NYOJ-216-A problem is easy-2013年10月17日13:57:33

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>int main(){int i,j,N,T,count;scanf("%d",&T);while(T--){count = 0;scanf("%d",&N);for(j=2;j*j<=N+1;j++)if((N+1)%j==0)count++;printf("%d\n",count);}return 0;}