An easy problem hdu 2601

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5055    Accepted Submission(s): 1249

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

 

 

Author

 

Teddy

 

 

 

 

 

这个题目是很经典的，其思想就是（1 + i )( 1 + j )==  n+1 , 所以要求这个问题时只需要去判断有多少个 （i+1)就可以了，我之前是想要找规律的，结果发现这个规律是很乱的，也就是没规律的。

 

 

 

 

#include<iostream>
#include<cstdio>
#include<math.h>
using namespace std;
int main()
{
    int t,i,j;
    __int64 n,f;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        n=n+1;
        j=0;
        f=sqrt((double)n);
        for(i=2;i<=f;i++)
        {
            if(n%i==0)
            {
              j++;
            }
        }
        printf("%d\n",j);
    }
    return 0;
}