hdu1019

Problem Description

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

23 5 7 156 4 10296 936 1287 792 1

 

Sample Output

10510296

解法：求最小公倍数。注意全局变量ans的使用

#include <iostream>

#include <cmath>
#include <cstdio>
#include <cstring>

using namespace std;
int ans;
void LCM(int m,int n)
{
    int m1=m;
    int n1=n;
    int temp;
    while(n1%m1!=0)
    {
        temp=n1%m1;
        n1=m1;
        m1=temp;
    }
    ans=n/m1*m;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m,k;
        cin>>n>>m;
        if(n==1)
        {
            printf("%d\n",m);
            continue;
        }
        cin>>k;
        LCM(k,m);
        for(int i=2;i<n;i++)
        {

            cin>>m;
            LCM(ans,m);
        }
        printf("%d\n",ans);
    }
    return 0;
}