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 ... nmwhere 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：

2
3 5 7 15
6 4 10296 936 1287 792 1

样例输出：

105
10296

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。