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;
  using ll = long long;
  using pii = pair<int, int>;
  constexpr int INF = 0x3f3f3f3f;
  void SolveTask() {
    int n, m;
    string a, b;
    cin >> n >> m >> a >> b;
    vector<vector<int>> pf(m + 1, vector<int>(2, 0));
    for (int i = 0; i <= m; ++i) {
      for (int r = min(i, m - 1); r >= 0; --r) {
        bool ok = true;
        for (int j = r - 1, h = i - 1; j >= 0; --j, --h) {
          if (b[j] != b[h]) {
            ok = false;
            break;
          }
        }
        if (ok) {
          pf[i][b[r] - '0'] = max(pf[i][b[r] - '0'], r + 1);
        }
      }
    }
    const auto SetMin = [](int& x, int y) { x = min(x, y); };
    vector<vector<vector<int>>> dp(
        n + 1, vector<vector<int>>(n + 2, vector<int>(m + 1, INF)));
    dp[0][0][0] = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j <= max(0, i - m + 2); ++j) {
        for (int k = 0; k <= min(i, m); ++k) {
          if (dp[i][j][k] == INF) {
            continue;
          }
          const int c = a[i] - '0';
          SetMin(dp[i + 1][j + (pf[k][c] == m)][pf[k][c]], dp[i][j][k]);
          SetMin(dp[i + 1][j + (pf[k][c ^ 1] == m)][pf[k][c ^ 1]],
                 dp[i][j][k] + 1);
        }
      }
    }
    vector<int> ans(n - m + 2, INF);
    for (int p = 0; p <= m; ++p) {
      for (int j = 0; j <= n - m + 2; ++j) {
        if (dp[n][j][p] != INF) {
          SetMin(ans[j], dp[n][j][p]);
        }
      }
    }
    for (int x : ans) {
      x = x == INF ? -1 : x;
      cout << x << ' ';
    }
    cout << "\n";
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int test_cases = 1;
    for (int yt = 1; yt <= test_cases; ++yt) {
      SolveTask();
    }
    return 0;
  }
  