Light, more light (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

3
6241
8191
0

Sample Output

no
yes
no

 

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(int argc, const char *argv[])
{
  unsigned int n, m;

  while (scanf("%d", &n) && n != 0) {

    m = (unsigned int)sqrt((double)n);

    if (m * m != n) {
      printf("no\n");
    } else {
      printf("yes\n");
    }

  }

  return 0;
}

总结：

　　做这一道题，可谓历尽波折。刚看到这题感到很简单，只需统计出（1,n）之间能够整除n的数的个数。但是提交上去TLE.没办法只能想其他办法，最后在网上找到完全平方数的性质，完全平方数的因子是奇数个，非完全平方数的因子是偶数个。利用这个性质可以轻松AC.