樣例輸入：

10 6
6 4 2 10 3 8 5 9 4 1

樣例輸出：

6500

將原問題轉化為： 判定 " 是否存在一個長度不小於L 的區間，區間的平均數不小於二分的值 "

精度問題，請注意！

AC程式碼：

//是否存在一個長度不小於L的區間,使得平均數不小於二分的值 
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define N 100007
#define inf 0x3f3f3f3f
using namespace std;

const double Eps = 1e-5 ; 

inline int wread(){
    char c=getchar ();int flag=1,wans=0;
    while (c<'0'||c>'9'){if (c=='-') flag=-1;c=getchar ();}
    while (c>='0'&&c<='9'){wans=wans*10+c-'0';c=getchar ();}
    return wans*=flag;
}

int n,a[N],m;
int l,r;
double b[N]={0};

bool check (int mid){
	double x=mid*1.0/1000;
	double cmp=inf;
	for (int i=1;i<=n;++i){
		if (i>=m)	cmp=min(cmp,b[i-m]);
		b[i]=b[i-1]-x+(double)a[i];
		double pr=b[i]-cmp;
		if (pr>= -Eps)	return true;
	}
	return false;	
}

int main (){
	n=wread();m=wread();
	l=inf;r=-inf;
	for (int i=1;i<=n;++i)
		a[i]=wread(),l=min (l,a[i]),r=max(r,a[i]);
	
	l*=1000,r*=1000;
	int ans=0;
	while (l<=r){
		int mid=(l+r)>>1;
		if (check (mid))	ans=mid,l=mid+1;
		else r=mid-1;
	}
	printf("%d\n",ans);	
	return 0;
}