Least Common Multiple

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26336    Accepted Submission(s): 9880

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

 

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#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int rem(int a,int b)
{
    int c=a%b;
    //printf("a %d a\n",a);
    //printf("b %d b\n",b);
    //printf("c %d c\n",c);
    if(c==0)
    return b;
    else
   {
    a=b;
    b=c;
    rem(a,b);
   }
}
int mul(int a,int b)
{
        //cout<<"rem"<<rem(a,b)<<"rem"<<endl;
        //printf("%d %d\n",a,b);
        return a/(rem(a,b))*b;//先除后乘，防止溢出，不过为什么改成__int64之后传错值了呢
}
int main()
{
    int s[100];
    int i,j,n,m,t;
    while(scanf("%d",&n)!=EOF)
    {
        while(n--)
        {
        scanf("%d",&m);
        for(j=0;j<m;j++)
        scanf("%d",&s[j]);
        t=s[0];
        /*for(i=0;i<m;i++)
        printf("m  %d   m\n",s[i]);*/
        for(i=1;i<m;i++)
        {
            //printf("%d gg %d\n",t,s[i]);
            t=mul(t,s[i]);
        }
        cout<<t<<endl;
        }
    }
    return 0;
}