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;
  char p[1000005], q[1000005];
  long long nxt[1000][2] = {0};
  long long dp[2][505][505];
  string x;
  int main() {
    long long t, n, m, i, j, k, l;
    cin >> n >> m;
    scanf("%s", p + 1);
    scanf("%s", q + 1);
    q[m + 1] = '2';
    nxt[0][q[1] - '0'] = 1;
    for (i = 1; i <= m; i++) {
      for (l = 0; l <= 1; l++) {
        for (j = 1; j <= i + 1; j++) {
          string z = "", y = "";
          for (k = j; k <= i; k++) z += q[k];
          z += l + '0';
          for (k = 1; k <= i - j + 2; k++) y += q[k];
          if (y == z) nxt[i][l] = max(nxt[i][l], i - j + 2);
        }
      }
    }
    for (i = 0; i <= 1; i++)
      for (j = 0; j <= 502; j++)
        for (k = 0; k <= 502; k++) dp[i][j][k] = 1e17;
    dp[0][0][0] = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j <= n; j++) {
        for (k = 0; k <= m; k++) {
          dp[!(i % 2)][j][k] = 1e17;
        }
      }
      for (j = 0; j <= n; j++) {
        for (k = 0; k <= m; k++) {
          if (dp[i % 2][j][k] != 1e17) {
            dp[!(i % 2)][j + (k == m)][nxt[k][0]] =
                min(dp[!(i % 2)][j + (k == m)][nxt[k][0]],
                    dp[i % 2][j][k] + (p[i + 1] == '0' ? 0 : 1));
            dp[!(i % 2)][j + (k == m)][nxt[k][1]] =
                min(dp[!(i % 2)][j + (k == m)][nxt[k][1]],
                    dp[i % 2][j][k] + (p[i + 1] == '1' ? 0 : 1));
          }
        }
      }
      for (j = 0; j <= n; j++) {
        for (k = 0; k <= m; k++) {
        }
      }
    }
    long long ans;
    for (i = 0; i <= n - m + 1; i++) {
      ans = 1e17;
      for (j = 0; j < m; j++) {
        ans = min(ans, dp[n % 2][i][j]);
      }
      if (i != 0) ans = min(ans, dp[n % 2][i - 1][m]);
      if (ans == 1e17) ans = -1;
      printf("%lld ", ans);
    }
    return 0;
  }
  