hdu2601 An easy problem（数论）

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5758    Accepted Submission(s): 1411

Problem Description

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 ?

 

Input

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10 ¹⁰).

 

Output

For each case, output the number of ways in one line.

 

Sample Input

  
  

   
   2
1
3
  
  

 

Sample Output

  
  

   
   0
1
  
  

 

i*j+i+j+1=(i+1)(j+1)

#include<stdio.h>
int main()
{
    int cas;
    __int64 i,n,ans;
    scanf("%d",&cas);
    while(cas--)
    {
        scanf("%I64d",&n);
        ans=0;
        for(i=2; i*i<=n+1; i++)
            if((n+1)%i==0) ans++;
        printf("%I64d\n",ans);
    }
    return 0;
}