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;
  vector<int> prefix_function(string s) {
    int n = (int)s.length();
    vector<int> pi(n);
    for (int i = 1; i < n; i++) {
      int j = pi[i - 1];
      while (j > 0 && s[i] != s[j]) j = pi[j - 1];
      if (s[i] == s[j]) j++;
      pi[i] = j;
    }
    return pi;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    string s;
    cin >> s;
    string t;
    cin >> t;
    const int inf = 1e9;
    vector<vector<int>> dp(m, vector<int>(n - m + 2, inf));
    vector<int> p = prefix_function(t);
    vector<vector<int>> jump(m, vector<int>(2));
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < 2; j++) {
        int pos = i;
        while (pos > 0 && t[pos] - '0' != j) {
          pos = p[pos - 1];
        }
        if (t[pos] - '0' == j)
          jump[i][j] = pos + 1;
        else
          jump[i][j] = 0;
      }
    }
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      vector<vector<int>> ndp(m, vector<int>(n - m + 2, inf));
      swap(dp, ndp);
      for (int j = 0; j < m; j++) {
        for (int k = 0; k < n - m + 2; k++) {
          if (ndp[j][k] == inf) continue;
          auto modify = [&](int i, int j, int c, int change) {
            int jj = jump[j][c];
            int kk = k;
            if (jj == m) jj = p[jj - 1], kk++;
            dp[jj][kk] = min(dp[jj][kk], ndp[j][k] + change);
          };
          modify(i, j, s[i] - '0', 0);
          modify(i, j, (s[i] - '0') ^ 1, 1);
        }
      }
    }
    for (int i = 0; i < n - m + 2; i++) {
      int ans = inf;
      for (int j = 0; j < m; j++) {
        ans = min(ans, dp[j][i]);
      }
      if (ans == inf) ans = -1;
      cout << ans << " ";
    }
  }
  