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    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

    题目意思:求所给数的最小公倍数;
    解题思路:两个两个的求,主要不要超范围了;


    #include <iostream>
    #include <stdio.h>
    using namespace std;
    
    typedef long long ll;
    
    ll pand(ll t,ll n1)
    {
        while(1)
        {
            ll temp = t % n1;
            if(temp == 0)
                return n1;
            t = n1;
            n1 = temp;
        }
    }
    
    int main()
    {
       int N;
       cin>>N;
       while(N--)
       {
           ll t,n1;
           ll n;
           cin>>n;
           cin>>t;
           n--;
           while(n--)
           {
               cin>>n1;
               if(t==0||n1==0)
               {
                   t=0;
                   break;
               }
    
               if(t < n1)
               {
                   ll temp = t;
                   t = n1;
                   n1 = temp;
               }
               ll sum = t*n1;
               ll x = pand(t,n1);
                t = sum/x;
           }
           cout<<t<<endl;
       }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/a2985812043/p/7289219.html
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