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;
  mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
  const int N = 505;
  const int INF = 0x3f3f3f3f;
  void upd(int &a, int b) { a = min(a, b); }
  int go[N][2];
  int dp[N][N];
  int ndp[N][N];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    string a, b;
    cin >> n >> m >> a >> b;
    for (int i = 0; i <= m; ++i) {
      for (int x = 0; x < 2; ++x) {
        string s = b.substr(0, i) + string({(char)('0' + x)});
        int mx = 0;
        for (int j = 0; j <= i + 1; ++j) {
          if (s.substr(i + 1 - j, j) == b.substr(0, j)) {
            mx = max(mx, j);
          }
        }
        go[i][x] = mx;
      }
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 0; i < n; ++i) {
      memset(ndp, 0x3f, sizeof(ndp));
      for (int cnt = 0; cnt <= max(0, i - m + 1); ++cnt) {
        for (int v = 0; v <= m; ++v) {
          for (int x = 0; x < 2; ++x) {
            int nv = go[v][x];
            int ncnt = cnt + (nv == m);
            upd(ndp[ncnt][nv], dp[cnt][v] + (x != a[i] - '0'));
          }
        }
      }
      memcpy(dp, ndp, sizeof(dp));
    }
    for (int i = 0; i <= n - m + 1; ++i) {
      int mn = INF;
      for (int v = 0; v <= m; ++v) mn = min(mn, dp[i][v]);
      cout << (mn == INF ? -1 : mn) << " ";
    }
    cout << "\n";
    return 0;
  }
  