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;
  short dp[503][503], tp[503][503];
  int N, M, trs[503][2], nxt[503];
  char s1[503], s2[503];
  int main() {
    scanf("%d %d %s %s", &N, &M, s1 + 1, s2 + 1);
    nxt[0] = -1;
    trs[0][s2[1] - '0'] = 1;
    for (int i = 1; i <= M; ++i) {
      int p = nxt[i - 1];
      while (p != -1 && s2[p + 1] != s2[i]) p = nxt[p];
      nxt[i] = p + 1;
      if (s2[i + 1] == '0')
        trs[i][0] = i + 1;
      else
        trs[i][0] = trs[nxt[i]][0];
      if (s2[i + 1] == '1')
        trs[i][1] = i + 1;
      else
        trs[i][1] = trs[nxt[i]][1];
    }
    memset(dp, 0x3f, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i <= N; ++i) {
      memset(tp, 0x3f, sizeof(tp));
      for (int j = 0; j <= M; ++j)
        for (int k = 0; k <= max(i - M, 0); ++k)
          if (dp[j][k] != 0x3f3f)
            for (int t = 0; t < 2; ++t) {
              int r = trs[j][t], nk = k + (r == M);
              tp[r][nk] = min(tp[r][nk], (short)(dp[j][k] + (s1[i] - '0' != t)));
            }
      memcpy(dp, tp, sizeof(dp));
    }
    for (int i = 0; i <= N - M + 1; ++i) {
      short mn = 0x3f3f;
      for (int j = 0; j <= M; ++j) mn = min(mn, dp[j][i]);
      cout << (mn == 0x3f3f ? -1 : mn) << ' ';
    }
  }
  