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;
  const int N = 5000 + 3;
  vector<vector<long long>> C(N, vector<long long>(N, 0));
  inline void init() {
    for (int n = 0; n < N; ++n) {
      C[n][0] = 1;
    }
    for (int n = 1; n < N; ++n) {
      for (int k = 1; k <= n; ++k) {
        C[n][k] = (C[n - 1][k] + C[n - 1][k - 1]) % MOD;
      }
    }
  }
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    init();
    int n, k;
    string s;
    while (cin >> n >> k >> s) {
      vector<int> pref(n + 1, 0);
      for (int i = 0; i < n; ++i) {
        s[i] -= '0';
        pref[i + 1] = pref[i] + s[i];
      }
      long long ans = 1;
      if (pref.back() >= k) {
        for (int i = 0; i < n; ++i) {
          for (int j = i + 1; j < n; ++j) {
            vector<int> cnt = {0, 0};
            cnt[1] = pref[j + 1] - pref[i];
            if (cnt[1] > k) {
              continue;
            }
            cnt[0] = j - i + 1 - cnt[1];
            if (cnt[s[i] ^ 1]-- == 0) {
              continue;
            }
            if (cnt[s[j] ^ 1]-- == 0) {
              continue;
            }
            int d = C[j - i - 1][cnt[0]];
            ans = (ans + d) % MOD;
          }
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  