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>
  #pragma warning(disable : 4996)
  #pragma comment(linker, "/STACK:16777216")
  using namespace std;
  const int INF = 1000000000 + 1e8;
  const long long LINF = 2000000000000000000;
  const int N = 503;
  int dp[2][N][N];
  vector<int> pf(string s) {
    int n = int(s.size());
    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;
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    string s, t;
    cin >> s >> t;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < N; j++) dp[0][i][j] = dp[1][i][j] = INF;
    dp[0][0][0] = 0;
    auto pi = pf(t);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= n - m + 1; k++) dp[1][j][k] = INF;
      for (int j = 0; j <= m; j++) {
        for (char c : {'0', '1'}) {
          int nj = j;
          if (nj == m) nj = pi[m - 1];
          while (nj > 0 && t[nj] != c) nj = pi[nj - 1];
          if (t[nj] == c) nj++;
          int cost = (c != s[i]);
          int add = (nj == m);
          for (int k = 0; k <= n - m + 1; k++) {
            dp[1][nj][k + add] = min(dp[1][nj][k + add], dp[0][j][k] + cost);
          }
        }
      }
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= n - m + 1; k++) {
          dp[0][j][k] = dp[1][j][k];
        }
    }
    for (int k = 0; k <= n - m + 1; k++) {
      int ans = INF;
      for (int j = 0; j <= m; j++) ans = min(ans, dp[1][j][k]);
      if (ans >= INF)
        cout << -1 << ' ';
      else
        cout << ans << ' ';
    }
  }
  int main() {
    srand(time(0));
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tst = 1;
    while (tst--) {
      solve();
    }
  }
  