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>
  using namespace std;
  const int MOD = (int)1e9 + 7;
  const int N = (int)5e7;
  int T;
  int n, m;
  char a[505], b[505];
  int pi[505], poz[505][2];
  int16_t dp[505][505][505];
  void precalc() {}
  void solve() {
    cin >> n >> m >> (a + 1) >> (b + 1);
    int p = 0;
    for (int i = 2; i <= m; i++) {
      while (p && b[p + 1] != b[i]) p = pi[p];
      if (b[p + 1] == b[i]) p++;
      pi[i] = p;
    }
    for (int i = 0; i <= m; i++) {
      for (char c = '0'; c <= '1'; c++) {
        p = i;
        while (p && b[p + 1] != c) p = pi[p];
        if (b[p + 1] == c) p++;
        poz[i][c - '0'] = p;
      }
    }
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k <= m; k++) dp[i][j][k] = static_cast<int16_t>(n + 5);
      }
    }
    dp[0][0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= i; j++) {
        for (int k = 0; k <= m; k++) {
          for (char c = '0'; c <= '1'; c++) {
            int k2 = poz[k][c - '0'];
            dp[i + 1][j + (k2 == m)][k2] =
                min(dp[i + 1][j + (k2 == m)][k2],
                    static_cast<int16_t>(dp[i][j][k] + (a[i + 1] != c)));
          }
        }
      }
    }
    for (int i = 0; i <= n - m + 1; i++) {
      int16_t ans = n + 5;
      for (int j = 0; j <= m; j++) ans = min(ans, dp[n][i][j]);
      cout << (ans == n + 5 ? -1 : ans) << " ";
    }
    cout << '\n';
  }
  int main() {
    ios_base ::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    T = 1;
    precalc();
    for (; T; T--) {
      solve();
    }
    return 0;
  }
  