拓展最小公倍数

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

 

<span style="color:#cc0000;">#include <iostream>
using namespace std;
int mmm(int a,int b)
{
 if(a<b) return mmm(b,a);
    if(b==0) return a;
    return mmm(b,a%b);
}
int main()
{
 int L,M,N,O,p;
 cin>>L;
 while(L--)
 {
  cin>>M>>N;
  M--;
  while(M--)
  {
   cin>>O;
   p=mmm(N,O);
   N=O*N/p;
  }
  cout<<N<<endl;
 }
 return 0;
}</span>

先输入组数，然后每组的个数，可以先输入每组第一个元素，以此用后面和前面一个最小公倍数作比较，这样比较简单的求出来整个的最小公倍数。

本来是个简答题，但是过于纠结于另一种解法 ，导致浪费很多时间，开始用后面一个数和1作比较，然后在用后面的来代替1，不知怎么结果老是出错。。。