51nod 1138 连续整数的和(数学公式)

1138 连续整数的和

#include <iostream>
#include <cmath>
#include <cstdio>
using namespace std;

int main()
{
    int n;
    while(cin>>n)
    {
        int m = (int)sqrt(n);
        int sum = 0;
        for(int i=m*2; i>=2; i--)
        {
            if( (2*n+i-i*i)%(2*i)==0 && (2*n+i-i*i)>0 )
            {
                cout<<(2*n+i-i*i)/(i*2)<<endl;
                sum++;
            }
        }
        if(!sum)
            puts("No Solution");
    }
    return 0;
}

  

原文地址:https://www.cnblogs.com/yoyo-sincerely/p/5656770.html