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 long long inf = 2e9;
  int32_t main() {
    cin.tie(0)->sync_with_stdio(0);
    long long n, m;
    string s, t;
    cin >> n >> m >> s >> t;
    vector<long long> pi(m, 0), nxt(m, 0);
    for (long long i = 1; i < m; i++) {
      {
        t[i] = char('0' + (t[i] - '0') ^ 1);
        long long j = pi[i - 1];
        while (j > 0 and t[i] != t[j]) {
          j = pi[j - 1];
        }
        if (t[i] == t[j]) j++;
        nxt[i] = j;
        t[i] = char('0' + (t[i] - '0') ^ 1);
      }
      {
        long long j = pi[i - 1];
        while (j > 0 and t[i] != t[j]) {
          j = pi[j - 1];
        }
        if (t[i] == t[j]) j++;
        pi[i] = j;
      }
    }
    s = '#' + s;
    long long dp[2][m][n - m + 2];
    fill(**dp, **dp + sizeof dp / sizeof(long long), inf);
    dp[0][0][0] = 0;
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < m; j++) {
        for (long long k = 0; k < n - m + 2; k++) {
          dp[(i & 1) ^ 1][j][k] = inf;
        }
      }
      for (long long j = 0; j < m; j++) {
        for (long long k = 0; k < n - m + 2; k++) {
          if (dp[i & 1][j][k] == inf) continue;
          if (j == m - 1) {
            dp[(i & 1) ^ 1][pi.back()][k + 1] =
                min(dp[(i & 1) ^ 1][pi.back()][k + 1],
                    dp[i & 1][j][k] + (s[i + 1] != t[j]));
          } else {
            dp[(i & 1) ^ 1][j + 1][k] = min(dp[(i & 1) ^ 1][j + 1][k],
                                            dp[i & 1][j][k] + (s[i + 1] != t[j]));
          }
          dp[(i & 1) ^ 1][nxt[j]][k] = min(dp[(i & 1) ^ 1][nxt[j]][k],
                                           dp[i & 1][j][k] + (s[i + 1] == t[j]));
        }
      }
    }
    for (long long i = 0; i < n - m + 2; i++) {
      long long mn = inf;
      for (long long j = 0; j < m; j++) {
        mn = min(mn, dp[n & 1][j][i]);
      }
      if (mn == inf) mn = -1;
      cout << mn << ' ';
    }
  }
  