UV problem solution: 846-steps

Let F (n) denote the furthest distance we can go with n steps, then we can derive this formula easily:
F (n) = (n + 1)/2) ^ 2 if n is odd, (1, 2, 3, 4, 3, 2, 1)
F (n) = (n/2) ^ 2 + n/2 If n is even. (1, 2, 3, 3, 2, 1)
Give the distance of F (N), we can find N by looking around the square root of F (n ).

Code:

1. /*************************************** ************************************
2. * Copyright (c) 2008 by Liu kaipeng *
3. * Liukaipeng at gmail dot com *
4. **************************************** ***********************************/
6. /* @ Judge_id 00000 846 C ++ "Steps "*/
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. /*
26. * Let F (n) denote the furthest distance we can go with n steps, then we can
27. * Derive this formula easily:
28. * F (n) = (n + 1)/2) ^ 2 if n is odd, (1, 2, 3, 4, 3, 2, 1)
29. * F (n) = (n/2) ^ 2 + n/2 If n is even. (1, 2, 3, 3, 2, 1)
30. * Give the distance of F (N), we can find N by looking around the square root
31. * Of F (n ).
32. */
33. Int steps (INT Dist)
34. {
35. If (Dist = 0) return 0;
36. Int step = SQRT (DIST );
37. If (Step * step = DIST) step = Step * 2-1;
38. Else if (Step * Step + step <Dist) step = Step * 2 + 1;
39. Else step = Step * 2;
40. Return step;
41. }
43. Int main (INT argc, char * argv [])
44. {
45. # Ifndef online_judge
46. Freopen (string (argv [0]) + ". In"). c_str (), "r", stdin );
47. Freopen (string (argv [0]) + ". Out"). c_str (), "W", stdout );
48. # Endif
50. Int ncases;
51. Cin> ncases;
52. For (int x, Y; ncases> 0 & CIN> x> Y; -- ncases)
53. Cout <steps (Y-x) <'/N ';
55. Return 0;
56. }
57.