hdu 1019Least Common Multiple(最大公约数）

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 50855    Accepted Submission(s): 19287

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

2
3 5 7 15
6 4 10296 936 1287 792 1

 

Sample Output

105
10296

 

Source

East Central North America 2003, Practice

 

两数之积 / 最大公约数 = 最小公倍数；

多个数的最小公倍数，就相当于不断求两个数间的最小公倍数

#include<iostream>
#include<string>
#include<string.h>
#include<sstream>
#include<algorithm>
#include<queue>
using namespace std;
int gy(long long int n,long long int m)//求两个数的最小公约数
{
	int t,r; 
	if(n<m) { t=n;n=m;m=t;}
	while(m!=0)
	{
		r=n%m; 
		n=m;
		m=r;
	} 
	return n;
}
int main()
{
	int num;
	cin>>num;

	while(num--)
	{
	    long long int m,i,j,k,n;
	    cin>>n;
	    k=1;
	    for(i=0;i<n;i++)
	    {
	    	cin>>m;
	    	k=(k*m)/gy(k,m);
		}
		cout<<k<<endl; 
	}
}