HDU——1019Least Common Multiple（多个数的最小公倍数）

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 42735    Accepted Submission(s): 16055

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

 

这题百度了下有出现n=1的情况，按我之前先取两个数得到第一个公倍数的做法会超时，n=1根本没无法输出。因此要重新写，顺便复习下gcd公式

代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
long long gcd(long long a,long long b)
{
  return b?gcd(b,a%b):a;//还是记这个吧，简单易用
} 
int main()
{
    int t;
    cin>>t;
    while (t--)
    {
    	int n;
		long long  a,tlcm,beg=1;//让beg初始化为1不影响结果并成为第0个数，这样一开始也就可以一个一个地求gcd
    	scanf("%d",&n);
    	for (int i=0; i<n; i++)
    	{
	    	scanf("%lld",&a);
	    	beg=(beg*a)/gcd(a,beg);
	    }
    	cout<<beg<<endl;
    }
    return 0;
}