf("%s", ptr);
plen = strlen(ptr);
slen = strlen(str);
getNext();
while (q--) {
x = in();
y = in();
int ans = KMP();
printf("%d\n", ans);
}
return 0;
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char str[1011];
char ptr[1011];
int Next[1011];
int plen, slen;
int n, m, q, x, y;
inline int in() {
int f = 1, ans, ch;
while ((ch = getchar()) < '0' || ch > '9')
if ('-' == ch) {
ch = getchar(), f = -1;
break;
}
ans = ch - '0';
while ((ch = getchar()) >= '0' && ch <= '9')
ans = (ans << 3) + (ans << 1) + ch - '0';
return f * ans;
}
void getNext() {
int i = 0, j = -1;
memset(Next, 0, sizeof(Next));
Next[0] = -1;
while (i < plen) {
if (j == -1 || ptr[i] == ptr[j]) {
i++, j++;
Next[i] = j;
} else
j = Next[j];
}
}
int KMP() {
int i = x - 1, j = 0;
int ans = 0;
while (i < y) {
if (j == -1 || str[i] == ptr[j]) {
i++, j++;
} else
j = Next[j];
if (j == plen) {
j = Next[j];
ans++;
}
}
return ans;
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", str);
scanf("%s", ptr);
plen = strlen(ptr);
slen = strlen(str);
getNext();
while (q--) {
x = in();
y = in();
int ans = KMP();
printf("%d\n", ans);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
string a, b;
cin >> n >> m >> q >> a >> b;
vector<int> v(n);
for (int i = 0; i < n; i++) {
bool good = true;
if (i) v[i] = v[i - 1];
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
good = false;
break;
}
}
if (good) v[i]++;
}
for (int i = 0, l, r; i < q; i++) {
cin >> l >> r;
l--, r--;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
r -= m - 1;
cout << (r >= 0 ? v[r] : 0) - (l > 0 ? v[l - 1] : 0) << endl;
}
}
| ### Prompt
Your challenge is to write a cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
int n, m, q;
string a, b;
cin >> n >> m >> q >> a >> b;
vector<int> v(n);
for (int i = 0; i < n; i++) {
bool good = true;
if (i) v[i] = v[i - 1];
for (int j = 0; j < m; j++) {
if (a[i + j] != b[j]) {
good = false;
break;
}
}
if (good) v[i]++;
}
for (int i = 0, l, r; i < q; i++) {
cin >> l >> r;
l--, r--;
if (r - l + 1 < m) {
cout << 0 << endl;
continue;
}
r -= m - 1;
cout << (r >= 0 ? v[r] : 0) - (l > 0 ? v[l - 1] : 0) << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
char a[1005], b[1005], c[1005];
int p[1005];
int n, m, q;
void pipei() {
int l, r, j = 0;
scanf("%d%d", &l, &r);
int lb = m;
int la = r - l + 1;
for (int i = l; i <= r; i++) c[i - l + 1] = a[i - 1];
for (int i = 2; i <= lb; i++) {
while (j > 0 && b[i] != b[j + 1]) j = p[j];
if (b[j + 1] == b[i]) j++;
p[i] = j;
}
j = 0;
int ans = 0;
for (int i = 1; i <= la; i++) {
while (j > 0 && b[j + 1] != c[i]) j = p[j];
if (b[j + 1] == c[i]) j++;
if (j == lb) ans++;
}
printf("%d\n", ans);
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", a, b + 1);
memset(p, 0, sizeof(p));
while (q--) {
pipei();
}
return 0;
}
| ### Prompt
Please provide a cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char a[1005], b[1005], c[1005];
int p[1005];
int n, m, q;
void pipei() {
int l, r, j = 0;
scanf("%d%d", &l, &r);
int lb = m;
int la = r - l + 1;
for (int i = l; i <= r; i++) c[i - l + 1] = a[i - 1];
for (int i = 2; i <= lb; i++) {
while (j > 0 && b[i] != b[j + 1]) j = p[j];
if (b[j + 1] == b[i]) j++;
p[i] = j;
}
j = 0;
int ans = 0;
for (int i = 1; i <= la; i++) {
while (j > 0 && b[j + 1] != c[i]) j = p[j];
if (b[j + 1] == c[i]) j++;
if (j == lb) ans++;
}
printf("%d\n", ans);
}
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s%s", a, b + 1);
memset(p, 0, sizeof(p));
while (q--) {
pipei();
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, l, r;
char s[1005], t[1005];
int sum[1005];
int flag[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", &s);
scanf("%s", &t);
sum[0] = 0;
for (int i = 0; i < n - m + 1; i++) {
int now = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
now = 0;
break;
}
}
flag[i] = now;
sum[i + 1] = sum[i] + flag[i];
}
for (int i = max(0, n - m + 1); i < n; i++) {
sum[i + 1] = sum[i];
}
for (int i = 0; i < q; i++) {
scanf("%d%d", &l, &r);
if (r - m + 1 >= l - 1)
printf("%d\n", sum[r - m + 1] - sum[l - 1]);
else
printf("0\n");
}
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, l, r;
char s[1005], t[1005];
int sum[1005];
int flag[1005];
int main() {
scanf("%d%d%d", &n, &m, &q);
scanf("%s", &s);
scanf("%s", &t);
sum[0] = 0;
for (int i = 0; i < n - m + 1; i++) {
int now = 1;
for (int j = 0; j < m; j++) {
if (s[i + j] != t[j]) {
now = 0;
break;
}
}
flag[i] = now;
sum[i + 1] = sum[i] + flag[i];
}
for (int i = max(0, n - m + 1); i < n; i++) {
sum[i + 1] = sum[i];
}
for (int i = 0; i < q; i++) {
scanf("%d%d", &l, &r);
if (r - m + 1 >= l - 1)
printf("%d\n", sum[r - m + 1] - sum[l - 1]);
else
printf("0\n");
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
long long n, m, q, l, r;
long long arr[10000];
char s[10000], t[10000];
int main() {
cin >> n >> m >> q;
cin >> s + 1 >> t + 1;
for (long long i = 1; i <= n - m + 1; i++) {
int acc = 1;
for (int j = 1; j <= m; j++)
if (acc == 1 && s[i + j - 1] != t[j]) acc = 0;
arr[i] = arr[i - 1] + acc;
}
while (q--) {
cin >> l >> r;
if (r >= l + m - 1)
cout << arr[r - m + 1] - arr[l - 1] << "\n";
else
cout << 0 << "\n";
}
return 0;
}
| ### Prompt
Your challenge is to write a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m, q, l, r;
long long arr[10000];
char s[10000], t[10000];
int main() {
cin >> n >> m >> q;
cin >> s + 1 >> t + 1;
for (long long i = 1; i <= n - m + 1; i++) {
int acc = 1;
for (int j = 1; j <= m; j++)
if (acc == 1 && s[i + j - 1] != t[j]) acc = 0;
arr[i] = arr[i - 1] + acc;
}
while (q--) {
cin >> l >> r;
if (r >= l + m - 1)
cout << arr[r - m + 1] - arr[l - 1] << "\n";
else
cout << 0 << "\n";
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s, t;
int st[1010], ed[1010];
int main() {
ios_base::sync_with_stdio(false);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
bool ok = true;
for (int j = 0; j < m; j++) {
if (t[j] != s[i + j]) {
ok = false;
break;
}
}
if (ok) {
ed[i + m - 1]++;
}
}
for (int i = 1; i < n; i++) {
st[i] += st[i - 1];
ed[i] += ed[i - 1];
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--, r--;
int R = r;
int L = l + m - 1;
if (L <= R) {
int res = ed[R];
if (L) res -= ed[L - 1];
cout << res << '\n';
} else
cout << 0 << '\n';
}
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
string s, t;
int st[1010], ed[1010];
int main() {
ios_base::sync_with_stdio(false);
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i < n - m + 1; i++) {
bool ok = true;
for (int j = 0; j < m; j++) {
if (t[j] != s[i + j]) {
ok = false;
break;
}
}
if (ok) {
ed[i + m - 1]++;
}
}
for (int i = 1; i < n; i++) {
st[i] += st[i - 1];
ed[i] += ed[i - 1];
}
for (int i = 0; i < q; i++) {
int l, r;
cin >> l >> r;
l--, r--;
int R = r;
int L = l + m - 1;
if (L <= R) {
int res = ed[R];
if (L) res -= ed[L - 1];
cout << res << '\n';
} else
cout << 0 << '\n';
}
}
``` |
#include <bits/stdc++.h>
const long long N = 1e6 + 10;
const long long mod = 998244353;
using namespace std;
long long w[1111];
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
for (long long i = 0; i < n - m + 1; i++) {
long long ff = 0;
if (a[i] == b[0]) {
for (long long j = i, k = 0; j < a.size() && k < b.size(); j++, k++) {
if (a[j] != b[k]) {
ff = 1;
break;
}
}
if (ff == 0) w[i + 1] = 1;
}
}
for (long long i = 1; i <= 1010; i++) w[i] += w[i - 1];
while (q--) {
long long l, r;
cin >> l >> r;
r -= m;
r++;
if (r < l)
cout << 0 << endl;
else
cout << w[r] - w[l - 1] << endl;
}
return 0;
}
| ### Prompt
Develop a solution in CPP to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
const long long N = 1e6 + 10;
const long long mod = 998244353;
using namespace std;
long long w[1111];
int32_t main() {
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
long long n, m, q;
cin >> n >> m >> q;
string a, b;
cin >> a >> b;
for (long long i = 0; i < n - m + 1; i++) {
long long ff = 0;
if (a[i] == b[0]) {
for (long long j = i, k = 0; j < a.size() && k < b.size(); j++, k++) {
if (a[j] != b[k]) {
ff = 1;
break;
}
}
if (ff == 0) w[i + 1] = 1;
}
}
for (long long i = 1; i <= 1010; i++) w[i] += w[i - 1];
while (q--) {
long long l, r;
cin >> l >> r;
r -= m;
r++;
if (r < l)
cout << 0 << endl;
else
cout << w[r] - w[l - 1] << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
int a[1000];
using namespace std;
int main() {
int n, m, q, l, r, i, x;
scanf("%d %d %d", &n, &m, &q);
string s1, t;
cin >> s1 >> t;
for (i = 0; i <= n - m; i++) {
if (s1.substr(i, m) == t) a[i] = 1;
}
while (q--) {
x = 0;
scanf("%d %d", &l, &r);
for (i = l - 1; i <= r - m; i++) {
if (a[i]) x++;
}
printf("%d\n", x);
}
return 0;
}
| ### Prompt
Create a solution in Cpp for the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
int a[1000];
using namespace std;
int main() {
int n, m, q, l, r, i, x;
scanf("%d %d %d", &n, &m, &q);
string s1, t;
cin >> s1 >> t;
for (i = 0; i <= n - m; i++) {
if (s1.substr(i, m) == t) a[i] = 1;
}
while (q--) {
x = 0;
scanf("%d %d", &l, &r);
for (i = l - 1; i <= r - m; i++) {
if (a[i]) x++;
}
printf("%d\n", x);
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long INF = (1LL << 45LL);
const long long MAXLL = 9223372036854775807LL;
const unsigned long long MAXULL = 18446744073709551615LLU;
const long long MOD = 1000000007;
const long double DELTA = 0.000000001L;
inline long long fmm(long long a, long long b, long long m = MOD) {
long long r = 0;
a %= m;
b %= m;
while (b > 0) {
if (b & 1) {
r += a;
r %= m;
}
a += a;
a %= m;
b >>= 1;
}
return r % m;
}
inline long long fme(long long a, long long b, long long m = MOD) {
long long r = 1;
a %= m;
while (b > 0) {
if (b & 1) {
r *= a;
r %= m;
}
a *= a;
a %= m;
b >>= 1;
}
return r % m;
}
inline long long sfme(long long a, long long b, long long m = MOD) {
long long r = 1;
a %= m;
while (b > 0) {
if (b & 1) r = fmm(r, a, m);
a = fmm(a, a, m);
b >>= 1;
}
return r % m;
}
std::vector<long long> primes;
long long primsiz;
std::vector<long long> fact;
std::vector<long long> invfact;
inline void sieve(long long n) {
long long i, j;
std::vector<bool> a(n);
a[0] = true;
a[1] = true;
for (i = 2; i * i < n; ++i) {
if (!a[i]) {
for (j = i * i; j < n; j += i) {
a[j] = true;
}
}
}
for (i = 2; i < n; ++i)
if (!a[i]) primes.push_back(i);
primsiz = primes.size();
}
inline void sieve() {
long long n = 1010000, i, j, k = 0;
std::vector<bool> a(n);
primes.resize(79252);
a[0] = a[1] = true;
for (i = 2; (j = (i << 1)) < n; ++i) a[j] = true;
for (i = 3; i * i < n; i += 2) {
if (!a[i]) {
k = (i << 1);
for (j = i * i; j < n; j += k) a[j] = true;
}
}
k = 0;
for (i = 2; i < n; ++i)
if (!a[i]) primes[k++] = i;
primsiz = k;
}
inline bool isPrimeSmall(unsigned long long n) {
if (((!(n & 1)) && n != 2) || (n < 2) || (n % 3 == 0 && n != 3)) return false;
for (unsigned long long k = 1; 36 * k * k - 12 * k < n; ++k)
if ((n % (6 * k + 1) == 0) || (n % (6 * k - 1) == 0)) return false;
return true;
}
bool _p(unsigned long long a, unsigned long long n) {
unsigned long long t, u, i, p, c = 0;
u = n / 2, t = 1;
while (!(u & 1)) {
u /= 2;
++t;
}
p = fme(a, u, n);
for (i = 1; i <= t; ++i) {
c = (p * p) % n;
if ((c == 1) && (p != 1) && (p != n - 1)) return 1;
p = c;
}
if (c != 1) return 1;
return 0;
}
inline bool isPrime(unsigned long long n) {
if (((!(n & 1)) && n != 2) || (n < 2) || (n % 3 == 0 && n != 3)) return 0;
if (n < 1373653) {
for (unsigned long long k = 1; (((36 * k * k) - (12 * k)) < n); ++k)
if ((n % (6 * k + 1) == 0) || (n % (6 * k - 1) == 0)) return 0;
return 1;
}
if (n < 9080191) {
if (_p(31, n) || _p(73, n)) return 0;
return 1;
}
if (_p(2, n) || _p(7, n) || _p(61, n)) return 0;
return 1;
}
unsigned long long nCk(long long n, long long k, unsigned long long m = MOD) {
if (k < 0 || k > n || n < 0) return 0;
if (k == 0 || k == n) return 1;
if (fact.size() >= (unsigned long long)n && isPrime(m)) {
return (((fact[n] * invfact[k]) % m) * invfact[n - k]) % m;
}
unsigned long long i = 0, j = 0, a = 1;
k = ((k) < (n - k) ? (k) : (n - k));
for (; i < (unsigned long long)k; ++i) {
a = (a * (n - i)) % m;
while (j < (unsigned long long)k && (a % (j + 1) == 0)) {
a = a / (j + 1);
++j;
}
}
while (j < (unsigned long long)k) {
a = a / (j + 1);
++j;
}
return a % m;
}
void nCkInit(unsigned long long m = MOD) {
long long i, mx = 1010000;
fact.resize(mx + 1);
invfact.resize(mx + 1);
fact[0] = 1;
for (i = 1; i <= mx; ++i) {
fact[i] = (i * fact[i - 1]) % m;
}
invfact[mx] = fme(fact[mx], m - 2, m);
for (i = mx - 1; i >= 0; --i) {
invfact[i] = (invfact[i + 1] * (i + 1)) % m;
}
}
template <class T>
T gcd(T a, T b) {
if (b == 0) return a;
return gcd(b, a % b);
}
void extGCD(long long a, long long b, long long &x, long long &y) {
if (b == 0) {
x = 1, y = 0;
return;
}
long long x1, y1;
extGCD(b, a % b, x1, y1);
x = y1;
y = x1 - (a / b) * y1;
}
inline void get(long long &x) {
int n = 0;
x = 0;
char c = getchar_unlocked();
if (c == '-') n = 1;
while (c < '0' || c > '9') {
c = getchar_unlocked();
if (c == '-') n = 1;
}
while (c >= '0' && c <= '9') {
x = (x << 3) + (x << 1) + c - '0';
c = getchar_unlocked();
}
if (n) x = -x;
}
inline int get(char *p) {
char c = getchar_unlocked();
int i = 0;
while (c != '\n' && c != '\0' && c != ' ' && c != '\r' && c != EOF) {
p[i++] = c;
c = getchar_unlocked();
}
p[i] = '\0';
return i;
}
inline void put(long long a) {
int n = (a < 0 ? 1 : 0);
if (n) a = -a;
char b[20];
int i = 0;
do {
b[i++] = a % 10 + '0';
a /= 10;
} while (a);
if (n) putchar_unlocked('-');
i--;
while (i >= 0) putchar_unlocked(b[i--]);
putchar_unlocked(' ');
}
inline void putln(long long a) {
int n = (a < 0 ? 1 : 0);
if (n) a = -a;
char b[20];
int i = 0;
do {
b[i++] = a % 10 + '0';
a /= 10;
} while (a);
if (n) putchar_unlocked('-');
i--;
while (i >= 0) putchar_unlocked(b[i--]);
putchar_unlocked('\n');
}
const int K = 3;
std::vector<std::vector<long long> > mul(std::vector<std::vector<long long> > a,
std::vector<std::vector<long long> > b,
unsigned long long m = MOD) {
std::vector<std::vector<long long> > c(K, std::vector<long long>(K));
for (long long ii = 0; ii < (K); ++ii)
for (long long jj = 0; jj < (K); ++jj)
for (long long kk = 0; kk < (K); ++kk)
c[ii][jj] = (c[ii][jj] + a[ii][kk] * b[kk][jj]) % m;
return c;
}
std::vector<std::vector<long long> > fme(std::vector<std::vector<long long> > a,
unsigned long long n,
unsigned long long m = MOD) {
if (n == 1) return a;
if (n & 1) return mul(a, fme(a, n - 1, m), m);
std::vector<std::vector<long long> > x = fme(a, n / 2, m);
return mul(x, x, m);
}
long long a[1010];
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
long long n = 0, m = 0, maxx = 0, minn = 0, curr = 0, k = 0, num = 0, siz = 0,
n1 = 0, n2 = 0, n3 = 0, n4 = 0, ind = 0;
long long root = 0, sum = 0, diff = 0, q = 0, choice = 0, d = 0, len = 0,
begg = 0, endd = 0, pos = 0, cnt = 0, lo = 0, hi = 0, mid = 0,
ans = 0;
bool flag = false;
std::string s, t, s1, s2, s3, str;
char ch, ch1, ch2, ch3, *ptr;
double dub = 0;
cin >> n >> m >> q;
cin >> s;
cin >> t;
for (long long i = 0; i < (n); ++i) {
a[i + 1] = a[i];
if (i < n - m + 1) {
flag = true;
for (long long j = 0; j < (m); ++j) {
if (s[i + j] != t[j]) flag = false;
}
if (flag) a[i + 1]++;
}
}
for (long long i = 0; i < (q); ++i) {
cin >> n1 >> n2;
cout << (a[max(n1 - 1, n2 - m + 1)] - a[n1 - 1]) << '\n';
}
return 0;
}
| ### Prompt
Develop a solution in Cpp to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long INF = (1LL << 45LL);
const long long MAXLL = 9223372036854775807LL;
const unsigned long long MAXULL = 18446744073709551615LLU;
const long long MOD = 1000000007;
const long double DELTA = 0.000000001L;
inline long long fmm(long long a, long long b, long long m = MOD) {
long long r = 0;
a %= m;
b %= m;
while (b > 0) {
if (b & 1) {
r += a;
r %= m;
}
a += a;
a %= m;
b >>= 1;
}
return r % m;
}
inline long long fme(long long a, long long b, long long m = MOD) {
long long r = 1;
a %= m;
while (b > 0) {
if (b & 1) {
r *= a;
r %= m;
}
a *= a;
a %= m;
b >>= 1;
}
return r % m;
}
inline long long sfme(long long a, long long b, long long m = MOD) {
long long r = 1;
a %= m;
while (b > 0) {
if (b & 1) r = fmm(r, a, m);
a = fmm(a, a, m);
b >>= 1;
}
return r % m;
}
std::vector<long long> primes;
long long primsiz;
std::vector<long long> fact;
std::vector<long long> invfact;
inline void sieve(long long n) {
long long i, j;
std::vector<bool> a(n);
a[0] = true;
a[1] = true;
for (i = 2; i * i < n; ++i) {
if (!a[i]) {
for (j = i * i; j < n; j += i) {
a[j] = true;
}
}
}
for (i = 2; i < n; ++i)
if (!a[i]) primes.push_back(i);
primsiz = primes.size();
}
inline void sieve() {
long long n = 1010000, i, j, k = 0;
std::vector<bool> a(n);
primes.resize(79252);
a[0] = a[1] = true;
for (i = 2; (j = (i << 1)) < n; ++i) a[j] = true;
for (i = 3; i * i < n; i += 2) {
if (!a[i]) {
k = (i << 1);
for (j = i * i; j < n; j += k) a[j] = true;
}
}
k = 0;
for (i = 2; i < n; ++i)
if (!a[i]) primes[k++] = i;
primsiz = k;
}
inline bool isPrimeSmall(unsigned long long n) {
if (((!(n & 1)) && n != 2) || (n < 2) || (n % 3 == 0 && n != 3)) return false;
for (unsigned long long k = 1; 36 * k * k - 12 * k < n; ++k)
if ((n % (6 * k + 1) == 0) || (n % (6 * k - 1) == 0)) return false;
return true;
}
bool _p(unsigned long long a, unsigned long long n) {
unsigned long long t, u, i, p, c = 0;
u = n / 2, t = 1;
while (!(u & 1)) {
u /= 2;
++t;
}
p = fme(a, u, n);
for (i = 1; i <= t; ++i) {
c = (p * p) % n;
if ((c == 1) && (p != 1) && (p != n - 1)) return 1;
p = c;
}
if (c != 1) return 1;
return 0;
}
inline bool isPrime(unsigned long long n) {
if (((!(n & 1)) && n != 2) || (n < 2) || (n % 3 == 0 && n != 3)) return 0;
if (n < 1373653) {
for (unsigned long long k = 1; (((36 * k * k) - (12 * k)) < n); ++k)
if ((n % (6 * k + 1) == 0) || (n % (6 * k - 1) == 0)) return 0;
return 1;
}
if (n < 9080191) {
if (_p(31, n) || _p(73, n)) return 0;
return 1;
}
if (_p(2, n) || _p(7, n) || _p(61, n)) return 0;
return 1;
}
unsigned long long nCk(long long n, long long k, unsigned long long m = MOD) {
if (k < 0 || k > n || n < 0) return 0;
if (k == 0 || k == n) return 1;
if (fact.size() >= (unsigned long long)n && isPrime(m)) {
return (((fact[n] * invfact[k]) % m) * invfact[n - k]) % m;
}
unsigned long long i = 0, j = 0, a = 1;
k = ((k) < (n - k) ? (k) : (n - k));
for (; i < (unsigned long long)k; ++i) {
a = (a * (n - i)) % m;
while (j < (unsigned long long)k && (a % (j + 1) == 0)) {
a = a / (j + 1);
++j;
}
}
while (j < (unsigned long long)k) {
a = a / (j + 1);
++j;
}
return a % m;
}
void nCkInit(unsigned long long m = MOD) {
long long i, mx = 1010000;
fact.resize(mx + 1);
invfact.resize(mx + 1);
fact[0] = 1;
for (i = 1; i <= mx; ++i) {
fact[i] = (i * fact[i - 1]) % m;
}
invfact[mx] = fme(fact[mx], m - 2, m);
for (i = mx - 1; i >= 0; --i) {
invfact[i] = (invfact[i + 1] * (i + 1)) % m;
}
}
template <class T>
T gcd(T a, T b) {
if (b == 0) return a;
return gcd(b, a % b);
}
void extGCD(long long a, long long b, long long &x, long long &y) {
if (b == 0) {
x = 1, y = 0;
return;
}
long long x1, y1;
extGCD(b, a % b, x1, y1);
x = y1;
y = x1 - (a / b) * y1;
}
inline void get(long long &x) {
int n = 0;
x = 0;
char c = getchar_unlocked();
if (c == '-') n = 1;
while (c < '0' || c > '9') {
c = getchar_unlocked();
if (c == '-') n = 1;
}
while (c >= '0' && c <= '9') {
x = (x << 3) + (x << 1) + c - '0';
c = getchar_unlocked();
}
if (n) x = -x;
}
inline int get(char *p) {
char c = getchar_unlocked();
int i = 0;
while (c != '\n' && c != '\0' && c != ' ' && c != '\r' && c != EOF) {
p[i++] = c;
c = getchar_unlocked();
}
p[i] = '\0';
return i;
}
inline void put(long long a) {
int n = (a < 0 ? 1 : 0);
if (n) a = -a;
char b[20];
int i = 0;
do {
b[i++] = a % 10 + '0';
a /= 10;
} while (a);
if (n) putchar_unlocked('-');
i--;
while (i >= 0) putchar_unlocked(b[i--]);
putchar_unlocked(' ');
}
inline void putln(long long a) {
int n = (a < 0 ? 1 : 0);
if (n) a = -a;
char b[20];
int i = 0;
do {
b[i++] = a % 10 + '0';
a /= 10;
} while (a);
if (n) putchar_unlocked('-');
i--;
while (i >= 0) putchar_unlocked(b[i--]);
putchar_unlocked('\n');
}
const int K = 3;
std::vector<std::vector<long long> > mul(std::vector<std::vector<long long> > a,
std::vector<std::vector<long long> > b,
unsigned long long m = MOD) {
std::vector<std::vector<long long> > c(K, std::vector<long long>(K));
for (long long ii = 0; ii < (K); ++ii)
for (long long jj = 0; jj < (K); ++jj)
for (long long kk = 0; kk < (K); ++kk)
c[ii][jj] = (c[ii][jj] + a[ii][kk] * b[kk][jj]) % m;
return c;
}
std::vector<std::vector<long long> > fme(std::vector<std::vector<long long> > a,
unsigned long long n,
unsigned long long m = MOD) {
if (n == 1) return a;
if (n & 1) return mul(a, fme(a, n - 1, m), m);
std::vector<std::vector<long long> > x = fme(a, n / 2, m);
return mul(x, x, m);
}
long long a[1010];
int main() {
std::ios_base::sync_with_stdio(false);
std::cin.tie(NULL);
std::cout.tie(NULL);
long long n = 0, m = 0, maxx = 0, minn = 0, curr = 0, k = 0, num = 0, siz = 0,
n1 = 0, n2 = 0, n3 = 0, n4 = 0, ind = 0;
long long root = 0, sum = 0, diff = 0, q = 0, choice = 0, d = 0, len = 0,
begg = 0, endd = 0, pos = 0, cnt = 0, lo = 0, hi = 0, mid = 0,
ans = 0;
bool flag = false;
std::string s, t, s1, s2, s3, str;
char ch, ch1, ch2, ch3, *ptr;
double dub = 0;
cin >> n >> m >> q;
cin >> s;
cin >> t;
for (long long i = 0; i < (n); ++i) {
a[i + 1] = a[i];
if (i < n - m + 1) {
flag = true;
for (long long j = 0; j < (m); ++j) {
if (s[i + j] != t[j]) flag = false;
}
if (flag) a[i + 1]++;
}
}
for (long long i = 0; i < (q); ++i) {
cin >> n1 >> n2;
cout << (a[max(n1 - 1, n2 - m + 1)] - a[n1 - 1]) << '\n';
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5, MOD = 1000000007;
long long n, m, q;
string s, t;
vector<long long> v;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> m >> q >> s >> t;
for (long long i = 0; i < n; i++) {
if ((m + i) > n) break;
string temp;
temp.clear();
for (long long j = i; j < (i + m); j++) {
temp += s[j];
}
if (temp == t) v.push_back(i);
}
v.push_back(1e7);
while (q--) {
long long l, r;
cin >> l >> r;
l--;
r--;
if ((r - l) < (m - 1)) {
cout << 0 << endl;
continue;
}
auto it1 = lower_bound(v.begin(), v.end(), l);
auto it2 = lower_bound(v.begin(), v.end(), r - m + 1);
if (it1 == v.end()) {
cout << 0;
continue;
}
if ((it2 == v.end()) || (*it2 != (r - m + 1))) it2--;
cout << it2 - it1 + 1 << endl;
}
return 0;
}
| ### Prompt
Please formulate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5, MOD = 1000000007;
long long n, m, q;
string s, t;
vector<long long> v;
signed main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cin >> n >> m >> q >> s >> t;
for (long long i = 0; i < n; i++) {
if ((m + i) > n) break;
string temp;
temp.clear();
for (long long j = i; j < (i + m); j++) {
temp += s[j];
}
if (temp == t) v.push_back(i);
}
v.push_back(1e7);
while (q--) {
long long l, r;
cin >> l >> r;
l--;
r--;
if ((r - l) < (m - 1)) {
cout << 0 << endl;
continue;
}
auto it1 = lower_bound(v.begin(), v.end(), l);
auto it2 = lower_bound(v.begin(), v.end(), r - m + 1);
if (it1 == v.end()) {
cout << 0;
continue;
}
if ((it2 == v.end()) || (*it2 != (r - m + 1))) it2--;
cout << it2 - it1 + 1 << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int nxt[100001];
char s[100001], t[100001];
int temp[100001];
int l, r;
bool check(int u) {
for (int i = 0; i < m; ++i) {
if (s[u + i] != t[i]) return 0;
}
return 1;
}
void pre_do() {
for (int i = 1; i <= n - m + 1; i++) {
temp[i] += check(i);
}
}
void putin() {
cin >> n >> m >> q;
scanf("%s%s", s + 1, t);
}
int main() {
putin();
pre_do();
for (int i = 1; i <= n; i++) {
temp[i] += temp[i - 1];
}
for (int i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else
cout << temp[r - m + 1] - temp[l - 1] << endl;
}
}
| ### Prompt
Please formulate a CPP solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q;
int nxt[100001];
char s[100001], t[100001];
int temp[100001];
int l, r;
bool check(int u) {
for (int i = 0; i < m; ++i) {
if (s[u + i] != t[i]) return 0;
}
return 1;
}
void pre_do() {
for (int i = 1; i <= n - m + 1; i++) {
temp[i] += check(i);
}
}
void putin() {
cin >> n >> m >> q;
scanf("%s%s", s + 1, t);
}
int main() {
putin();
pre_do();
for (int i = 1; i <= n; i++) {
temp[i] += temp[i - 1];
}
for (int i = 1; i <= q; i++) {
cin >> l >> r;
if (r - l + 1 < m)
cout << 0 << endl;
else
cout << temp[r - m + 1] - temp[l - 1] << endl;
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int maxx = 1000004;
struct XXX {
long long x, y;
} xx[maxx];
bool my(XXX a, XXX b) { return a.x < b.x; }
bool mt(long long a, long long b) { return a > b; }
long long a[maxx], b[maxx];
long long read() {
long long k;
scanf("%lld", &k);
return k;
}
long long n, m;
int f[maxx];
int ans;
char z[maxx], c[maxx];
void getfail(char* s) {
f[0] = f[1] = 0;
;
int len = strlen(s);
for (int i = 1; i < len; i++) {
int j = f[i];
while (j && s[j] != s[i]) j = f[j];
if (s[j] == s[i])
f[i + 1] = j + 1;
else
f[i + 1] = 0;
}
}
int Ans[maxx];
void Kmp(char* T, char* S) {
int n = strlen(T), m = strlen(S);
getfail(S);
int j = 0;
for (int i = 0; i <= n; i++) {
while (j && S[j] != T[i]) j = f[j];
if (S[j] == T[i]) j++;
if (j == m) {
ans++;
j = f[j];
}
Ans[i] = ans;
}
}
int main() {
int q;
n = read();
m = read();
q = read();
scanf("%s", z);
scanf("%s", c);
Kmp(z, c);
for (int i = 1; i <= q; i++) {
int x = read() - 1;
int y = read() - 1;
x = x + m - 1;
if (x > y)
cout << 0 << endl;
else
cout << max(Ans[y] - Ans[x - 1], 0) << endl;
}
return 0;
}
| ### Prompt
Please provide a CPP coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxx = 1000004;
struct XXX {
long long x, y;
} xx[maxx];
bool my(XXX a, XXX b) { return a.x < b.x; }
bool mt(long long a, long long b) { return a > b; }
long long a[maxx], b[maxx];
long long read() {
long long k;
scanf("%lld", &k);
return k;
}
long long n, m;
int f[maxx];
int ans;
char z[maxx], c[maxx];
void getfail(char* s) {
f[0] = f[1] = 0;
;
int len = strlen(s);
for (int i = 1; i < len; i++) {
int j = f[i];
while (j && s[j] != s[i]) j = f[j];
if (s[j] == s[i])
f[i + 1] = j + 1;
else
f[i + 1] = 0;
}
}
int Ans[maxx];
void Kmp(char* T, char* S) {
int n = strlen(T), m = strlen(S);
getfail(S);
int j = 0;
for (int i = 0; i <= n; i++) {
while (j && S[j] != T[i]) j = f[j];
if (S[j] == T[i]) j++;
if (j == m) {
ans++;
j = f[j];
}
Ans[i] = ans;
}
}
int main() {
int q;
n = read();
m = read();
q = read();
scanf("%s", z);
scanf("%s", c);
Kmp(z, c);
for (int i = 1; i <= q; i++) {
int x = read() - 1;
int y = read() - 1;
x = x + m - 1;
if (x > y)
cout << 0 << endl;
else
cout << max(Ans[y] - Ans[x - 1], 0) << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
const int N = 1021;
int n, m, q, ans[N];
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i + m <= n; i++)
if (s.substr(i, m) == t) ans[i + 1]++;
for (int i = 2; i <= n; i++) ans[i] += ans[i - 1];
while (q--) {
int a, b;
cin >> a >> b;
b = b - m + 1;
printf("%d\n", max(0, ans[max(0, b)] - ans[a - 1]));
}
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1021;
int n, m, q, ans[N];
string s, t;
int main() {
cin >> n >> m >> q;
cin >> s >> t;
for (int i = 0; i + m <= n; i++)
if (s.substr(i, m) == t) ans[i + 1]++;
for (int i = 2; i <= n; i++) ans[i] += ans[i - 1];
while (q--) {
int a, b;
cin >> a >> b;
b = b - m + 1;
printf("%d\n", max(0, ans[max(0, b)] - ans[a - 1]));
}
}
``` |
#include <bits/stdc++.h>
using namespace std;
int n, m, q, a[2010], sum[2010];
string s, t;
vector<pair<int, int> > v;
int main() {
cin >> n >> m >> q;
cin >> s;
cin >> t;
for (int i = 0; i < n && i + m - 1 < n; i++) {
bool found = true;
for (int j = 0; j < m && j + i < n; j++) {
if (s[i + j] != t[j]) {
found = false;
break;
}
}
if (found) {
v.push_back({i + 1, i + m});
}
}
for (int i = 1; i <= q; i++) {
int l, r;
cin >> l >> r;
int ans = 0;
for (auto interval : v) {
if (interval.first >= l && interval.second <= r) ans++;
}
cout << ans << endl;
}
return 0;
}
| ### Prompt
Generate a Cpp solution to the following problem:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
int n, m, q, a[2010], sum[2010];
string s, t;
vector<pair<int, int> > v;
int main() {
cin >> n >> m >> q;
cin >> s;
cin >> t;
for (int i = 0; i < n && i + m - 1 < n; i++) {
bool found = true;
for (int j = 0; j < m && j + i < n; j++) {
if (s[i + j] != t[j]) {
found = false;
break;
}
}
if (found) {
v.push_back({i + 1, i + m});
}
}
for (int i = 1; i <= q; i++) {
int l, r;
cin >> l >> r;
int ans = 0;
for (auto interval : v) {
if (interval.first >= l && interval.second <= r) ans++;
}
cout << ans << endl;
}
return 0;
}
``` |
#include <bits/stdc++.h>
using namespace std;
char s1[1005];
char s2[1005];
int ans[1005];
int main() {
int n, m, q, r, l, len1, len2, an, count = 0;
bool flag;
scanf("%d %d %d", &n, &m, &q);
scanf("%s", s1);
scanf("%s", s2);
len1 = strlen(s1);
len2 = strlen(s2);
for (int i = 0; i < len1 - len2 + 1; i++) {
flag = 1;
for (int j = 0; j < len2; j++) {
if (s1[i + j] != s2[j]) {
flag = 0;
break;
}
}
if (flag) {
count++;
}
ans[i + 1] = count;
}
for (int i = 0; i < 10; i++) {
}
while (q--) {
scanf("%d %d", &l, &r);
if (r - l + 1 < len2) {
cout << 0 << endl;
} else {
cout << ans[r - len2 + 1] - ans[l - 1] << endl;
}
}
return 0;
}
| ### Prompt
Please provide a Cpp coded solution to the problem described below:
You are given two strings s and t, both consisting only of lowercase Latin letters.
The substring s[l..r] is the string which is obtained by taking characters s_l, s_{l + 1}, ..., s_r without changing the order.
Each of the occurrences of string a in a string b is a position i (1 ≤ i ≤ |b| - |a| + 1) such that b[i..i + |a| - 1] = a (|a| is the length of string a).
You are asked q queries: for the i-th query you are required to calculate the number of occurrences of string t in a substring s[l_i..r_i].
Input
The first line contains three integer numbers n, m and q (1 ≤ n, m ≤ 10^3, 1 ≤ q ≤ 10^5) — the length of string s, the length of string t and the number of queries, respectively.
The second line is a string s (|s| = n), consisting only of lowercase Latin letters.
The third line is a string t (|t| = m), consisting only of lowercase Latin letters.
Each of the next q lines contains two integer numbers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the arguments for the i-th query.
Output
Print q lines — the i-th line should contain the answer to the i-th query, that is the number of occurrences of string t in a substring s[l_i..r_i].
Examples
Input
10 3 4
codeforces
for
1 3
3 10
5 6
5 7
Output
0
1
0
1
Input
15 2 3
abacabadabacaba
ba
1 15
3 4
2 14
Output
4
0
3
Input
3 5 2
aaa
baaab
1 3
1 1
Output
0
0
Note
In the first example the queries are substrings: "cod", "deforces", "fo" and "for", respectively.
### Response
```cpp
#include <bits/stdc++.h>
using namespace std;
char s1[1005];
char s2[1005];
int ans[1005];
int main() {
int n, m, q, r, l, len1, len2, an, count = 0;
bool flag;
scanf("%d %d %d", &n, &m, &q);
scanf("%s", s1);
scanf("%s", s2);
len1 = strlen(s1);
len2 = strlen(s2);
for (int i = 0; i < len1 - len2 + 1; i++) {
flag = 1;
for (int j = 0; j < len2; j++) {
if (s1[i + j] != s2[j]) {
flag = 0;
break;
}
}
if (flag) {
count++;
}
ans[i + 1] = count;
}
for (int i = 0; i < 10; i++) {
}
while (q--) {
scanf("%d %d", &l, &r);
if (r - l + 1 < len2) {
cout << 0 << endl;
} else {
cout << ans[r - len2 + 1] - ans[l - 1] << endl;
}
}
return 0;
}
``` |