description: The Winter holiday will be here soon. Mr. Chanek wants to decorate his house's wall with ornaments. The wall can be represented as a binary string a of length n. His favorite nephew has another binary string b of length m (m ≤ n).
  
  Mr. Chanek's nephew loves the non-negative integer k. His nephew wants exactly k occurrences of b as substrings in a. 
  
  However, Mr. Chanek does not know the value of k. So, for each k (0 ≤ k ≤ n - m + 1), find the minimum number of elements in a that have to be changed such that there are exactly k occurrences of b in a.
  
  A string s occurs exactly k times in t if there are exactly k different pairs (p,q) such that we can obtain s by deleting p characters from the beginning and q characters from the end of t.
  
  Input
  
  The first line contains two integers n and m (1 ≤ m ≤ n ≤ 500) — size of the binary string a and b respectively.
  
  The second line contains a binary string a of length n.
  
  The third line contains a binary string b of length m.
  
  Output
  
  Output n - m + 2 integers — the (k+1)-th integer denotes the minimal number of elements in a that have to be changed so there are exactly k occurrences of b as a substring in a.
  
  Example
  
  Input
  
  
  9 3
  100101011
  101
  
  
  Output
  
  
  1 1 0 1 6 -1 -1 -1
  
  Note
  
  For k = 0, to make the string a have no occurrence of 101, you can do one character change as follows.
  
  100101011 → 100100011
  
  For k = 1, you can also change a single character.
  
  100101011 → 100001011
  
  For k = 2, no changes are needed.
solution: #include <bits/stdc++.h>
  const int inf = 1 << 29;
  inline void upd(int &v, int x) {
    if (v > x) v = x;
  }
  char s[510], t[510];
  int f[510], dp[2][510][510];
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    scanf("%s%s", s, t);
    f[0] = -1;
    for (int i = 1; i < m; i++) {
      int j = f[i];
      while (j != -1 && t[i] != t[j]) {
        j = f[j];
      }
      f[i + 1] = j + 1;
    }
    int u = n - m + 1;
    for (int i = 0; i <= u; i++) {
      for (int j = 0; j < m; j++) {
        dp[0][i][j] = inf;
      }
    }
    dp[0][0][0] = 0;
    int y = 0, z = 1;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= u; j++) {
        for (int k = 0; k < m; k++) {
          dp[z][j][k] = inf;
        }
      }
      for (int j = 0; j <= u; j++) {
        for (int k = 0; k < m; k++) {
          int x = dp[y][j][k];
          if (x == inf) continue;
          for (int l = 0; l < 2; l++) {
            int p = k, nj = j;
            while (p != -1 && (s[i] == t[p]) == l) {
              p = f[p];
            }
            if (++p == m) {
              nj++;
              p = f[k];
              while (p != -1 && (s[i] == t[p]) == l) {
                p = f[p];
              }
              ++p;
            }
            upd(dp[z][nj][p], x + l);
          }
        }
      }
      y ^= 1;
      z ^= 1;
    }
    for (int j = 0; j <= u; j++) {
      int ans = inf;
      for (int k = 0; k < m; k++) {
        upd(ans, dp[y][j][k]);
      }
      if (ans == inf) ans = -1;
      printf("%d ", ans);
    }
    return 0;
  }
  