B - 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 <= 1010).
 
Output
For each case, output the number of ways in one line.
 
Sample Input
2
1
3
 
Sample Output

0

1

输入n，计算有多少种i*j+i+j==n

#include <iostream>#include <cstring>#include <cstdio>#include<cmath>#include<algorithm>using namespace std;int main(){    int t;    scanf("%d",&t);    long long n;    for(int k=0;k<t;k++)    {        cin>>n;        n++;        int m=(int)sqrt(n*1.0);        int count=0;        for(int i=2;i<=m;i++)        {            if(n%i==0)            count++;        }        cout<<count<<endl;    }    return 0;}