hdu 1019 Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42968    Accepted Submission(s): 16146

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

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define N 100000
long long gcd(long long m,long long n)
{
    if(m<n) return gcd(n,m);
    if(n==0) return m;
    return gcd(m%n,n);
}


int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int m;
        long long a=1,b;
        cin>>m;
        for(int i=0;i<m;i++)
        {
            cin>>b;
            a=a*b/gcd(a,b);
        }
        cout<<a<<endl;
    }
    return 0;
}

注：

这里没必要建立数组，直接在接收数据的时候求出a和b的最小公倍数即可

在一开始，记a=1，求其最小公倍数