HDU1019 Least Common Multiple

题目链接：HDU1019

CODE:

#include<iostream>
#include<cmath>
using namespace std;
int gcd(int m,int n)
{
    if(m%n==0)return n;
    else return gcd(n,m%n);
}
long long LCM(int m,int n)
{
    return m/gcd(m,n)*n;
}
int main()
{
    int kase;cin>>kase;
    while(kase--)
    {
        int n;cin>>n;
        int ret=1;
        for(int i=0;i<n;i++)
        {
            int m;cin>>m;
            ret=LCM(m,ret);
        }
        cout<<ret<<endl;
    }
}