HDU 2601：An easy problem【数学】

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 9414    Accepted Submission(s): 2326


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

 
   本题是求整数的因子的个数， i * j + i + j =n,即为(i+1)*(j+1)=n+1,转化为了求解n+1的因子的个数

#include<cstdio>
#include<cmath>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
long long n,m,ans=0;
scanf("%lld",&n);
n++;
m=sqrt(n);
for(int i=2;i<=m;i++)
{
if(n%i==0)
ans++;
}
printf("%lld\n",ans);
}
return 0;
}