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;
  template <typename T1, typename T2>
  inline void chkmin(T1 &a, T2 b) {
    if (a > b) a = b;
  }
  template <typename T1, typename T2>
  inline void chkmax(T1 &a, T2 b) {
    if (a < b) a = b;
  }
  using ll = long long;
  using ld = long double;
  const string FILENAME = "input";
  const int MAXN = 508;
  int n, m;
  string a;
  string b;
  int next228[MAXN][2];
  short dp[MAXN][MAXN][MAXN];
  void solve() {
    cin >> n >> m;
    cin >> a >> b;
    string c;
    for (int i = 0; i <= m; i++) {
      for (int t = 0; t < 2; t++) {
        c.push_back('0' + t);
        for (int j = 0; j <= (int)(c).size(); j++) {
          bool bad = false;
          for (int k = j; k < (int)(c).size(); k++) {
            if (k - j > (int)(b).size() || b[k - j] != c[k]) {
              bad = true;
              break;
            }
          }
          if (!bad) {
            next228[i][t] = (int)(c).size() - j;
            break;
          }
        }
        c.pop_back();
      }
      c += b[i];
    }
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= m; j++) {
        for (int f = 0; f <= i; f++) {
          dp[i][j][f] = 10000;
        }
      }
    }
    dp[0][0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= m; j++) {
        for (int f = 0; f <= i; f++) {
          for (int k = 0; k < 2; k++) {
            chkmin(dp[i + 1][next228[j][k]][f + (next228[j][k] == m)],
                   dp[i][j][f] + (k != a[i] - '0'));
          }
        }
      }
    }
    for (int f = 0; f <= n - m + 1; f++) {
      int res = 1e9;
      for (int t = 0; t <= m; t++) {
        chkmin(res, dp[n][t][f]);
      }
      if (res > 5000) {
        cout << -1;
      } else {
        cout << res;
      }
      cout << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  