HDOJ---2601 An easy problem[水题]

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4072    Accepted Submission(s): 968

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

 

 

Source

HDU 1st “Vegetable-Birds Cup” Programming Open Contest

 

 

Recommend

lcy

 

 

 

 

 

 

水题了，i*j+i+j=(i+1)*(j+1)-1=n

 

code:

#include<iostream>
using namespace std;
int main()
{
    int t;
    __int64 n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%I64d",&n);
        int flag=0;
        for(__int64 i=1;(i+1)*(i+1)<=(n+1);i++)
        {
            if((n+1)%(i+1)==0)
                flag++;
        }
        printf("%d\n",flag);
    }
    return 0;
}