计算机挑战赛5

有N个正整数，求这N个正整数两两之间的最大公约数之积 输入: 第一行输入正整数N(N<=100)第二行有N个正整数(<10000)输出: 输出这N个正整数两两之间的最大公约数之积，结果对1000000007取模样例

输入:

4

6 8 9 10

样例输出:

24

代码：
C++：

面向对象：

#include<iostream>
#include<algorithm>
using namespace std;
class Soulution {
public:int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;}
};
bool cmp(int A, int B) {return A < B;
}
int main() {int n;cin >> n;Soulution s = Soulution();int res = 1;int arr[100];for (int i = 0; i < n; i++) {int num; cin >> num;arr[i] = num;}sort(arr, arr + n, cmp);for (int i = 0; i < n - 1; i++) {for (int j = i + 1; j < n; j++) {res *= s.gcd(arr[i], arr[j]);}}cout << res << endl;return 0;
}

#include<iostream>
#include<algorithm>
using namespace std;
int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;
}
int cmp(int A, int B) {return A < B;
}
int main() {int n;cin >> n;int res = 1;int arr[100];for (int i = 0; i < n; i++) {int num; cin >> num;arr[i] = num;}sort(arr, arr + n, cmp);for (int i = 0; i < n - 1; i++) {for (int j = i + 1; j < n; j++) {res *= gcd(arr[i], arr[j]);}}cout << res << endl;return 0;
}

Python：

class Solution:def gcd(self, a, b):while b != 0:temp = bb = a % ba = tempreturn adef main(self):n = int(input())arr = []num = list(input().split())res = 1for i in range(n):arr.append(int(num[i]))arr = sorted(arr)for i in range(0, n - 1, 1):for j in range(i + 1, n, 1):res *= self.gcd(arr[i], arr[j])print(res)
if __name__ == "__main__":s = Solution()s.main()

Java：

package com.my.gududu;import java.util.*;class Solution{int gcd(int a, int b) {while (b != 0) {int temp = b;b = a % b;a = temp;}return a;}
}
public class Main {public static void main(String[] args) {Solution s =new Solution();Scanner input = new Scanner(System.in);int n = input.nextInt();int arr[] = new int[100];for (int i = 0; i < n; i++) {int num;num = input.nextInt();arr[i] = num;}Arrays.sort(arr, 0, n);int res = 1;for (int i = 0; i < n - 1; i++) {for (int j = i + 1; j < n; j++) {res *= s.gcd(arr[i], arr[j]);}}System.out.println(res);}
}