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 MAXN = 502;
  const int INF = 0x3F3F3F3F;
  int dp[2][MAXN][MAXN];
  int aut[MAXN][2], pi[MAXN];
  bool a[MAXN], b[MAXN];
  int n, m;
  char ch;
  void fast_io() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
  }
  int main() {
    fast_io();
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> ch;
      a[i] = (ch - '0');
    }
    for (int i = 0; i < m; i++) {
      cin >> ch;
      b[i] = (ch - '0');
    }
    for (int i = 1; i < m; i++) {
      int at = pi[i - 1];
      while (at and b[at] != b[i]) at = pi[at - 1];
      at += (b[at] == b[i]);
      pi[i] = at;
    }
    for (int i = 0; i <= m; i++) {
      for (int c = 0; c < 2; c++) {
        if ((i and b[i] != c) or (i == m))
          aut[i][c] = aut[pi[i - 1]][c];
        else
          aut[i][c] = i + (b[i] == c);
      }
    }
    memset(dp, INF, sizeof(dp));
    dp[0][0][0] = 0;
    for (int i = 1; i <= n; i++) {
      bool at = (i & 1);
      memset(dp[at], INF, sizeof(dp[at]));
      for (int j = 0; j <= (n - m + 1); j++) {
        for (int k = 0; k <= m; k++) {
          if (dp[!at][j][k] >= INF) continue;
          int to = aut[k][a[i]];
          dp[at][j + (to == m)][to] =
              min(dp[at][j + (to == m)][to], dp[!at][j][k]);
          to = aut[k][!a[i]];
          dp[at][j + (to == m)][to] =
              min(dp[at][j + (to == m)][to], 1 + dp[!at][j][k]);
        }
      }
    }
    for (int i = 0; i <= (n - m + 1); i++) {
      int opt = INF;
      for (int j = 0; j <= m; j++) opt = min(opt, dp[n & 1][i][j]);
      printf("%d ", (opt != INF ? opt : -1));
    }
    return 0;
  }
  