CODEFORCES 645E Intellectual Inquiry

Posted 2020-07-14

// CODEFORCES 645E Intellectual Inquiry

http://codeforces.com/problemset/problem/645/E

字母表只有前面 k 个小写字母。给出一个字符串 s，你可以往 s 后添加 n 个字母，使得，总的字符串的不同子序列最多。

Input

n k
s

1 ≤ n ≤ 1,000,000
1 ≤ k ≤ 26
1 ≤ |s| ≤ 1,000,000

Output

输出答案模 1,000,000,007。

Solution

看了题解才会的。

f[i] 表示以 i 为结尾的子序列（不含空串）有多少个，往字符串后面加一个字母 c，则只有 f[c] 会改变。而且是改成 ∑f[] + 1。重点在这个值跟 c 是没有关系的，改变之后的所以子序列数目的和是 ∑f[] - f[c] + ∑f[] + 1，要使这个值最大，应该使得 f[c] 最小，也就是说，添加一个使得 f[c] 最小的 c 在字符串末尾是最佳的方案。

Code

1. /** 

   ---

2. * Author: +7w 

   ---

3. */ 

   ---

4. #include <assert.h> 

   ---

5. #include <ctype.h> 

   ---

6. #include <inttypes.h> 

   ---

7. #include <math.h> 

   ---

8. #include <stdio.h> 

   ---

9. #include <stdlib.h> 

   ---

10. #include <string.h> 

    ---

11. #include <time.h> 

    ---

12.  

    ---

13. #include <algorithm> 

    ---

14. #include <bitset> 

    ---

15. #include <functional> 

    ---

16. #include <list> 

    ---

17. #include <iostream> 

    ---

18. #include <map> 

    ---

19. #include <queue> 

    ---

20. #include <set> 

    ---

21. #include <stack> 

    ---

22. #include <sstream> 

    ---

23. #include <string> 

    ---

24. #include <vector> 

    ---

25.  

    ---

26. const int INF3 = 0x3f3f3f3f; 

    ---

27. const int INF5 = 0x5f5f5f5f; 

    ---

28. const double PI = acos(-1.0); 

    ---

29. const double EPS9 = 1e-9; 

    ---

30. const int TEN3 = 1000; 

    ---

31. const int TEN6 = TEN3 * TEN3; 

    ---

32. const int TEN9 = TEN6 * TEN3; 

    ---

33.  

    ---

34. typedef long long llong; 

    ---

35. typedef unsigned long long ullong; 

    ---

36. typedef std::pair<int, int> pii; 

    ---

37.  

    ---

38. #define FST first 

    ---

39. #define SEC second 

    ---

40. #define PB push_back 

    ---

41. #define MP make_pair 

    ---

42. #define CLEAR(a) memset(a, 0, sizeof(a)) 

    ---

43. #define FILL(a, c) memset(a, c, sizeof(a)) 

    ---

44. #define COPY(a, b) memcpy(a, b, sizeof(b)) 

    ---

45.  

    ---

46. const int MOD = TEN9 + 7; 

    ---

47. const int N = TEN6 + 10; 

    ---

48. const int K = 30; 

    ---

49. const int M = TEN6 + 10; 

    ---

50.  

    ---

51. char str[M]; 

    ---

52. pii cv[N]; 

    ---

53.  

    ---

54. int main(int argc, char const *argv[]) 

    ---

55. { 

    ---

56. int n, k; 

    ---

57. scanf("%d %d", &n, &k); 

    ---

58. scanf("%s", str); 

    ---

59. for (int i = 0; i < k; ++i) { 

    ---

60. cv[i].FST = i; // cmper 

    ---

61. cv[i].SEC = 0; // value 

    ---

62. } 

    ---

63. int sum = 0; 

    ---

64. for (int i = 0; str[i]; ++i) { 

    ---

65. int j = str[i] - ‘a‘; 

    ---

66. int tmp_sec = cv[j].SEC; 

    ---

67. cv[j].SEC = (sum + 1) % MOD; 

    ---

68. sum = (sum - tmp_sec + MOD) % MOD; 

    ---

69. sum = (sum + cv[j].SEC) % MOD; 

    ---

70. cv[j].FST = i + k; 

    ---

71. } 

    ---

72. std::sort(cv, cv + k); 

    ---

73. for (int i = 0; i < n; ++i) { 

    ---

74. int j = i % k; 

    ---

75. int tmp_sec = cv[j].SEC; 

    ---

76. cv[j].SEC = (sum + 1) % MOD; 

    ---

77. sum = (sum - tmp_sec + MOD) % MOD; 

    ---

78. sum = (sum + cv[j].SEC) % MOD; 

    ---

79. } 

    ---

80. printf("%d\n", (sum + 1) % MOD); 

    ---

81. return 0; 

    ---

82. } 

    ---