UVAOJ 10110 基础题 灯光 数论
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
3 6241 8191 0
Sample Output
no yes no
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<iostream> 9 #include<string> 10 #include<cstdio> 11 #include<cmath> 12 #include<algorithm> 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