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;
  using i64 = long long;
  const int N = 5050, MOD = 998244353;
  i64 inv[N], fac[N], inv_fac[N];
  void precalc() {
    inv[1] = 1;
    for (int n = 2; n < N; n++) {
      inv[n] = (MOD - (MOD / n) * inv[MOD % n] % MOD) % MOD;
    }
    fac[0] = 1;
    for (int n = 1; n < N; n++) {
      fac[n] = n * fac[n - 1] % MOD;
    }
    inv_fac[0] = 1;
    for (int n = 1; n < N; n++) {
      inv_fac[n] = inv[n] * inv_fac[n - 1] % MOD;
    }
  }
  i64 binom(i64 n, i64 k) {
    if (k < 0 || k > n) return 0;
    return fac[n] * inv_fac[k] % MOD * inv_fac[n - k] % MOD;
  }
  int main() {
    int n, k;
    cin >> n >> k;
    precalc();
    string a;
    cin >> a;
    i64 res = 1;
    for (int i = 0; i < n; i++) {
      int o = 0, j = i;
      for (; j < n; j++) {
        if (o == k && a[j] == '1') break;
        o += a[j] == '1';
      }
      if (i == 0 && o < k) break;
      int z = j - i - o;
      res = (res + binom(j - i - 1, z - (a[i] == '1'))) % MOD;
      cerr << j - i - 1 << " " << z - (a[i] == '1') << endl;
    }
    cout << res << '\n';
  }
  