南阳216 A problem is easy_南阳a problem is easy-优快云博客

本文讲述了Teddy小时候喜欢思考简单的数学问题，并在梦境中遇到了名为‘RuLai’的老者，老者给出了一个关于N的数学问题，即计算N可以分解成i*j+i+j的方式有多少种。通过将问题转换为（i+1）*（j+1）-1的形式，文章提供了一个解决方法，并通过实例展示了如何求解不同数值的N对应的答案。

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

#include<stdio.h>
int main()
{
	 int n,i,t;
	 scanf("%d",&t);
	 while(t--)
	 {
		 int s=0;
		 scanf("%d",&n);
		 for(i=1;(i+1)*(i+1)<=n+1;i++)
		 {
		 	if((n+1)%(i+1)==0)
		 		s++;
		 }
	 printf("%d\n",s);
	 }
	 return 0; 
}
 //吧i*j+i+j变形一下=（i+1）*（j+1）-1