A problem is easy

难度：3

描述

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 ?

输入

The first line contain a T(T <= 2000) . followed by T lines ,each line contain an integer N (0<=N <= 10^11).

输出

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

样例输入

213

样例输出

0
1

import java.util.*;public class Main { public static void main(String[] args) {  Scanner in=new Scanner(System.in);  int k=in.nextInt();       while(k-->0){       int n=in.nextInt();           int cnt=0;           for (int i =2; i <=Math.sqrt(n+1); i++) { if((n+1)%i==0)cnt++;      }   System.out.println(cnt);       }    } }         

突破思路，数学的配凑，N+1=(i+1)*(j+1)，这样只要一个循环即可，两个循环会超时，0和1都是0不用考虑直接输出默认值0，所以从2开始循环寻找