UV problem solution: 10110-light, more light

The final state of the last bulb is determined by the number of factors of N. only the number which is the square of a number has an odd number of factors, thus it has a state of "yes ".

Code:

1. /*************************************** ************************************
2. * Copyright (c) 2008 by Liu kaipeng *
3. * Liukaipeng at gmail dot com *
4. **************************************** ***********************************/
6. /* @ Judge_id 00000 10110 C ++ "light, more light "*/
8. # Include <algorithm>
9. # Include <cmath>
10. # Include <cstdio>
11. # Include <cstring>
12. # Include <deque>
13. # Include <fstream>
14. # Include <iostream>
15. # Include <list>
16. # Include <map>
17. # Include <queue>
18. # Include <set>
19. # Include <stack>
20. # Include <string>
21. # Include <vector>
23. Using namespace STD;
25. Int main (INT argc, char * argv [])
26. {
27. # Ifndef online_judge
28. Freopen (string (argv [0]) + ". In"). c_str (), "r", stdin );
29. Freopen (string (argv [0]) + ". Out"). c_str (), "W", stdout );
30. # Endif
32. /* The final state of the last bulb is determined by the number of factors
33. Of N. Only the number which is the square of a number has odd number
34. Factors, thus it has a state of "yes ".*/
35. For (unsigned int N; CIN >>n & n! = 0 ;){
36. Unsigned int r = SQRT (N );
37. If (R * r = N) cout <"Yes/N ";
38. Else cout <"No/N ";
39. }
41. Return 0;
42. }
43.