hdu 2715 Herd Sums

简单数学题目，输入n,用sum计数，若存在2个正整数x,y使（x+y)*（y-x+1）=2*n,则sum++;因此我们就把2*n分解成2个数m,n的乘积，若(m+n-1)/2为正整数，则满足题意

#include<iostream>#include<cmath>using namespace std;int main(){int n;while(cin>>n){int i,sum=0;for(i=1;i<=sqrt(2*n);i++){if(i*(2*n/i)==2*n){if((i+2*n/i-1)%2==0){sum++;}}}cout<<sum<<endl;}return 0;}