HDOJ 2028 Lowest Common Multiple Plus

Lowest Common Multiple Plus

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 49459    Accepted Submission(s): 20445

Problem Description

求n个数的最小公倍数。

 

Input

输入包含多个测试实例，每个测试实例的开始是一个正整数n，然后是n个正整数。

 

Output

为每组测试数据输出它们的最小公倍数，每个测试实例的输出占一行。你可以假设最后的输出是一个32位的整数。

 

Sample Input

2 4 63 2 5 7

 

Sample Output

1270

#include <cstdio>using namespace std;long long gcd(long long a, long long b){    if (b == 0)        return a;    else        gcd(b, a % b);}long long lcm(long long a, long long b){    return a * b / gcd(a, b);}int main(){    long long n, a, b, res;    while (scanf("%I64d", &n) != EOF){        scanf("%I64d", &a);        for (int i = 1; i < n; i++){            scanf("%I64d", &b);            a = lcm(a, b);        }        printf("%I64d\n", a);    }    return 0;}