NYOJ - A problem is easy

A  problem is easy

时间限制：1000 ms  |           内存限制：65535 KB

难度：3

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

213

样例输出

01

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