A problem is easy

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 ?

2
1
3

0
1

程序主要目的是怎么简化时间复杂度，想到原式等价于（m+1）=（i+1）*（j+1），所以问题就归结于求（m+1）大于等于2的因子的个数.

 1 #include <stdio.h>
 2 #include <math.h>
 3
 4 int main(){
 5     int T;
 6     int n;
 7     int amount;
 8     int i;
 9
10
11     scanf("%d",&T);
12
13     while(T--){
14         scanf("%d",&n);
15
16         amount=0;
17         for(i=2;i<=sqrt(n+1);i++){
18             if((n+1)%i==0)
19                 amount++;
20         }
21
22
23         printf("%d\n",amount);
24     }
25
26     return 0;
27 }