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 = 510;
  int n, m, tot = 1, ch[maxn][2], fail[maxn];
  int f[maxn][maxn], g[maxn][maxn];
  char s[maxn], t[maxn];
  void chk(int &x, int y) {
    if (x > y) x = y;
  }
  int main() {
    scanf("%d %d %s %s", &n, &m, s + 1, t + 1);
    for (int i = 1; i <= m; i++) {
      ch[i][t[i] - '0'] = i + 1;
    }
    fill(ch[0], ch[0] + 2, 1), fail[0] = 1;
    queue<int> Q;
    Q.push(1);
    while (!Q.empty()) {
      int v = Q.front();
      Q.pop();
      for (int c : {0, 1}) {
        if (ch[v][c])
          fail[ch[v][c]] = ch[fail[v]][c], Q.push(ch[v][c]);
        else
          ch[v][c] = ch[fail[v]][c];
      }
    }
    memset(f, 0x3f, sizeof(f)), f[0][1] = 0;
    for (int i = 1; i <= n; i++) {
      memset(g, 0x3f, sizeof(g));
      for (int j = 0; j <= n - m + 1; j++) {
        for (int k = 1; k <= m + 1; k++)
          for (int c : {0, 1}) {
            chk(g[j + (ch[k][c] == m + 1)][ch[k][c]],
                f[j][k] + (s[i] != c + '0'));
          }
      }
      swap(f, g);
    }
    for (int i = 0; i <= n - m + 1; i++) {
      int ans = INT_MAX;
      for (int j = 1; j <= m + 1; j++) chk(ans, f[i][j]);
      printf("%d ", ans > 1e9 ? -1 : ans);
    }
    return 0;
  }
  