UV problem solution: 701-the archaeologist's Dilemma

Let p denotes the prefix, T Denotes the # of lost digits. We are searching for N, that the prefix of 2 ^ N is P.
We have an inequlity
P * 10 ^ t <= 2 ^ n <(p + 1) * 10 ^ t
Thus
Log2 (p * 10 ^ t) <= log2 (2 ^ N) <log2 (p + 1) * 10 ^ t ),
Which is
Log2 (p) + T * log2 (10) <= n <log2 (p + 1) + T * log2 (10 ).
Also, we know that
P <10 ^ (T-1 ),
That is
T> log10 (p) + 1.
Then, we can brute force on T and find the minmum n.

Code:

1. /*************************************** **********************************
2. * Copyright (c) 2008 by liukaipeng *
3. * Liukaipeng at gmail dot com *
4. **************************************** *********************************/
6. /* @ Judge_id 00000 701 C ++ "the archaeologist's Dilemma "*/
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 p denotes the prefix, T Denotes the # of lost digits. We are searching
27. For N, that the prefix of 2 ^ N is P.
28. We have an inequlity
29. P * 10 ^ t <= 2 ^ n <(p + 1) * 10 ^ t
30. Thus
31. Log2 (p * 10 ^ t) <= log2 (2 ^ N) <log2 (p + 1) * 10 ^ t ),
32. Which is
33. Log2 (p) + T * log2 (10) <= n <log2 (p + 1) + T * log2 (10 ).
34. Also, we know that
35. P <10 ^ (T-1 ),
36. That is
37. T> log10 (p) + 1.
38. Then, we can brute force on T and find the minmum n.
39. */
40. Long Double minexpof2with (Long Double prefix)
41. {
42. Long double lower = log2l (prefix );
43. Long Double upper = log2l (prefix + 1 );
44. Long double F = log2l (10 );
45. Long Double T = ceill (log10l (prefix + 0.5) + 1; // avoid log10l (1) = 0
46. For (; ceill (lower + T * f )! = Floorl (upper + T * F); t + = 1 ){}
47. Return ceill (lower + T * F );
48. }
50. Int main (INT argc, char * argv [])
51. {
52. # Ifndef online_judge
53. Freopen (string (argv [0]) + ". In"). c_str (), "r", stdin );
54. Freopen (string (argv [0]) + ". Out"). c_str (), "W", stdout );
55. # Endif
57. Long Double prefix;
58. While (CIN> prefix)
59. Cout <(long) minexpof2with (prefix) <'/N ';
61. Return 0;
62. }
63.