HDU-2601-An easy problem

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2601

An easy problem

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 <= 1010).

 

Output

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

 

Sample Input

2
1
3

 

Sample Output

0
1

将一个数 N 分解成 i * j + i + j 有多少种分解方式。n+1=(i+1)(j+1).

代码如下：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int T;
cin>>T;
while(T--){
long long int n;
cin>>n;
long long int m=sqrt(n+1);
int js=0;
for(int i=m;i>=2;i--)
if((n+1)%i==0)
js++;
cout<<js<<endl;
}
return 0;
}