nyoj216A problem is easy（数学题）

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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>
#include<math.h>
int main()
{int n,i,j,sum,T;scanf("%d",&T);while(T--){sum=0;scanf("%d",&n);for(i=2;i<=sqrt(n+1);i++){if((n+1)%i==0)sum++;}printf("%d\n",sum);}return 0;
}
//n=i*j+i+j等价于(n+1)=(i+1)*(j+1)
//所以这道题就是求 n+1 大于2的因子的个数 

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