hduoj1019 Least Common Multiple

View Code
#include<stdio.h>
int gcd(int a,int b)
{
    int t,aa=a,bb=b;
    if(a<b)
    {t=a;a=b;b=t;}
    while(b)
    {t=a%b;a=b;b=t;}
    //printf("%d\n",(int)((double)aa/a*bb));//
    return (int)((double)aa/a*bb);
}
int main()
{
    int ncase,m,t,x;
    scanf("%d",&ncase);
    while(ncase--)
    {
        scanf("%d",&m);
        t=1;
        while(m--)
        {
            scanf("%d",&x);
            t=gcd(t,x);
        }
        printf("%d\n",t);
    }
    return 0;
}
 

Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16870    Accepted Submission(s): 6275


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
原文地址:https://www.cnblogs.com/zhaojiedi1992/p/zhaojiedi_2012_07_180.html