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 ll = long long;
  using pi = pair<ll, ll>;
  const ll inf = 0x3f3f3f3f3f3f3f3f;
  const ll minf = 0xc0c0c0c0c0c0c0c0;
  const ll mod = 998244353;
  const ll MAXN = 5050;
  ll dp[5050];
  ll mx[5050];
  ll fact[MAXN] = {1, 1}, inv[MAXN] = {1, 1}, fact_inv[MAXN] = {1, 1};
  void init() {
    for (ll i = 2; i < MAXN; i++) {
      fact[i] = ll(fact[i - 1]) * i % mod;
    }
    for (ll i = 2; i < MAXN; i++) {
      inv[i] = ll(inv[mod % i]) * (mod - mod / i) % mod;
    }
    for (ll i = 2; i < MAXN; i++) {
      fact_inv[i] = ll(fact_inv[i - 1]) * inv[i] % mod;
    }
  }
  ll nCr(ll n, ll r) {
    return n >= r && r >= 0
               ? ll(fact[n]) * fact_inv[n - r] % mod * fact_inv[r] % mod
               : 0;
  }
  ll nPr(ll n, ll r) {
    return n >= r && r >= 0 ? ll(fact[n]) * fact_inv[n - r] % mod : 0;
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout << fixed << setprecision(20);
    init();
    ll n, k;
    cin >> n >> k;
    string s;
    cin >> s;
    if (k == 0) {
      cout << 1 << '\n';
      return 0;
    }
    memset(mx, -1, sizeof(mx));
    for (ll l = 0, r = -1, c = 0;;) {
      while (r < n - 1 && c + (s[r + 1] == '1') <= k) {
        ++r;
        c += s[r] == '1';
      }
      if (c < k) {
        break;
      }
      mx[l] = r;
      c -= s[l] == '1';
      ++l;
    }
    for (ll i = n - 1; i >= 0; i--) {
      if (mx[i] == -1) {
        dp[i] = 1;
      } else if (mx[i] == n - 1) {
        dp[i] = nCr(n - i, k);
      } else {
        dp[i] = dp[i + 1];
        ll R = mx[i];
        if (s[i] == '0') {
          dp[i] += nCr(R - i, k - 1);
          if (dp[i] >= mod) {
            dp[i] -= mod;
          }
        } else {
          dp[i] += nCr(R - i, k);
          if (dp[i] >= mod) {
            dp[i] -= mod;
          }
        }
      }
    }
    cout << dp[0] << '\n';
    return 0;
  }
  