[Digital DP] spoj 10738 ra-one numbers

Given X and Y, the number of even digits between [x, y] is equal to or equal to one minus odd digits.

One digit is the first.

Example: 10 = 1-0 = 1 so 10 is such a number

Idea: digit DP [I] [Sum] [OK] I bit and sum whether there is a leading 0.

Then, because there is a negative number, 0 is set to 100 Based on the range, and then the last and equal to 101 are the numbers.

Code:

[CPP]View plaincopyprint?

1. # Include "cstdlib"
2. # Include "cstdio"
3. # Include "cstring"
4. # Include "cmath"
5. # Include "stack"
6. # Include "algorithm"
7. # Include "iostream"
8. Using namespace STD;
9. Int DP [12] [200] [2], num [12];
10. Int fuck [2] = {1,-1 };
11. Int DFS (INT site, int sum, int OK, int F)
12. {
13. If (Site = 0)
14. {
15. If (OK = 0) return 0;
16. Return sum = 101? 1:0; // The sum of small processing and 101
17. }
18. If (! F & DP [site] [Sum] [OK]! =-1) return DP [site] [Sum] [OK];
19. Int Len = f? Num [site]: 9;
20. Int ans = 0;
21. For (INT I = 0; I <= Len; I ++)
22. {
23. If (OK = 0)
24. {
25. If (I = 0) ans + = DFS (site-1, sum, OK | I! = 0, F & I = Len );
26. Else ans + = DFS (site-1, sum + I * fuck [Site % 2], OK | I! = 0, F & I = Len );
27. }
28. Else
29. {
30. Ans + = DFS (site-1, sum + I * fuck [Site % 2], OK | I! = 0, F & I = Len );
31. }
32. }
33. If (! F) DP [site] [Sum] [OK] = ans;
34. Return ans;
35. }
36. Int solve (int x)
37. {
38. If (x <0) return 0;
39. Int CNT = 0;
40. While (X)
41. {
42. Num [++ CNT] = x % 10;
43. X/= 10;
44. }
45. Return DFS (CNT, 100, 0, 1); // sum =
46. }
47. Int main ()
48. {
49. Int T;
50. Scanf ("% d", & T );
51. Memset (DP,-1, sizeof (DP ));
52. While (t --)
53. {
54. Int X, Y;
55. Scanf ("% d", & X, & Y );
56. Printf ("% d \ n", solve (y)-solve (x-1 ));
57. }
58. Return 0;
59. }

[Digital DP] spoj 10738 ra-one numbers