UVAOJ 10110 基础题 灯光 数论
发布日期:2022-03-30 20:19:25 浏览次数:39 分类:博客文章

本文共 2039 字,大约阅读时间需要 6 分钟。

 

 http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1051

Light, more light

The Problem

There is man named "mabu" for switching on-off light in our University. He switches on-off the lights in a corridor. Every bulb has its own toggle switch. That is, if it is pressed then the bulb turns on. Another press will turn it off. To save power consumption (or may be he is mad or something else) he does a peculiar thing. If in a corridor there is `n' bulbs, he walks along the corridor back and forth `n' times and in i'th walk he toggles only the switches whose serial is divisable by i. He does not press any switch when coming back to his initial position. A i'th walk is defined as going down the corridor (while doing the peculiar thing) and coming back again.

Now you have to determine what is the final condition of the last bulb. Is it on or off? 

 

The Input

The input will be an integer indicating the n'th bulb in a corridor. Which is less then or equals 2^32-1. A zero indicates the end of input. You should not process this input.

The Output

Output "yes" if the light is on otherwise "no" , in a single line.

Sample Input

3624181910

Sample Output

noyesno

Sadi Khan 

Suman Mahbub 
01-04-2001

1 /************************************************************************* 2     > File Name: 12345.cpp 3     > Author: acmicpcstar 4     > Mail: acmicpcstar@gmail.com 5     > Created Time: 2014年04月24日 星期四 11时46分18秒 6  ************************************************************************/ 7  8 #include
9 #include
10 #include
11 #include
12 #include
13 using namespace std;14 const double pi=atan(1.0)*4.0;15 int main()16 {long long k,root;17 double n;18 while(cin>>k&&k!=0)19 {if(k==1) {printf("yes\n");continue;}20 n=sqrt(k);21 root=int(n);22 if (root*root==k||(root+1)*(root+1)==k) printf("yes\n");23 else printf("no\n");24 }25 return 0;26 }

总结。。。完全平方数。。。因子为。。。奇数个。。。只有1特例。。。。然后。。。要用LONG LONG

转载地址:https://www.cnblogs.com/acmicpcstar/p/3687234.html 如侵犯您的版权,请留言回复原文章的地址,我们会给您删除此文章,给您带来不便请您谅解!

上一篇:C# Aes加密算法
下一篇:UVAOJ 550 基础题 移位乘法 数论

发表评论

最新留言

路过,博主的博客真漂亮。。
[***.116.15.85]2024年04月03日 20时33分48秒

关于作者

    喝酒易醉,品茶养心,人生如梦,品茶悟道,何以解忧?唯有杜康!
-- 愿君每日到此一游!

推荐文章