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

這個題是求幾個數的最小公倍數，但是本質上是求這幾個數的最大公約數。因為最小公倍數就是兩個數的乘積除以這兩個數的最大公約數。記得用long long 型別，在這裡wa了兩次。 程式碼如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#define ll long long
using namespace std;

const int maxx=1e6+10;
ll gcd(ll x,ll y)//歐幾里得求最大公約數（輾轉相除法）
{
	if(y==0) return x;
	else return gcd(y,x%y);
}
ll n;
ll a[maxx];

int main()
{
	int t;
	while(cin>>t)
	{
		while(t--)
		{
			cin>>n;
			for(int i=0;i<n;i++)
			{
				cin>>a[i];
			}
			int ans=a[0];
			for(int i=1;i<n;i++)
			{
				int cnt=gcd(ans,a[i]);
				ans=ans*a[i]/cnt;
			}
			cout<<ans<<endl;
		}	
	}
}
 

努力加油a啊，(^o)/~