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;
  namespace KMP {
  vector<int> nxt;
  void get_next(string& s) {
    nxt.resize(s.length());
    nxt[0] = -1;
    for (int i = 1, j = -1; i < s.length(); i++) {
      while (j != -1 && s[j + 1] != s[i]) j = nxt[j];
      if (s[j + 1] == s[i]) ++j;
      nxt[i] = j;
    }
  }
  }  // namespace KMP
  using namespace KMP;
  class Solution {
   private:
    const int inf = 0x3f3f3f3f;
    int n, m;
    string a, b;
    ;
  
   public:
    void solveOne() {
      cin >> n >> m;
      cin >> a >> b;
      get_next(b);
      vector<vector<vector<int>>> dp(
          n + 1, vector<vector<int>>(m + 1, vector<int>(n - m + 2, inf)));
      dp[0][0][0] = 0;
      for (int i = 0; i <= n - 1; i++) {
        for (int j = 0; j <= m; j++) {
          for (int k = 0; k <= n - m + 1; k++) {
            if (dp[i][j][k] == inf) continue;
            int tj = (j == m ? nxt[m - 1] : j - 1);
            while (tj != -1 && b[tj + 1] != a[i]) tj = nxt[tj];
            if (b[tj + 1] == a[i]) ++tj;
            if (tj + 1 == m) {
              if (k != n - m + 1)
                dp[i + 1][tj + 1][k + 1] =
                    min(dp[i + 1][tj + 1][k + 1], dp[i][j][k]);
            } else {
              dp[i + 1][tj + 1][k] = min(dp[i + 1][tj + 1][k], dp[i][j][k]);
            }
            char v = '0';
            if (a[i] == '0') v = '1';
            tj = (j == m ? nxt[m - 1] : j - 1);
            while (tj != -1 && b[tj + 1] != v) tj = nxt[tj];
            if (b[tj + 1] == v) ++tj;
            if (tj + 1 == m) {
              if (k != n - m + 1)
                dp[i + 1][tj + 1][k + 1] =
                    min(dp[i + 1][tj + 1][k + 1], dp[i][j][k] + 1);
            } else {
              dp[i + 1][tj + 1][k] = min(dp[i + 1][tj + 1][k], dp[i][j][k] + 1);
            }
          }
        }
      }
      for (int k = 0; k <= n - m + 1; k++) {
        int ans = inf;
        for (int j = 0; j <= m; j++) {
          ans = min(ans, dp[n][j][k]);
        }
        cout << (ans == inf ? -1 : ans) << " ";
      }
      cout << endl;
    }
    void solve() {
      int t = 1;
      for (int i = 0; i < t; ++i) {
        solveOne();
      }
    }
  };
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    Solution solver;
    solver.solve();
    return 0;
  }
  