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the least common multiplier

时间:2015-07-17 22:24:13      阅读:255      评论:0      收藏:0      [点我收藏+]

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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
 
技术分享
#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

 

the least common multiplier

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原文地址:http://www.cnblogs.com/demodemo/p/4655719.html

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