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;
  string a, b;
  int n, m;
  int p[510];
  void kmppreprocess() {
    int i = 0, j = -1;
    p[0] = -1;
    while (i < m) {
      while (j >= 0 && b[i] != b[j]) j = p[j];
      i++, j++;
      p[i] = j;
    }
  }
  int kmpsearch(string s) {
    int n = s.size();
    int i = 0, j = 0, tot = 0;
    while (i < n) {
      while (j >= 0 && s[i] != b[j]) j = p[j];
      i++, j++;
      if (j == m) {
        tot++;
        j = p[j];
      }
    }
    return tot;
  }
  int dp[510][510], tmp[510][510];
  int go[510][2];
  int v[510];
  int main() {
    cin >> n >> m >> a >> b;
    kmppreprocess();
    for (int i = 0; i < n; i++) v[i] = a[i] - '0';
    for (int i = 0; i < m; i++) {
      int j = i;
      if (b[i] == '0') {
        go[i][0] = i + 1;
        while (j >= 0 && b[j] != '1') j = p[j];
        go[i][1] = j + 1;
      } else {
        go[i][1] = i + 1;
        while (j >= 0 && b[j] != '0') j = p[j];
        go[i][0] = j + 1;
      }
    }
    int j = p[m];
    if (b[j] == '0') {
      go[m][0] = j + 1;
      while (j >= 0 && b[j] != '1') j = p[j];
      go[m][1] = j + 1;
    } else {
      go[m][1] = j + 1;
      while (j >= 0 && b[j] != '0') j = p[j];
      go[m][0] = j + 1;
    }
    for (int i = 0; i < 510; i++)
      for (int j = 0; j < 510; j++) dp[i][j] = 1000000000;
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      for (int i = 0; i < 510; i++)
        for (int j = 0; j < 510; j++) tmp[i][j] = 1000000000;
      for (int j = 0; j <= m; j++) {
        int to[2];
        to[0] = go[j][v[i]];
        to[1] = go[j][1 - v[i]];
        for (int k = 0; k < 510 - 10; k++) {
          for (int pl = 0; pl < 2; pl++) {
            int lel = to[pl] == m;
            tmp[to[pl]][k + lel] = min(tmp[to[pl]][k + lel], dp[j][k] + pl);
          }
        }
      }
      for (int i = 0; i < 510; i++)
        for (int j = 0; j < 510; j++) dp[i][j] = tmp[i][j];
    }
    for (int i = 1; i <= m; i++)
      for (int j = 0; j < 510; j++) dp[i][j] = min(dp[i][j], dp[i - 1][j]);
    for (int i = 0; i <= n - m + 1; i++)
      cout << (dp[m][i] == 1000000000 ? -1 : dp[m][i]) << " ";
  }
  