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  • the least common multiplier

    Description

    There are many unsolvable problem in the world.It could be about one or about zero.But this time it is about bigger number. 
    Given an integer n(2 <= n <= 10 9).We should find a pair of positive integer a, b so that a + b = n and [a, b] is as large as possible. [a, b] denote the least common multiplier of a, b.
     

    Input

    The first line contains integer T(1<= T<= 10000),denote the number of the test cases. 
    For each test cases,the first line contains an integer n.
     

    Output

    For each test cases, print the maximum [a,b] in a line.
     

    Sample Input

    3 2 3 4
     

    Sample Output

    1 2 3
     
    #include <iostream>
    using namespace std;
    int gcd(int a,int b){
        int c=b%a;
        while(c>0){
            b=a;
            a=c;
            c=b%a;
        }
    return a;
    }
    
    int main()
    {
        int t;cin>>t;int n;
        while(t--){
            __int64 sum=0,i;
            cin>>n;
            for(i=n/2;i>0;i--){
                if(gcd(i,n-i)==1)
                {
                    sum=i*(n-i);
                    break;
                }
            }
            cout<<sum<<endl;
        }
     return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/demodemo/p/4655719.html
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