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 n, m;
  int a[510], b[510], c[510];
  int dp[2][510][510];
  int nxt[510][2], nxtm;
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; ++i) {
      scanf("%1d", &a[i]);
    }
    for (int i = 1; i <= m; ++i) {
      scanf("%1d", &b[i]);
    }
    if (b[1] == 0)
      nxt[0][0] = 1;
    else
      nxt[0][1] = 1;
    for (int j = 2; j <= m; ++j) {
      bool flag = 0;
      for (int k = j; k <= m; ++k) {
        if (b[k] != b[k - j + 1]) {
          flag = 1;
        }
      }
      if (!flag) {
        nxtm = m - j + 1;
        break;
      }
    }
    for (int i = 1; i < m; ++i) {
      c[i] = b[i];
      if (b[i + 1] == 0) {
        nxt[i][0] = i + 1;
        c[i + 1] = 1;
        for (int j = 1; j <= i + 1; ++j) {
          bool flag = 0;
          for (int k = j; k <= i + 1; ++k) {
            if (c[k] != b[k - j + 1]) {
              flag = 1;
            }
          }
          if (!flag) {
            nxt[i][1] = i + 2 - j;
            break;
          }
        }
      } else {
        nxt[i][1] = i + 1;
        c[i + 1] = 0;
        for (int j = 1; j <= i + 1; ++j) {
          bool flag = 0;
          for (int k = j; k <= i + 1; ++k) {
            if (c[k] != b[k - j + 1]) {
              flag = 1;
            }
          }
          if (!flag) {
            nxt[i][0] = i + 2 - j;
            break;
          }
        }
      }
    }
    int now = 0, per = 1;
    for (int j = 0; j <= n; ++j) {
      for (int k = 0; k < m; ++k) {
        dp[now][j][k] = dp[per][j][k] = 0x3f3f3f3f;
      }
    }
    dp[per][0][0] = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) {
        for (int k = 0; k < m; ++k) {
          dp[now][j][k] = 0x3f3f3f3f;
        }
      }
      for (int j = 0; j <= n; ++j) {
        for (int k = 0; k < m; ++k) {
          if (dp[per][j][k] != 0x3f3f3f3f) {
            int nxtk = nxt[k][a[i]];
            int nxtj = j;
            if (nxtk == m) {
              nxtj++;
              nxtk = nxtm;
            }
            dp[now][nxtj][nxtk] = min(dp[now][nxtj][nxtk], dp[per][j][k]);
            nxtk = nxt[k][!a[i]];
            nxtj = j;
            if (nxtk == m) {
              nxtj++;
              nxtk = nxtm;
            }
            dp[now][nxtj][nxtk] = min(dp[now][nxtj][nxtk], dp[per][j][k] + 1);
          }
        }
      }
      per++;
      now++;
      per %= 2;
      now %= 2;
    }
    for (int i = 0; i <= n - m + 1; ++i) {
      int minn = 0x3f3f3f3f;
      for (int j = 0; j < m; ++j) {
        minn = min(minn, dp[per][i][j]);
      }
      if (minn == 0x3f3f3f3f) {
        printf("-1 ");
      } else
        printf("%d ", minn);
    }
    return 0;
  }
  