AcWing 1052. 设计密码

题目链接：

https://www.acwing.com/problem/content/description/1054/

题解：

kmp + dp
f[i][j]: 在第i个字母已经选取的情况下，状态为j的所有方案的数量，这里的j表示kmp中已经匹配的模板串的数量
状态机的状态解释很难想到，建议背过

AC代码：

import java.util.*;

public class Main {
    static int N = 55;
    static int[][] f = new int[N][N];
    static int[] ne = new int[N];
    static int MOD = (int) 1e9 + 7;
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = Integer.parseInt(sc.nextLine());
        String p = " " + sc.nextLine();
        int m = p.length() - 1;
        
        for (int i = 2, j = 0; i <= m; i ++) {
            while (j != 0 && p.charAt(i) != p.charAt(j + 1)) j = ne[j];
            if (p.charAt(i) == p.charAt(j + 1)) j ++;
            ne[i] = j;
        }
        
        f[0][0] = 1;
        
        for (int i = 1; i <= n; i ++) {
            for (int j = 0; j < m; j ++) {
                for (char k = 'a'; k <= 'z'; k ++) {
                    int u = j;
                    while (u != 0 && k != p.charAt(u + 1)) u = ne[u];
                    if (k == p.charAt(u + 1)) u ++;
                    if (u < m) f[i][u] = (f[i][u] + f[i - 1][j]) % MOD;
                }
            }
        }
        
        int res = 0;
        for (int i = 0; i < m; i ++) res = (res + f[n][i]) % MOD;
        
        System.out.println(res);
    }
}