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

程序主要目的是怎么简化时间复杂度，想到原式等价于（m+1）=（i+1）*（j+1），所以问题就归结于求（m+1）大于等于2的因子的个数.

 1 #include <stdio.h> 2 #include <math.h> 3  4 int main(){ 5     int T; 6     int n; 7     int amount; 8     int i; 9 10     11     scanf("%d",&T);12     13     while(T--){14         scanf("%d",&n);15         16         amount=0;17         for(i=2;i<=sqrt(n+1);i++){18             if((n+1)%i==0)19                 amount++;20         }21         22         23         printf("%d\n",amount);24     }25     26     return 0;27 }

 

A problem is easy