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 int maxn = 505;
  inline void down(int &a, const int &b) {
    if (a == -1 || a > b) a = b;
  }
  int n, m;
  char A[maxn], B[maxn];
  int f[2][maxn][maxn];
  int ans[maxn];
  struct AC {
    int root, tot;
    int son[maxn][2], fail[maxn], ed[maxn];
    int newnode() {
      ++tot;
      memset(son[tot], 0, sizeof son[tot]);
      fail[tot] = 0;
      ed[tot] = 0;
      return tot;
    }
    void init() {
      tot = 0;
      root = newnode();
    }
    queue<int> q;
    void build(char str[], int len) {
      int x = root;
      for (int i = 0; i < len; i++) {
        int c = str[i] - '0';
        if (!son[x][c]) son[x][c] = newnode();
        x = son[x][c];
        if (i == len - 1) ed[x] = 1;
      }
      q.push(root);
      while (!q.empty()) {
        x = q.front();
        q.pop();
        int fl = fail[x];
        if (ed[fl]) ed[x] = 1;
        for (int i = 0; i < 2; i++) {
          int &y = son[x][i];
          if (y) {
            if (fl)
              fail[y] = son[fl][i];
            else
              fail[y] = root;
            q.push(y);
          } else {
            if (fl)
              y = son[fl][i];
            else
              y = root;
          }
        }
      }
    }
    void dp(char str[], int len, int uj) {
      memset(f, -1, sizeof f);
      int now = 0;
      f[now][root][0] = 0;
      for (int i = 0; i < len; i++) {
        now = !now;
        int cc = str[i] - '0';
        for (int x = 1; x <= tot; x++)
          for (int j = 0; j <= uj; j++)
            if (f[!now][x][j] != -1) {
              int &temp = f[!now][x][j];
              int y = son[x][cc];
              down(f[now][y][j + ed[y]], temp + 0);
              y = son[x][!cc];
              down(f[now][y][j + ed[y]], temp + 1);
              temp = -1;
            }
      }
      for (int j = 0; j <= uj; j++) ans[j] = -1;
      for (int x = 1; x <= tot; x++)
        for (int j = 0; j <= uj; j++)
          if (f[now][x][j] != -1) {
            down(ans[j], f[now][x][j]);
          }
    }
  } ac;
  int main() {
    ios_base::sync_with_stdio(false);
    cin >> n >> m;
    cin >> (A) >> (B);
    ac.init();
    ac.build(B, m);
    ac.dp(A, n, n - m + 1);
    for (int i = 0; i <= n - m + 1; i++) cout << ans[i] << " \n"[i == n - m + 1];
    return 0;
  }
  