HDU 2601：An easy problem【数学】

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8382    Accepted Submission(s): 2044

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
   
   
   
   

 if n=i*j+i+j,then n+1=(i+1)*(j+1)

AC-code:

#include<cstdio>
#include<cmath>
int main()
{
	int T;
	long long n,m,i,k;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%lld",&n);
		n++;
		k=0;
		m=sqrt(n);
		for(i=2;i<=m;i++)
			if(n%i==0)
				k++;
		printf("%lld\n",k);
	}
	return 0;
}