hdu 2601 An easy problem

 

An easy problem

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1505    Accepted Submission(s): 302

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

 

说Easy 却不 Easy  的题目。。。郁闷， 要把n = i*j + i + j 化成 n + 1 = (i + 1)* (j + 1) ;

n + 1 取余i + 1 或者 j + 1 就可以了, 遇到 i ， j 不要双循环。。。不然就TLE 到死。。。。

#include <iostream> #include <cmath> #include <cstring> #include <map> #include <algorithm> using namespace std; int main() { freopen("in4.txt","r", stdin); int t; long long n; long long sqr; long long count = 0; cin>>t; while (t--) { cin>>n; n++; count = 0; sqr = (int)sqrt(n); for (int i = 2; i <= sqr; i++) { if (n%i == 0) count++; } cout<<count<<endl; } }