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