HDU 2601 An easy problem

An easy problem

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

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

这道题题意是找出在1~n中满足n=i*j+i+j（0<i<=j）的数的个数
n=i*j+i+j可变为n+1=(i+1)*(j+1)
则i，j最大情况为i=j

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
using namespace std;
int main()
{
long long int T,n,res,temp;
scanf("%d",&T);
while(T--)
{
res=0;
scanf("%lld",&n);
n++;
temp=sqrt(n*1.0);
for(long long int i=2;i<=temp;i++)
{
if(n%i==0)
res++;
}
printf("%lld\n",res);
}
return 0;
}