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 long long mod = 998244353;
  long long n, k;
  long long fac[5010] = {1, 1};
  long long inv[5010] = {1, 1};
  char s[5010];
  vector<long long> v;
  long long qpow(long long a, long long b, long long res = 1) {
    for (; b; b >>= 1, (a *= a) %= mod)
      if (b & 1) (res *= a) %= mod;
    return res;
  }
  void init(void) {
    for (long long i = 2; i <= 5000; i++) fac[i] = fac[i - 1] * i % mod;
    for (long long i = 2; i <= 5000; i++) inv[i] = qpow(fac[i], mod - 2);
    return;
  }
  long long C(long long a, long long b) {
    return fac[a] * inv[b] % mod * inv[a - b] % mod;
  }
  void solve(long long res = 0) {
    scanf("%lld %lld %s", &n, &k, s + 1);
    if (!k) {
      puts("1");
      return;
    }
    s[0] = s[n + 1] = '1';
    v.clear();
    for (long long i = 0; i <= n + 1; i++)
      if (s[i] == '1') v.push_back(i);
    if (v.size() - 2 < k) {
      puts("1");
      return;
    }
    res = C(v[k + 1] - 1, k);
    for (long long i = k + 1; i < (long long)v.size() - 1; i++) {
      long long len = v[i + 1] - v[i - k] - 1;
      long long cov = v[i] - v[i - k] - 1;
      (res += C(len, k)) %= mod;
      (res -= C(cov, k - 1)) %= mod;
      (res += mod) %= mod;
    }
    printf("%lld\n", res);
    return;
  }
  signed main() {
    init();
    solve();
    return 0;
  }
  