hdu 2601 An easy problem

好坑爹的一题，刚开始暴力DFS，果断超时。

其实可以做一个变形，i×j+i+j可以变形为i×（j+1）+j+1=n+1;进而可以变化为（i+1）×（j+1）=n;

即统计一下n%（i+1）==0可以出现多少次即可;

写这一题目是在Ubuntu的codeblocks交了不下十多次吧！总是wrong!

最后，不管了编译器，把long long怒改写成了__int64,ac了，无语啊！

至今仍然不知道why？

View Code

 1 #include<stdio.h>

 2 #include<math.h>

 3 int main()

 4 {

 5 

 6     int t;

 7     scanf("%d",&t);

 8     while(t--)

 9     {

10         __int64 n;

11         scanf("%I64d",&n);

12         n++;

13        __int64 i;

14         __int64 sum=0;

15         for(i=2;i*i<=n;i++)

16             if(n%i==0)

17             sum++;

18          printf("%I64d\n",sum);

19 

20     }

21     return 0;

22 }

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 <= 10 ¹⁰).

 

 

Output

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

 

 

Sample Input

2

1

3

 

 

Sample Output

0

1