HDU 2601

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 <= 10 ¹⁰).

 

Output

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

 

Sample Input

 2
1
3 

 

Sample Output

 0
1 

 

#include<stdio.h>
#include<math.h>
int main()
{


__int64 p,q;
int t;
int i;
scanf("%d",&t);
while(t--)
{
    int  count=0;
    scanf("%I64d",&p);
    p=p+1;
    q=sqrt(p);
    for(i=2;i<=q;i++)
    {
        if(p%i==0)
        count=count+1;
    }
    printf("%d\n",count);
}
}

I*J+I+j=n    得（i+1)*(j+1)=(n +1)

所以只需要求n+1的因子个数

根据题意 i>0;，i<=j又因为（i+1)*(j+1)=(n +1)，，所以i+1>=2 所以for循环里从2开始