化简一下即可

描述

When Teddy was a child , he was always thinking about some simple math problems ,such as “What it’s 1 cup of water plus 1 pile of dough ..” , “100 yuan buy 100 pig” .etc..

One day Teddy met a old man in his dream , in that dream the man whose name was“RuLai” gave Teddy a problem :

Given an N , can you calculate how many ways to write N as i * j + i + j (0 < i <= j) ?

Teddy found the answer when N was less than 10…but if N get bigger , he found it was too difficult for him to solve.
Well , you clever ACMers ,could you help little Teddy to solve this problem and let him have a good dream ?

输入

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

输出

For each case, output the number of ways in one line

样例输入

2
1
3

样例输出

0
1

上传者

苗栋栋

#include<stdio.h>
int main()
{
	int i,j,s,m,sum;
	scanf("%d",&s);
	while(s--)
	{
		sum=0;
		scanf("%d",&m);
		for(i=1;(i+1)*(i+1)-1<=m;i++)
		{
			if((m+1)%(i+1)==0)
				sum++;
		}
		printf("%d\n",sum);
	}
	return 0;
}

NYOJ 216 A problem is easy【数学题】