Snapper Chain!! - Google codejam-2010

http://code.google.com/codejam/contest/dashboard?c=433101#s=p0

You can find the problem description above! Believe me it's so much of fun to solve this!!
Go ahead and try!!!! I could not believe after I got the solution!.....Ufffff.......

Hint :: Think in bits!!

Approach ::
Lets say we have 2 snappers (N =2)-
Initially both are in OFF state

Remember : With a snap, only the snapper receiving power changes state!!!

OFF	OFF	Initial State
ON	OFF	After snap k =1
OFF	ON	After snap k =2
ON	ON	After snap k =3
OFF	OFF	Snap k =4

See the bit pattern ?? Lets say OFF = 0; and ON = 1

0	0	Initial State K=0
1	0	K=1
0	1	K=2
1	1	K=3
0	0	K=4

So, if there are N snappers, we will turn the light on when K = [2 (power) N ] - 1
In the above case, N = 2, K = pow(2,2)-1 = 3. , After that, a snap will get us to the initial state!!

Isn't it simple and great!!!!!!!!

===================================================

int main () {
   int n,k,num_cases;
   int factor,i;
   FILE *fp;

   fp = fopen("A-large-practice.in","r");
   if (!fp) {
      exit(0);
   }
   fscanf(fp,"%d",&num_cases);
   for (i = num_cases ; i > 0; i--) {
      fscanf(fp,"%d %d", &n, &k);
      factor = pow(2,n);

      printf("Case #%d: ",num_cases-i+1);
      if (k%factor == (factor-1))
          printf("ON\n");
      else
          printf("OFF\n");

   }
 fclose(fp);

}