Least Common Multiple最小公倍数
Problem Description
The least common multiple (LCM) of a set of positiveintegers is the smallest positive integer which is divisible by all the numbersin the set. For example, the LCM of 5, 7 and 15 is 105.
一组正整数的最小公倍数（LCM）的最小正整数整除在该组的所有数字。例如，第五，第七和15的LCM是105。
Input
Input will consistof multiple problem instances. The first line of the input will contain asingle integer indicating the number of problem instances. Each instance willconsist of a single line of the form m n1 n2 n3 ... nm where m is the number ofintegers in the set and n1 ... nm are the integers. All integers will bepositive and lie within the range of a 32-bit integer.
输入将包含多个问题的实例。输入的第一行包含一个单一的整数，表示问题的实例的数量。每个实例将包括形式米单行N1 N2 N3... nm其中，m是在该组和N1 ......纳米整数的数目是整数。所有整数将为正和位于32位整数的范围内。
Output
For each probleminstance, output a single line containing the corresponding LCM. All resultswill lie in the range of a 32-bit integer.
对于每个问题的实例，包含输出一行相应的LCM。所有结果将位于32位整数的范围
Sample Input
2
3 5 7 15
6 4 10296 936 1287 792 1
Sample Output
105
10296
代码如下：
#include <stdio.h>
#include <stdlib.h>
int gcd(int a,int b)
{if(b==0)return a;return gcd(b,a%b);
}
int main()
{int group;int n,a,b,i;int cnt;scanf("%u",&group);while(group--){scanf("%d",&n);cnt=a=1;for(i=1;i<=n;i++){scanf("%d",&b);cnt=a/gcd(a,b)*b;//换一下，先除后乘，免得数据溢出 a=cnt;} printf("%d\n",cnt); } return 0;
}