NYOJ 216:An easy 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=n等价于i*j+i+j+1=n+1

分离变量得：（i+1）*（j+1）=n+1

也就是i*j=n+1

i从2开始到sqrt(n+1);

#include<cstdio>
#include<cmath>
int main()
{
	__int64 t;
	while(scanf("%d",&t)!=EOF)
 {
	while(t--)
	{
		__int64 n,cnt=0;
		scanf("%I64d",&n);
		for(int i=2;i<=sqrt(n+1);i++)
		{
			if((n+1)%i==0)
			{
				cnt++; 
			}
		}
		printf("%I64d\n",cnt);
	}
 }

	return 0;
}