题目1439：Least Common Multiple

题目描述：

The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105.

输入：

Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer.

输出：

For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer.

样例输入：

23 5 7 156 4 10296 936 1287 792 1

样例输出：

10510296

代码：

#include <stdio.h>int gcd(int a,int b) {    if(b==0) return a;    else return gcd(b,a%b);}int num[1000];int main() {    int n,i;    while(scanf("%d",&n)!=EOF) {        for(i=0;i<n;i++) {            int x,j,m;            scanf("%d",&m);            for(j=0;j<m;j++)                scanf("%d",&num[j]);            int ans = num[0] * num[1] / gcd(num[0],num[1]);            for(j=2;j<m;j++)                ans = ans * num[j] / gcd(ans,num[j]);            printf("%d\n",ans);        }    }    return 0;}

这个题在自己机器上测试正常，不知道为什么提交了为WA。

怎么求三个数的最小公倍数呢？

先算两个数的最小公倍数，再算这个最小公倍数与第三个数的最小公倍数，所得结果即为答案。