UVAOJ 10110 基础题 灯光 数论
发布日期:2022-03-30 20:19:25 浏览次数:34 分类:博客文章

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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


Sample Output


Sadi Khan 

Suman Mahbub 

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

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