HDOJ2601 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:

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5     int t;
 6     __int64 n;
 7     scanf("%d",&t);
 8     while(t--)
 9     {
10         scanf("%I64d",&n);
11         int flag=0;
12         for(__int64 i=1;(i+1)*(i+1)<=(n+1);i++)
13         {
14             if((n+1)%(i+1)==0)
15                 flag++;
16         }
17         printf("%d\n",flag);
18     }
19     return 0;
20 }