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;
  short dp[505][505][505], nxt[505][2];
  string a, b;
  int main(int argc, char** argv) {
    int n, m;
    cin >> n >> m;
    cin >> a >> b;
    a = ' ' + a, b = ' ' + b;
    for (int i = 0; i <= m; i++) {
      string ling = b.substr(1, i) + '0';
      nxt[i][0] = 0;
      for (int j = min(m, (int)ling.size()); j >= 1; j--) {
        if (b.substr(1, j) == ling.substr(ling.size() - j)) {
          nxt[i][0] = j;
          break;
        }
      }
      string yi = b.substr(1, i) + '1';
      nxt[i][1] = 0;
      for (int j = min(m, (int)yi.size()); j >= 1; j--) {
        if (b.substr(1, j) == yi.substr(yi.size() - j)) {
          nxt[i][1] = j;
          break;
        }
      }
    }
    for (int i = 0; i <= n; i++)
      for (int j = 0; j <= n; j++)
        for (int k = 0; k <= m; k++) dp[i][j][k] = 20000;
    dp[0][0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j <= n; j++) {
        for (int k = 0; k <= m; k++) {
          dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]] =
              min(dp[i + 1][j + (nxt[k][0] == m)][nxt[k][0]],
                  (short)(dp[i][j][k] + (a[i + 1] != '0')));
          dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]] =
              min(dp[i + 1][j + (nxt[k][1] == m)][nxt[k][1]],
                  (short)(dp[i][j][k] + (a[i + 1] != '1')));
        }
      }
    }
    for (int i = 0; i <= n - m + 1; i++) {
      short ans = 20000;
      for (int k = 0; k <= m; k++) ans = min(ans, dp[n][i][k]);
      if (ans == 20000) ans = -1;
      cout << ans << " ";
    }
    return 0;
  }
  