和大家的思路一样，二分平均值，check 函数利用单调队列 $\def\timeO#1{\text{O(#1)}} \timeO{n}$ 解决，总复杂度为 $\def\timeO#1{\text{O(#1)}} \timeO{n log n}$.

代码：

using namespace std;
const int N = 1e7+5;
int n, m;
int s[N],a[N],S,T;
deque<int> q,emp;
bool check(double x) {
	q = emp;
    q.push_back(0)	;
    for(int i = 1; i <= n; i++) {
        while(q.front() + T < i) { q.pop_front();}
		if(i - q.front() >= S && i - q.front() <= T) {
			if((s[i]-s[q.front()])*1.0/(i-q.front()) >= x) return 1;
		}
        while(!q.empty() && s[q.back()] >= s[i]) { q.pop_back(); }
        q.push_back(i);
    }
    return 0;
}
int main() {
    scanf("%d%d%d", &n, &S, &T);
    for(int i = 1; i <= n; i++) {
        scanf("%d",a + i);
        s[i] = s[i - 1] + a[i];
    }
    double l=-1e5,r=1e5;
    while(l+0.0001<r) {
		double mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid-1;
    }
    printf("%.3lf",l);
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
const int N = 1e7+5;
int n, m;
int s[N],a[N],S,T;
deque<int> q,emp;
bool check(double x) {
	q = emp;
    q.push_back(0)	;
    for(int i = 1; i <= n; i++) {
        while(q.front() + T < i) { q.pop_front();}
		if(i - q.front() >= S && i - q.front() <= T) {
			if((s[i]-s[q.front()])*1.0/(i-q.front()) >= x) return 1;
		}
        while(!q.empty() && s[q.back()] >= s[i]) { q.pop_back(); }
        q.push_back(i);
    }
    return 0;
}
int main() {
    scanf("%d%d%d", &n, &S, &T);
    for(int i = 1; i <= n; i++) {
        scanf("%d",a + i);
        s[i] = s[i - 1] + a[i];
    }
    double l=-1e5,r=1e5;
    while(l+0.0001<r) {
		double mid=(l+r)/2;
		if(check(mid)) l=mid;
		else r=mid-1;
    }
    printf("%.3lf",l);
    return 0;
}