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;
  int i, j, k, n, m, t, f[505][505], g[505][505], nxt[666], it, res[666], sb,
      nxt2[666][2];
  bool a[666], b[666];
  int main() {
    memset(f, 1, sizeof(f));
    memset(g, 1, sizeof(g));
    memset(res, 1, sizeof(res));
    f[0][0] = 0;
    scanf("%d%d", &n, &m);
    for (i = 1; i <= n; i++) scanf("%1d", &a[i]);
    for (i = 1; i <= m; i++) scanf("%1d", &b[i]);
    for (i = 2; i <= m; i++) {
      while (it && b[i] != b[it + 1]) {
        it = nxt[it];
      }
      if (b[i] == b[it + 1]) it++;
      nxt[i] = it;
    }
    for (i = 0; i <= m; i++) {
      j = i;
      while (j && b[j + 1] != 1) {
        j = nxt[j];
      }
      if (b[j + 1] == 1) j++;
      nxt2[i][1] = j;
      j = i;
      while (j && b[j + 1] != 0) {
        j = nxt[j];
      }
      if (b[j + 1] == 0) j++;
      nxt2[i][0] = j;
    }
    for (i = 1; i <= n; i++) {
      for (j = 0; j < n - m + 2; j++) {
        for (k = 0; k < m; k++) {
          if (a[i] == b[k + 1]) {
            if (k == m - 1) {
              g[j + 1][nxt[m]] = min(g[j + 1][nxt[m]], f[j][k]);
            } else {
              g[j][k + 1] = min(g[j][k + 1], f[j][k]);
            }
            sb = nxt2[k][a[i] ^ 1];
            g[j][sb] = min(g[j][sb], f[j][k] + 1);
          } else {
            if (k == m - 1) {
              g[j + 1][nxt[m]] = min(g[j + 1][nxt[m]], f[j][k] + 1);
            } else {
              g[j][k + 1] = min(g[j][k + 1], f[j][k] + 1);
            }
            sb = nxt2[k][a[i]];
            g[j][sb] = min(g[j][sb], f[j][k]);
          }
        }
      }
      memcpy(f, g, sizeof(g));
      memset(g, 1, sizeof(g));
    }
    for (i = 0; i < n - m + 2; i++) {
      for (j = 0; j < m; j++) {
        if (f[i][j] < 114514) {
          res[i] = min(res[i], f[i][j]);
        }
      }
    }
    for (i = 0; i < n - m + 2; i++) {
      printf("%d ", (res[i] >= 114514) ? -1 : res[i]);
    }
  }
  