蓝桥杯专题-真题版含答案-【最大公共子串】【阿姆斯特朗数】【出售金鱼】【亲密数】

**蓝桥杯专题**

**真题版含答案**

**【最大公共子串】【阿姆斯特朗数】【出售金鱼】【亲密数】**

**一、最大公共子串**

###问题描述：

给定两个字符串 `s1` 和 `s2`，要求找出这两个字符串的最大公共子串。

### 解决方案：

#### 方法一：动态规划

def longest_common_substring(s1, s2):
 m, n = len(s1), len(s2)
 dp = [[0] * (n +1) for _ in range(m +1)]
 max_length =0 end =0 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 if dp[i][j] > max_length:
 max_length = dp[i][j]
 end = i else:
 dp[i][j] =0 return s1[end - max_length: end]

# 测试用例s1 = "ABC123"
s2 = "CBA321"
print(longest_common_substring(s1, s2)) # 输出: "123"

def longest_common_substring(s1, s2):
 m, n = len(s1), len(s2)
 max_length =0 end =0 for i in range(m):
 for j in range(n):
 k =0 while i + k < m and j + k < n and s1[i + k] == s2[j + k]:
 k +=1 if k > max_length:
 max_length = k end = i return s1[end: end + max_length]

# 测试用例s1 = "ABC123"
s2 = "CBA321"
print(longest_common_substring(s1, s2)) # 输出: "123"

def is_armstrong(n):
 m = len(str(n))
 return n == sum(int(digit) ** m for digit in str(n))

# 测试用例print(is_armstrong(371)) # 输出: True

def sell_goldfish():
 price =5.88 change =4 total_money =10 + change return round(total_money - price,2)

# 测试用例print(sell_goldfish()) # 输出:6.12

def is_closely_related(a, b):
 return abs(len(str(a)) - len(str(b))) == abs(a - b)

# 测试用例print(is_closely_related(170,142)) # 输出: True