NYOJ 216 A problem is easy

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

我第一次写的算法提交上去时间超限，= =，然后参考了讨论区的算法，发现这又是一个需要运用数学技巧的题！

我第一次写的代码

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

经过数学技巧处理后的算法

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

加油吧！妹砸！你要学习的还有很多很多~