[原]NYOJ-216-A problem is easy

大学生程序代写

/*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

2

1

3

样例输出

0

1

上传者

苗栋栋*/





/*

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>//不需要long long int

int main(){

   int m,i;

   scanf("%d",&m);

   while(m--){

int n;

   scanf("%d",&n);

int count = 0;

int s=sqrt(n+1);

for(i=2;i<=s;++i)

{

 if((n +1)%i==0)

 count++;

}

printf("%d\n",count);

}



return 0;

}

作者：chao1983210400 发表于2013-7-20 12:33:26 原文链接

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