开始所有的灯是灭的，不过我们只关心最后一个灯。
在第i次走动时，只有编号为i的倍数的灯的状态才会改变。
也就是说n有偶数个约数的时候，最后一个灯的状态不会改变，也就是灭的。
n有奇数个约数的时候也就是n为完全平方数的时候，最后一个灯会是亮的。
最后抽象出来，就是判断输入的数是否为完全平方数。
| 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
AC代码：
1 //#define LOCAL 2 #include <iostream> 3 #include <cstdio> 4 #include <cstring> 5 #include <cmath> 6 using namespace std; 7 8 int main(void) 9 { 10 #ifdef LOCAL 11 freopen("10110in.txt", "r", stdin); 12 #endif 13 14 unsigned int N; 15 while(scanf("%d", &N) == 1 && N) 16 { 17 int n = (int)sqrt(N); 18 if(n * n == N) 19 cout << "yes" << endl; 20 else 21 cout << "no" << endl; 22 } 23 return 0; 24 }