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  • HDU1019 Least Common Multiple(多个数的最小公倍数)

    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. 

    InputInput 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. 
    OutputFor 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

    这道题挺简单的,但我居然想着去分解质因数然后乘一遍,又果断的TLE了.但其实真的很简单,不断地读入一个数,然后求它和之前所有数的lcm,这个怎么求呢?除去两数的gcd即可.说起来还是太菜了....

    代码如下:
    #include<cstdio>
    using namespace std;
    
    long long ans=1,n,i,x;
    
    long long gcd(long long a,long long b)
    {
        if(b>a)
        {
            long long w=a;a=b;b=w;
        }
        if(a%b==0)
        {
            return b;
        }
        else
        {
            return gcd(b,a%b);
        }
    }
    
    int main()
    {
        long long t;
        scanf("%lld",&t);
        while(t--)
        {
            ans=1;
            scanf("%lld",&n);
            for(i=1;i<=n;i++)
            {
                scanf("%lld",&x);
                ans*=x/gcd(ans,x);
            }
            printf("%lld
    ",ans);
        }
        return 0;
    }

    每日刷题身体棒棒!

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  • 原文地址:https://www.cnblogs.com/stxy-ferryman/p/7470758.html
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