# A problem is easy

## A problem is easy

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 ?

```2
1
3```

```0
1```
```
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n;
int a,b;
cin>>n;
while(n--)
{
cin>>a;
b=0;
for(int i=2;i*i<=(a+1);i++)
{
int j = (a+1)/i;
if(i*j==a+1 && i<=j)

b++;
}
cout<<b<<endl;
}
return 0;
}        ```

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