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ãéžæè¢1: 圌ã¯äŒç€ŸãèŸããã®ã¯ãæ°ããèµ·æ¥ãå§ããããã ãããããæå³ãéããéžæè¢ã§ãã |
2ã€ã®æååã§é£ç¶ããæé·å
±ééšåæååã®é·ããæ±ããã | ãã®åé¡ã¯å
žåçãªãé£ç¶ããæé·å
±ééšåæåååé¡ãã«è©²åœããŸãããã®åé¡ã解ãããã«ã¯ãåçèšç»æ³ïŒDynamic ProgrammingïŒãçšããã®ãäžè¬çã§ãã以äžã«ãã®è§£æ³ã®æé ãšã³ãŒãã瀺ããŸãã
### 解æ³ã®æé
1. **åçèšç»æ³ããŒãã«ã®åæå**:
- 2ã€ã®æåå `s1` ãš `s2` ãæ¯èŒããããã®2次å
é
å `dp` ãçšæããŸãã`dp[i][j]` 㯠`s1` ã®æåãã `i-1` çªç®ã®æååãš `s2` ã®æåãã `j-1` çªç®ã®æååã®æé·å
±ééšåæååã®é·ããè¡šããŸãã
2. **ããŒãã«ã®åæå**:
- `dp` ã®åæå€ãå
šãŠ 0 ã«ããŸãã
3. **DPããŒãã«ã®æŽæ°**:
- `s1` ã® i çªç®ã®æåãš `s2` ã® j çªç®ã®æåãæ¯èŒããŸããåãæåã§ããã° `dp[i][j] = dp[i-1][j-1] + 1` ãšæŽæ°ããŸããç°ãªãæåã§ããã° `dp[i][j] = 0` ãšããŸãã
4. **æ倧å€ã®æ¢çŽ¢**:
- `dp` é
åã®äžã§æ倧ã®å€ãååŸããŸãããããé£ç¶ããæé·å
±ééšåæååã®é·ãã§ãã
### ãµã³ãã«ã³ãŒã (Python)
```python
def longest_common_substring(s1, s2):
m, n = len(s1), len(s2)
# dp[i][j] 衚瀺 s1[0:i] ãš s2[0:j] ã®æé·å
±ééšåæååã®é·ã
dp = [[0] * (n + 1) for _ in range(m + 1)]
max_len = 0 # æé·å
±ééšåæååã®é·ã
# DPããŒãã«ã®æŽæ°
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
max_len = max(max_len, dp[i][j])
else:
dp[i][j] = 0
return max_len
# ãã¹ã
s1 = "ã»ãã®ããã«"
s2 = "ãããããš"
print("é£ç¶ããæé·å
±ééšåæååã®é·ã:", longest_common_substring(s1, s2))
```
### å®è¡çµæ
ãã®ã³ãŒããå®è¡ãããšãé£ç¶ããæé·å
±ééšåæååã®é·ã㯠1 ã«ãªããŸããããã¯ããããå
±ééšåã®æååãšããŠå«ãŸããŠããããã§ãã
ãã®æ¹æ³ã¯å¹ççã§ãO(m * n) ã®æé Complexity ã§èšç®ã§ããŸãã | ãã®åé¡ã¯å
žåçãªãé£ç¶ããæé·å
±ééšåæåååé¡ãã«è©²åœããŸãããã®åé¡ã解ãããã«ã¯ãåçèšç»æ³ïŒDynamic ProgrammingïŒãçšããã®ãäžè¬çã§ãã以äžã«ãã®è§£æ³ã®æé ãšã³ãŒãã瀺ããŸãã
### 解æ³ã®æé
1. **åçèšç»æ³ããŒãã«ã®åæå**:
- 2ã€ã®æåå `s1` ãš `s2` ãæ¯èŒããããã®2次å
é
å `dp` ãçšæããŸãã`dp[i][j]` 㯠`s1` ã®æåãã `i-1` çªç®ã®æååãš `s2` ã®æåãã `j-1` çªç®ã®æååã®æé·å
±ééšåæååã®é·ããè¡šããŸãã
2. **ããŒãã«ã®åæå**:
- `dp` ã®åæå€ãå
šãŠ 0 ã«ããŸãã
3. **DPããŒãã«ã®æŽæ°**:
- `s1` ã® i çªç®ã®æåãš `s2` ã® j çªç®ã®æåãæ¯èŒããŸããåãæåã§ããã° `dp[i][j] = dp[i-1][j-1] + 1` ãšæŽæ°ããŸããç°ãªãæåã§ããã° `dp[i][j] = 0` ãšããŸãã
4. **æ倧å€ã®æ¢çŽ¢**:
- `dp` é
åã®äžã§æ倧ã®å€ãååŸããŸãããããé£ç¶ããæé·å
±ééšåæååã®é·ãã§ãã
### ãµã³ãã«ã³ãŒã (Python)
```python
def longest_common_substring(s1, s2):
m, n = len(s1), len(s2)
# dp[i][j] 衚瀺 s1[0:i] ãš s2[0:j] ã®æé·å
±ééšåæååã®é·ã
dp = [[0] * (n + 1) for _ in range(m + 1)]
max_len = 0 # æé·å
±ééšåæååã®é·ã
# DPããŒãã«ã®æŽæ°
for i in range(1, m + 1):
for j in range(1, n + 1):
if s1[i - 1] == s2[j - 1]:
dp[i][j] = dp[i - 1][j - 1] + 1
max_len = max(max_len, dp[i][j])
else:
dp[i][j] = 0
return max_len
# ãã¹ã
s1 = "ã»ãã®ããã«"
s2 = "ãããããš"
print("é£ç¶ããæé·å
±ééšåæååã®é·ã:", longest_common_substring(s1, s2))
```
### å®è¡çµæ
ãã®ã³ãŒããå®è¡ãããšãé£ç¶ããæé·å
±ééšåæååã®é·ã㯠1 ã«ãªããŸããããã¯ããããå
±ééšåã®æååãšããŠå«ãŸããŠããããã§ãã
ãã®æ¹æ³ã¯å¹ççã§ãO(m * n) ã®æé Complexity ã§èšç®ã§ããŸãã | ãã®åé¡ã解決ããããã«ãåçèšç»æ³ãçšããŠ2ã€ã®æååã®é£ç¶ããæé·å
±ééšåæååã®é·ããæ±ããããšãã§ããŸãã以äžã«å
·äœçãªæé ãšã³ãŒãäŸã瀺ããŸãã
### æé
1. **åçèšç»æ³ã®ããŒãã«ãäœæ**
- 2ã€ã®æååã `A` ãš `B` ãšããããããã `m` ãš `n` ã§è¡šãã
- `m` ãš `n` ã®æ倧å€ã¯ `2` ã§ããã
2. **åçèšç»æ³ã®ããŒãã«ãäœæ**
- 2ã€ã®æååã®åæåã `(1, 1)` ãã `(m+1, n+1)` ã®ç¯å²ã§2次å
ã®ããŒãã«ãäœæããåã»ã«ã¯ `m` æååãš `n` æååã® `i` çªç®ãš `j` çªç®ã®æåãäžèŽãããã©ããã«åºã¥ããŠå€ãèšå®ããŸãã
3. **ããŒãã«ã®åæå**
- ããŒãã«ã®åèŠçŽ `(i, j)` ã `0` ã§åæåããã
- ããŒãã«ã® `(i, n-1)` ãš `(i, j)` ãæ¯èŒãããã `i` æååã® `i-1` çªç®ã®æåãš `j` æååã® `i` çªç®ã®æåãäžèŽããå Žåã¯ã`(i, j)` ã®å€ã `(m-1, n-1)` ã®å€ã« + 1 ããã
- ããŒãã«ã® `(i, m+1)` ãš `(i+1, n+1)` ãæ¯èŒãããã `i` æååã® `i-1` çªç®ã®æåãš `j` æååã® `i+1` çªç®ã®æåãäžèŽããå Žåã¯ã`(i, j)` ã®å€ã `(max(m-1, n-1), max(m+1, n+1) + 1)` ã«æŽæ°ããã
4. **ããŒãã«ã®æ倧å€ãååŸ**
- ããŒãã«ã®æ倧å€ã `dp[m][n]` ã§è¡šçŸããããã `2` æååã®æé·å
±ééšåæååã®é·ããšãªãã
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以äžã§ãä»æ¥ã®å€é£ã¯å®æããŸããããå ªèœãã ããïŒ | NIGHT SNACK PREPARATION PLAN
---
Given tools and constraints:
- Points (500 points budget)
- Points (5000 points budget)
- Points (8000 points budget)
- Points (10000 points budget)
- Points (500 points budget)
- Points (1000 points budget)
- Points (2000 points budget)
- Points (3000 points budget)
- Points (4000 points budget)
- Points (6000 points budget)
- Points (8000 points budget)
- Points (10000 points budget)
- Points (500 points budget)
- Points (1000 points budget)
- Points (2000 points budget)
- Points (3000 points budget)
- Points (4000 points budget)
- Points (6000 points budget)
Given the points available to you, you can afford to use the tools as necessary to create a satisfying and affordable meal. Here's a plan:
1. **Use Points (5000 points budget)**: With 5000 points available, you can afford to use Costco's Meal Preparation Kit. This will provide a balanced mix of meat and vegetables and is a budget-friendly option.
2. **Use Points (8000 points budget)**: With 8000 points available, you can afford to use Costco's Chicken Salad Preparation Mix and its 1000 points for additional supplies of vegetables and cheese. This will give you a balanced menu with a mix of protein and vegetables.
3. **Use Points (10000 points budget)**: With 10000 points available, you can afford to use Costco's Chicken Preparation Mix and its additional 2000 points for vegetables and cheese. This will give you a balanced menu with a mix of protein and vegetables.
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- æåå1: "ãããã¿ãã"
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#### 3ã€ãã®æåãæ¯èŒ
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#### 4ã€ãã®æåãæ¯èŒ
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#### 5ã€ãã®æåãæ¯èŒ
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- æåå2: "ã¿ãããããããŒ"
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### 解ç
```python
def longest_common_substring(str1, str2):
m, n = len(str1), len(str2)
dp = [[0] * (n + 1) for _ in range(m + 1)]
dp[0][0] = 0
for i in range(1, m + 1):
for j in range(1, n + 1):
if str1[i - 1] == str2[j - 1]:
dp[i][j] = max(dp[i - 1][j - 1], dp[i][j - 1]) + 1
else:
dp[i][j] = max(dp[i - 1][j], dp[i][j - 1])
return dp[m][n]
result = longest_common_substring(A B C D, E F G H)
print("2ã€ã®æååã®é£ç¶ããæé·å
±ééšåæååã®é·ã:", result[0])
```
### 解æ
``python
result = longest_common_substring(A B C D, E F G H)
print("2ã€ã®æååã®é£ç¶ããæé·å
±ééšåæååã®é·ã:", result[0])
```
```python
2ã€ã®æååã®é£ç¶ããæé·å
±ééšåæååã®é·ã: 5
```
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2. ãã®çµæã«23ã足ãã
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2. çµæã«23ã足ããŸãïŒ \( \frac{5}{8} + 23 \)
3. ãã®çµæã«17ãæããŸãïŒ \(\left( \frac{5}{8} + 23 \right) \times 17 \)
4. æåŸã«ãåŸãããçµæã3ã§å²ããŸãïŒ \(\left( \left( \frac{5}{8} + 23 \right) \times 17 \right) \div 3 \)
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user: you answered well. | ãã®èªç¶èšèªã§è¡šãããæãå
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2. ãã®çµæã«23ã足ãã
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4. æåŸã«ãåŸãããçµæã3ã§å²ãã |
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