『算法-ACM 竞赛-数学-数论』 HDU 2601 An easy problem（约束和）

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

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
using namespace std;

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