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

Source: Internet
Author: User

[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. }




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