CODEFORCES 645E Intellectual Inquiry

// 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. /**

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2. * Author: +7w 
   

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3. */ 
   

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4. #include <assert.h> 
   

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5. #include <ctype.h> 
   

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6. #include <inttypes.h> 
   

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7. #include <math.h> 
   

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8. #include <stdio.h> 
   

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9. #include <stdlib.h> 
   

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10. #include <string.h> 
    

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11. #include <time.h> 
    

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12. ---

13. #include <algorithm> 
    

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14. #include <bitset> 
    

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15. #include <functional> 
    

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16. #include <list> 
    

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17. #include <iostream> 
    

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18. #include <map> 
    

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19. #include <queue> 
    

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20. #include <set> 
    

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21. #include <stack> 
    

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22. #include <sstream> 
    

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23. #include <string> 
    

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24. #include <vector> 
    

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25. ---

26. const int INF3 = 0x3f3f3f3f; 
    

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27. const int INF5 = 0x5f5f5f5f; 
    

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28. const double PI = acos(-1.0); 
    

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29. const double EPS9 = 1e-9; 
    

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30. const int TEN3 = 1000; 
    

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31. const int TEN6 = TEN3 * TEN3; 
    

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32. const int TEN9 = TEN6 * TEN3; 
    

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33. ---

34. typedef long long llong; 
    

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35. typedef unsigned long long ullong; 
    

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36. typedef std::pair<int, int> pii; 
    

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37. ---

38. #define FST first 
    

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39. #define SEC second 
    

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40. #define PB push_back 
    

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41. #define MP make_pair 
    

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

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

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

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45. ---

46. const int MOD = TEN9 + 7; 
    

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47. const int N = TEN6 + 10; 
    

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48. const int K = 30; 
    

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49. const int M = TEN6 + 10; 
    

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50. ---

51. char str[M]; 
    

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52. pii cv[N]; 
    

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53. ---

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

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55. { 
    

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56. int n, k; 
    

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57. scanf("%d %d", &n, &k); 
    

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58. scanf("%s", str); 
    

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59. for (int i = 0; i < k; ++i) { 
    

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60. cv[i].FST = i; // cmper 
    

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61. cv[i].SEC = 0; // value 
    

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62. } 
    

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63. int sum = 0; 
    

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64. for (int i = 0; str[i]; ++i) { 
    

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65. int j = str[i] - ‘a‘; 
    

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66. int tmp_sec = cv[j].SEC; 
    

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67. cv[j].SEC = (sum + 1) % MOD; 
    

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

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69. sum = (sum + cv[j].SEC) % MOD; 
    

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70. cv[j].FST = i + k; 
    

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71. } 
    

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72. std::sort(cv, cv + k); 
    

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73. for (int i = 0; i < n; ++i) { 
    

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74. int j = i % k; 
    

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75. int tmp_sec = cv[j].SEC; 
    

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76. cv[j].SEC = (sum + 1) % MOD; 
    

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

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78. sum = (sum + cv[j].SEC) % MOD; 
    

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79. } 
    

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80. printf("%d\n", (sum + 1) % MOD); 
    

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81. return 0; 
    

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82. } 
    

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