hdoj 2601（判断N=i*j+i+j）

Problem E

Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 16   Accepted Submission(s) : 9

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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

易知用两个循环嵌套判断n=i*j+i+j必然超时，可以推得n+1=i*j+i+j+1,即为n+1=(i+1)*(j+1);
题中提示j>=i>0;那么j+1>=i+1,用x代替（i+1）,将x从2遍历到sqrt(n+1),在此条件下判断(n+1)%x,若为0，说明
存在（j+1），使得(i+1)*(j+1)=n+1,由于是在(i+1)<=sqrt(n+1)的条件下的结果，必然有(j+1)>=(i+1)成立。
时间复杂度由O(n)变为O(sqrt(n)).