NYOJ：A problem is easy（数学）

A problem is easy

时间限制：1000 ms  |  内存限制：65535 KB

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

01

上传者

苗栋栋

思路：双重循环肯定超时的 直接用一个判断条件能减少时间 
i*j+i+j =N 经过观察，可以变形为i*j+i+j+1=N+1,也就是说，可以进一步变形为(i+1)*(j+1)=N+1
所以i从2判断到sqrt(n+1)即可 

#include <stdio.h>#include <math.h>int main(){int t;scanf("%d",&t);while(t--){int n;scanf("%d",&n);int cnt=0;for(int i=2;i<=sqrt(n+1);i++){if((n+1)%i==0)cnt++;}printf("%d\n",cnt);}return 0;}