Least Common Multiple （HDU 1019）

  额，这题我刚开始看错题目了%>_<%，，后面每行的第一个数是指个数。。

                     Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28673    Accepted Submission(s): 10770

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
   
   
   
   

程序如下：

//先用碾转相除法求最小公约数，再求公倍数
#include<stdio.h>
int gcd(int a,int b) //碾转相除
{
    if(b==0) return a;
    return gcd(b,a%b);
}

int main()
{
    int t,n,i,a,b;

    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        if(n==1)  //只给出一个数的情况
        {
            scanf("%d",&a);
            printf("%d\n",a);
            continue;
        }
        scanf("%d%d",&a,&b);
        a=a/gcd(a,b)*b;  //这样做比较保险，如用a*b/gcd(a,b)可能会超出int范围
        for(i=0; i<n-2; i++)
        {
            scanf("%d",&b);
            a=a/gcd(a,b)*b;
        }
        printf("%d\n",a);
    }
    return 0;
}