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;
  inline char nc() {
    static char buf[100000], *p1 = buf, *p2 = buf;
    return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
               ? EOF
               : *p1++;
  }
  inline int read() {
    register int x = 0, f = 1;
    register char ch = nc();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = nc();
    }
    while (ch >= '0' && ch <= '9') x = (x << 3) + (x << 1) + ch - '0', ch = nc();
    return x * f;
  }
  inline void write(register int x) {
    if (!x) putchar('0');
    if (x < 0) x = -x, putchar('-');
    static int sta[20];
    register int tot = 0;
    while (x) sta[tot++] = x % 10, x /= 10;
    while (tot) putchar(sta[--tot] + 48);
  }
  int n, m;
  int s[505], t[505], nxt[505], trans[505][2];
  int f[2][505][505], ans[505];
  void upd(int &x, int y) {
    if (x == -1) x = y;
    x = min(x, y);
  }
  int main() {
    n = read(), m = read();
    for (int i = 1; i <= n; ++i) {
      char ch = nc();
      while (ch != '0' && ch != '1') ch = nc();
      s[i] = ch - '0';
    }
    for (int i = 1; i <= m; ++i) {
      char ch = nc();
      while (ch != '0' && ch != '1') ch = nc();
      t[i] = ch - '0';
    }
    for (int i = 2, p = 0; i <= m; ++i) {
      while (p && t[p + 1] != t[i]) p = nxt[p];
      if (t[p + 1] == t[i]) ++p;
      nxt[i] = p;
    }
    for (int i = 0; i < m; ++i) {
      if (i < m - 1) trans[i][t[i + 1]] = i + 1;
      for (int j = 0; j <= 1; ++j)
        trans[i][j] = max(trans[i][j], trans[nxt[i]][j]);
    }
    int las = 0, cur = 1;
    memset(f[las], -1, sizeof(f[las]));
    f[las][0][0] = 0;
    for (int i = 1; i <= n; ++i) {
      memset(f[cur], -1, sizeof(f[cur]));
      for (int j = 0; j < m; ++j)
        for (int k = 0; k <= n - m + 1; ++k)
          if (f[las][j][k] != -1) {
            for (int l = 0; l <= 1; ++l) {
              if (j == m - 1 && l == t[m])
                upd(f[cur][trans[j][l]][k + 1], f[las][j][k] + (s[i] != l));
              else
                upd(f[cur][trans[j][l]][k], f[las][j][k] + (s[i] != l));
            }
          }
      cur ^= 1, las ^= 1;
    }
    memset(ans, -1, sizeof(ans));
    for (int i = 0; i < m; ++i)
      for (int j = 0; j <= n - m + 1; ++j)
        if (f[las][i][j] != -1) upd(ans[j], f[las][i][j]);
    for (int i = 0; i <= n - m + 1; ++i) printf("%d ", ans[i]);
    return 0;
  }
  