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;
  int n, k;
  char s[5005];
  long long C[5005][5005];
  int main() {
    scanf("%d%d", &n, &k);
    C[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      C[i][0] = 1;
      for (int j = 1; j <= i; j++) {
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 998244353;
      }
    }
    scanf("%s", s + 1);
    int cur = 0, x, y;
    for (int i = 1; i <= n; i++)
      if (s[i] == '1') cur++;
    if (cur < k) {
      printf("1\n");
      return 0;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      cur = s[i] - '0';
      for (int j = i + 1; j <= n; j++) {
        cur += s[j] - '0';
        if (cur > k) break;
        x = cur;
        y = j - i + 1 - cur;
        if (s[i] == '0')
          x--;
        else
          y--;
        if (s[j] == '0')
          x--;
        else
          y--;
        if (x < 0 || y < 0) continue;
        ans += C[x + y][x];
        ans %= 998244353;
      }
    }
    printf("%lld\n", (ans + 1) % 998244353);
    return 0;
  }
  