description: You are given a binary string (i. e. a string consisting of characters 0 and/or 1) s of length n. You can perform the following operation with the string s at most once: choose a substring (a contiguous subsequence) of s having exactly k characters 1 in it, and shuffle it (reorder the characters in the substring as you wish).
  
  Calculate the number of different strings which can be obtained from s by performing this operation at most once.
  
  Input
  
  The first line contains two integers n and k (2 ≤ n ≤ 5000; 0 ≤ k ≤ n).
  
  The second line contains the string s of length n, consisting of characters 0 and/or 1.
  
  Output
  
  Print one integer — the number of different strings which can be obtained from s by performing the described operation at most once. Since the answer can be large, output it modulo 998244353.
  
  Examples
  
  Input
  
  
  7 2
  1100110
  
  
  Output
  
  
  16
  
  
  Input
  
  
  5 0
  10010
  
  
  Output
  
  
  1
  
  
  Input
  
  
  8 1
  10001000
  
  
  Output
  
  
  10
  
  
  Input
  
  
  10 8
  0010011000
  
  
  Output
  
  
  1
  
  Note
  
  Some strings you can obtain in the first example:
  
    * to obtain 0110110, you can take the substring from the 1-st character to the 4-th character, which is 1100, and reorder its characters to get 0110; 
    * to obtain 1111000, you can take the substring from the 3-rd character to the 7-th character, which is 00110, and reorder its characters to get 11000; 
    * to obtain 1100101, you can take the substring from the 5-th character to the 7-th character, which is 110, and reorder its characters to get 101. 
  
  
  
  In the second example, k = 0 so you can only choose the substrings consisting only of 0 characters. Reordering them doesn't change the string at all, so the only string you can obtain is 10010.
solution: #include <bits/stdc++.h>
  using namespace std;
  const int mod = 998244353;
  long long dp[5000 + 10], cnt[5000 + 10], c[5000 + 10][5000 + 10];
  int main() {
    ios_base::sync_with_stdio(false);
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    for (int i = 0; i <= 5000; i++) {
      for (int j = 0; j <= i; j++) {
        if (j == 0 | j == i) {
          c[i][j] = 1;
        } else {
          c[i][j] = c[i - 1][j - 1] % mod + c[i - 1][j] % mod;
          c[i][j] %= mod;
        }
      }
    }
    int ind = 0;
    int f = 0;
    cnt[0] = -1;
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        f++;
        cnt[f] = i;
      }
      if (f == k) {
        break;
      }
      ind++;
      dp[i] = 1;
    }
    dp[ind] = c[ind + 1][k];
    for (int i = ind + 1; i < n; i++) {
      dp[i] = dp[i - 1];
      if (s[i] == '1') {
        f++;
        cnt[f] = i;
        int x = cnt[f - k] + 1;
        dp[i] += c[i - x][k] % mod;
      } else {
        int x = cnt[f - k] + 1;
        dp[i] += c[i - x][k - 1] % mod;
      }
      dp[i] %= mod;
    }
    cout << dp[n - 1] % mod;
  }
  