hdu 1019 n个数的最小公倍数

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.







Input

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.





Output

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





Sample Input

2

3 5 7 15

6 4 10296 936 1287 792 1





Sample Output

105

10296

题目的意思：有多组测试，每组有n个数，求这n个数的最小公倍数；

代码：

#include <iostream>
using namespace std;
long long n,m,i,j,k,a;
int main() {
    cin>>n;
    while(n--){
        cin>>m;
        cin>>k;m=m-1;
        while(m--)
		{
            cin>>a;
            j=k*a;
            while(k!=0)
			{
             i=a%k;
             a=k;
             k=i;
            }
             k=j/a;
        }
        cout<<k<<endl;
    }
    return 0;
}

思想是：是求两个数的最小公倍数延伸；比如现在有三个数，a，b，c；先求出a和b的最小公倍数d；然后再求d和a的最小公倍数e；e就是a，b，c的最小公倍数；