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>
  const int N = 505;
  using namespace std;
  int n;
  int m;
  int a[N];
  int b[N];
  int p[N];
  int f[N][2];
  int d[N][N];
  int d2[N][N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      char c;
      cin >> c;
      a[i] = c - '0';
    }
    for (int i = 0; i < m; i++) {
      char c;
      cin >> c;
      b[i] = c - '0';
    }
    for (int i = 1; i < m; i++) {
      p[i] = p[i - 1];
      while (p[i] > 0 && b[p[i]] != b[i]) {
        p[i] = p[p[i] - 1];
      }
      if (b[p[i]] == b[i]) {
        p[i] += 1;
      }
    }
    for (int i = 0; i < m; i++) {
      for (int x = 0; x < 2; x++) {
        if (i > 0 && x != b[i]) {
          f[i][x] = f[p[i - 1]][x];
        } else {
          f[i][x] = i + (x == b[i]);
        }
      }
    }
    f[m][0] = f[p[m - 1]][0];
    f[m][1] = f[p[m - 1]][1];
    int inf = 1e9;
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < N; j++) {
        d[i][j] = d2[i][j] = inf;
      }
    }
    d[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= m; j++) {
        for (int h = 0; h <= n; h++) {
          for (int x = 0; x < 2; x++) {
            d2[f[j][x]][(f[j][x] == m) + h] =
                min(d2[f[j][x]][(f[j][x] == m) + h], d[j][h] + (x ^ a[i]));
          }
        }
      }
      for (int j = 0; j < N; j++) {
        for (int h = 0; h < N; h++) {
          d[j][h] = d2[j][h];
          d2[j][h] = inf;
        }
      }
    }
    for (int i = 0; i <= n - m + 1; i++) {
      int res = inf;
      for (int j = 0; j < m + 1; j++) {
        res = min(res, d[j][i]);
      }
      if (res == inf) {
        res = -1;
      }
      cout << res << " ";
    }
  }
  