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 INF = 1000;
  int main() {
    int n, m;
    cin >> n >> m;
    string a;
    cin >> a;
    string b;
    cin >> b;
    vector<vector<int>> nxt(m + 1, vector<int>(2));
    for (int i = 0; i <= m; i++) {
      string p = b.substr(0, i);
      for (int j = 0; j < 2; j++) {
        string p2 = p + (char)('0' + j);
        int cnt = p2.size();
        for (int k = m; k >= 0; k--) {
          if (k <= cnt) {
            if (p2.substr(cnt - k) == b.substr(0, k)) {
              nxt[i][j] = k;
              break;
            }
          }
        }
      }
    }
    vector<vector<int>> dp(m + 1, vector<int>(n - m + 2, INF));
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      vector<vector<int>> dp2(m + 1, vector<int>(n - m + 2, INF));
      for (int j = 0; j <= m; j++) {
        for (int k = 0; k <= n - m; k++) {
          if (dp[j][k] != INF) {
            if (a[i] == '0') {
              if (nxt[j][0] == m) {
                dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k]);
              } else {
                dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k]);
              }
              if (nxt[j][1] == m) {
                dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k] + 1);
              } else {
                dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k] + 1);
              }
            }
            if (a[i] == '1') {
              if (nxt[j][0] == m) {
                dp2[nxt[j][0]][k + 1] = min(dp2[nxt[j][0]][k + 1], dp[j][k] + 1);
              } else {
                dp2[nxt[j][0]][k] = min(dp2[nxt[j][0]][k], dp[j][k] + 1);
              }
              if (nxt[j][1] == m) {
                dp2[nxt[j][1]][k + 1] = min(dp2[nxt[j][1]][k + 1], dp[j][k]);
              } else {
                dp2[nxt[j][1]][k] = min(dp2[nxt[j][1]][k], dp[j][k]);
              }
            }
          }
        }
      }
      swap(dp, dp2);
    }
    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) {
        cout << -1;
      } else {
        cout << ans;
      }
      if (i < n - m + 1) {
        cout << ' ';
      }
    }
    cout << endl;
  }
  