Light, more light UVA10110

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
少し考えなければならない数学の問題は、大体n個の明かりがあることを意味して、一人でn個の明かりを歩いて、i回目で明かりの状態を逆にして、最後のn個目の明かりの状態を求めます.
考え方:i番目のiがnで割り切れると仮定すると、j=n/iとなり、j番目のiでは逆を取り、状態は変化しないが、2つの数が等しいと変化し、この数が平方数であるかどうかを求めることになる.

#include<iostream>
#include<cmath>

using namespace std;

int main()
{
    long long n;
    while(cin>>n&&n)
    {
        long long i=static_cast<long long>(sqrt(n));
        if(i*i==n) cout<<"yes"<<endl;
        else cout<<"no"<<endl;
    }
    return 0;
}