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

2
1
3

样例输出

0
1

N+1=(i+1)*（j+1）

#include<stdio.h>

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

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NYOJ a problem is easy

A problem is easy

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