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): 59749    Accepted Submission(s): 22797

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

//例2，3，5，7；
//先求出2，3最小公倍数6，
//再求6，5最小公倍数30
//再求30，7的最小公倍数210就是结果 
#include<stdio.h>
#include<string.h>
#include<vector>
#include<algorithm>
using namespace std;
int gcd(int a,int b){//最大公约数 
	if(b==0) return a;
	else return gcd(b,a%b);
}
int main()
{
	int n;
	scanf("%d",&n);
	while(n--)
	{
		vector<long long>v;
		long long a,m;
		scanf("%lld",&m);
		for(int i=0;i<m;i++)
		{
			scanf("%lld",&a);
			v.push_back(a);
		}
		sort(v.begin(),v.end());
		long long sum=v[0];
		for(int i=1;i<v.size();i++)
		{
			sum=(sum*v[i]/gcd(v[i],sum));//两两求出最小公倍数 
		}
		printf("%lld\n",sum);//最终结果 
	}
}