n^2的简单DP，用单调队列优化掉n即可。。。
#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 400005
#define eps 1e-10
#define mod 1000000007
#define INF 999999999
#define lowbit(x) (x&(-x))
#define mp mark_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
//typedef vector<int>::iterator IT;
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head
int q[maxn], sum[maxn], n, num[maxn], k;
void read(void)
{
sum[0] = 0;
for(int i = 1; i <= n; i++) scanf(num[i]);
for(int i = 1; i <= n; i++) sum[i] = sum[i-1] + num[i];
}
bool check(int a, int b, int c)
{
LL t1 = (LL)(sum[b] - sum[a]) * (c - b);
LL t2 = (LL)(sum[c] - sum[b]) * (b - a);
if(t1 > t2) return true;
else return false;
}
bool solve(int a, int b, int c)
{
LL t1 = (LL)(sum[c] - sum[b]) * (c - a);
LL t2 = (LL)(sum[c] - sum[a]) * (c - b);
if(t1 > t2) return true;
else return false;
}
void work(void)
{
int l = 0, r = 0;
double t = 0, mx = 0;
for(int i = 0; i + k <= n; i++) {
while(r - l >= 2 && i <= n - k && check(q[r-2], q[r-1], i)) r--;
q[r++] = i;
while(r - l >= 2 && solve(q[l], q[l+1], i + k)) l++;
t = 1.0 * (sum[i + k] - sum[q[l]]) / (i + k - q[l]);
mx = max(mx, t);
}
printf("%.2f\n", mx);
}
int main(void)
{
while(scanf("%d%d", &n, &k)!=EOF) {
read();
work();
}
return 0;
}
