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;
  const int N = 510;
  int dp[2][N][N];
  char s[N], str[N], ss[N];
  int nxt[N][2];
  int fail_all;
  int get_lcp(int cnt) {
    for (int l = cnt - 1; l >= 1; l--) {
      bool flag = true;
      for (int j = 0; j < l; j++) {
        if (str[l - j] != ss[cnt - j]) {
          flag = false;
          break;
        }
      }
      if (flag) return l;
    }
    return 0;
  }
  void get_next(int n) {
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < 2; j++)
        if (str[i + 1] - '0' == j)
          nxt[i][j] = i + 1;
        else {
          int cnt = 0;
          for (int k = 1; k <= i; k++) {
            ss[++cnt] = str[k];
          }
          ss[++cnt] = '0' + j;
          nxt[i][j] = get_lcp(cnt);
        }
    }
    for (int i = 1; i <= n; i++) ss[i] = str[i];
    fail_all = get_lcp(n);
  }
  void update(int &x, int y) {
    if (x == -1 || x > y) x = y;
  }
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    scanf("%s", s + 1);
    scanf("%s", str + 1);
    get_next(m);
    memset(dp, -1, sizeof(dp));
    dp[0][0][0] = 0;
    for (int i = 0; i < n; i++) {
      int flag = i & 1;
      memset(dp[flag ^ 1], -1, sizeof(dp[flag ^ 1]));
      for (int j = 0; j < m; j++) {
        for (int k = 0; k <= n - m + 1; k++) {
          if (dp[flag][j][k] == -1) continue;
          for (int r = 0; r < 2; r++) {
            int l = nxt[j][r];
            int carry = 0;
            if (l == m) {
              l = fail_all;
              carry++;
            }
            update(dp[flag ^ 1][l][k + carry],
                   dp[flag][j][k] + (s[i + 1] - '0' != r));
          }
        }
      }
    }
    for (int i = 0; i <= n - m + 1; i++) {
      int ans = -1;
      for (int j = 0; j < m; j++)
        if (dp[n & 1][j][i] != -1) {
          update(ans, dp[n & 1][j][i]);
        }
      printf("%d ", ans);
    }
    printf("\n");
    return 0;
  }
  