rt

#include <bits/stdc++.h>
#define maxn 1000010
using namespace std;
double sum[maxn];
double a[maxn];
deque<double> q;
int n;
int s, t;
double all;

//参考题解
bool can(double mid) {
	for (int i = 1; i <= n; i++) {
		sum[i] = sum[i - 1] + a[i] - mid;
	}
	int t1 = 0;
	cout << "ok1" << endl;
	for (int i = s; i <= n; i++) {
		while (q.back() > sum[i - s] && !q.empty()) { //数组不空的情况下看队尾元素是否大于sum，是则出队
			q.pop_back();
		}
		q.push_back(i - s);a
		while (!q.empty() && q.front() < i - t) {
			q.pop_front();
		}
		if (sum[i] - q.front() >= 0 && !q.empty()) {
			cout << "ok2" << endl;
			return true;
		}
	}
	cout << "ok2" << endl;
	return false;
}


int main() {
	cin >> n;
	cin >> s >> t;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		sum[i] = sum[i - 1] + a[i];
		all += a[i];
	}
	all /= n; //把平均值求出来
	double l = 0, r = all;
	while (r - l >= 0.001) {
		double mid = (l + r) / 2;
		if (can(mid)) {
			l = mid;
			all = mid;
		} else {
			r = mid + 1;
		}
	}
	cout << all;
	return 0;
}

属于连题解都看不懂的fw了