UVA - 10110 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题意：给出N，可以知道其因子数目，如果数目为奇数就可以打开，yes如果数目为偶数就不可以打开，no多写几个数，可以知道，可以开方的数其因子为奇数；不可以开方的数其因子为偶数，题意也就转化为判断该数是否可以开方； 
#include<stdio.h>#include<math.h>#include<iostream>#include<string.h>using namespace std;int main(){long long int n;while(scanf("%lld",&n)){if(n == 0)break;long long int m;m = (int)sqrt(n);if(n == m*m)cout << "yes"<<endl;else cout << "no"<<endl;}return 0;}