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;
  void solve() {
    int n, k;
    int mod = 998244353;
    cin >> n >> k;
    string S;
    cin >> S;
    vector<vector<int>> c(n + 1, vector<int>(n + 1, 1));
    for (int i = 2; i <= n; ++i) {
      for (int j = 1; j < i; ++j) {
        c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % mod;
      }
    }
    int f = 0;
    for (auto s : S) {
      if (s == '1') {
        ++f;
      }
    }
    if (f < k) {
      cout << 1 << endl;
      return;
    }
    int r = 1;
    for (int i = 0; i < n; ++i) {
      if (S[i] == '1') {
        int j = i + 1, z = 0;
        while (j < n) {
          z += (S[j] == '1');
          if (z >= k) {
            --z;
            break;
          }
          ++j;
        }
        int x = j - i - 1;
        if (x == z || x == 0) {
          continue;
        }
        r += c[x][z + 1];
        r %= mod;
      } else {
        int j = i + 1, z = 0;
        while (j < n) {
          z += (S[j] == '1');
          if (z > k) {
            --z;
            break;
          }
          ++j;
        }
        int x = j - i - 1;
        if (z == 0) {
          continue;
        }
        r += c[x][z - 1];
        r %= mod;
      }
    }
    cout << r << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T;
    solve();
    return 0;
  }
  