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  • HDU-1019 Least Common Multiple

    http://acm.hdu.edu.cn/showproblem.php?pid=1019

    Least Common Multiple

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 25035    Accepted Submission(s): 9429

    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
     
    #include "stdio.h"
    long int gb(long m,long n)
    {
        if(m>0&&n>0)
        {
            long a,b,r;
            if(n>m)
            {
                r=m;
                m=n;
                n=r;
            }
            a=m;
            b=n;
            while(r!=0)
            {
                r=a%b;
                a=b;
                b=r;
            }
            return m/a*n;
        }
        else
            return 0;
    }
    int main()
    {
        int i,n,a,b,t;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d",&n);
            scanf("%d",&a);
            for(i=1;i<n;i++)
            {
                scanf("%d",&b);
                a=gb(a,b);
            }
            printf("%d
    ",a);
        }
    }
    #include"stdio.h"
    int mcm( long x, long y)                  
    {
         long  t;
        while(y)
        {
            t=x%y;
            x=y;
            y=t;
        }
        return x;
    }
    int main()
    {
         long  n,m;
         long  x,y,t1;
        while(scanf("%lld",&n)!=EOF)
        {
            while(n--)
            {
                scanf("%ld",&m);
                scanf("%ld",&x);
                
                while(--m)
                {
                    scanf("%ld",&y);  
                    t1=mcm(x,y);
                    x=x*y/t1;              
                }
                printf("%ld
    ",x);    
            }
        }
        return 0;
    }
                
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  • 原文地址:https://www.cnblogs.com/cancangood/p/3431548.html
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