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;
  int N, M;
  string A, B;
  int to[500], ed;
  int dp[2][505][501];
  int main() {
    cin >> N >> M >> A >> B;
    for (int i = 0; i < M; i++) {
      string now = B.substr(0, i);
      now += B[i] == '0' ? '1' : '0';
      for (int k = 0; k <= now.size(); k++) {
        if (now.substr(k) == B.substr(0, now.size() - k)) {
          to[i] = now.size() - k;
          break;
        }
      }
    }
    for (int k = 1; k <= M; k++)
      if (B.substr(k) == B.substr(0, M - k)) {
        ed = M - k;
        break;
      }
    int now = 0;
    int mx = N - M + 1;
    for (int i = 0; i <= mx; i++)
      for (int j = 0; j < M; j++) dp[now][i][j] = 1e9;
    dp[now][0][0] = 0;
    for (int i = 0; i < N; i++) {
      int nxt = 1 - now;
      for (int j = 0; j <= mx; j++)
        for (int k = 0; k < M; k++) dp[nxt][j][k] = 1e9;
      for (int j = 0; j <= mx; j++)
        for (int k = 0; k < M; k++) {
          if (A[i] == B[k]) {
            if (k + 1 == M)
              dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k]);
            else
              dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k]);
            dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k] + 1);
          } else {
            if (k + 1 == M)
              dp[nxt][j + 1][ed] = min(dp[nxt][j + 1][ed], dp[now][j][k] + 1);
            else
              dp[nxt][j][k + 1] = min(dp[nxt][j][k + 1], dp[now][j][k] + 1);
            dp[nxt][j][to[k]] = min(dp[nxt][j][to[k]], dp[now][j][k]);
          }
        }
      now = nxt;
    }
    for (int k = 0; k <= mx; k++) {
      int ans = 1e9;
      for (int j = 0; j < M; j++) ans = min(ans, dp[now][k][j]);
      if (ans == (int)1e9) ans = -1;
      cout << ans << (k == mx ? "\n" : " ");
    }
  }
  