[Digital dp] spoj 10738 Ra-One Numbers, spoj10738

[Digital dp] spoj 10738 Ra-One Numbers, spoj10738

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. }