和老钟方法一样

#include<bits/stdc++.h>
using namespace std;
int n,L;
double ans,i[100005],p[100005],j[100005],k[100005];
int main()
{
	scanf("%d%d",&n,&L);
	double l=0,r=2000,jin=1e-6,mid;
	for(int c=0;c<n;c++) scanf("%lf",&i[c]);
	while(r-l>jin){
		mid=(l+r)/2;
		for(int c=0;c<n;c++){
			p[c]=i[c]-mid;
			if(c){
				k[c]=max(k[c-1]+p[c],p[c]);
				j[c]=j[c-1]+p[c];
			}else{
				j[c]=k[c]=p[c];
			}
		}
		ans=j[L-1];
		for(int c=L;c<n;c++){
			double ans1=j[c]-j[c-L+1]+k[c-L+1];
			ans=max(ans,ans1);
		}
		ans>=0?l=mid:r=mid;
	}
	printf("%d",(int)(r*1000));
}

难以言喻的一题，再一次感觉到了我的愚蠢

#include<iostream>
using namespace std;
#define MAX 100007
#define INF 0x3f3f3f3f

double l=1e5+7,r=0;
int n,f;
int a[MAX],s[MAX];

bool check(double ave){
	double sum1=s[f-1]-ave*(f-1);
	for(int i=f;i<=n;i++){
		double sum2=s[i]-s[i-f]-ave*f;
		sum1=sum1+a[i]-ave;
		sum1=max(sum1,sum2);
		if(sum1>-1e-5)return false;
	}
	return true;
}

double search(double L,double R){
	double mid;
	while(R-L>=1e-5){
		mid=(L+R)/2;
		if(check(mid))R=mid;
		else L=mid;
	}
	return R;
}

int main(){
	ios::sync_with_stdio(false);
	cin>>n>>f;
	for(int i=1;i<=n;i++){
		cin>>a[i];
		if(a[i]>r)r=a[i];
		if(a[i]<l)l=a[i];
		s[i]=s[i-1]+a[i]; 
	}
	
	double ans=search(l,r);
	cout<<(int)(ans*1000);
	return 0;
}

https://blog.csdn.net/justidle/article/details/104318076

#include <bits/stdc++.h>
using namespace std;
 
const int MAXN = 1e5+2;
double data[MAXN] = {};//保存数组A
double a[MAXN];//减去枚举值后的数组
double sum[MAXN];//前缀和数组
double maxx[MAXN];// i 为结尾的最大连续区间和
double dp[MAXN];// i 为结尾区间连续长度大于等于 L 的最大连续区间和
 
int main() {
    //读入n和l
    int n,l;
    scanf("%d%d", &n, &l);
    //读入数组A
    int i;
    double left = 2000;//左边界
    double right = 1;//右边界
    for (i=1; i<=n; i++) {
        //注意由于需要使用前缀和，所以第一个数据要设置为0
        scanf("%lf", &data[i]);
        left = min(left, data[i]);//更新左边界
        right = max(right, data[i]);//更新右边界
    }
 
    double eps = 1e-6;//精度
    while (right-left > eps) {
        //计算中间值
        double mid = (left+right)/2;
 
        //数组A的每个数据减去这个枚举值，并计算前缀和
        for (i=1; i<=n; i++) {
            a[i]=data[i]-mid;
            sum[i]=sum[i-1]+a[i];
            maxx[i] = max(a[i], maxx[i-1]+a[i]);
        }
 
        //遍历长度为L到n所有子序列
        double ans=sum[l];
        for (i=l+1; i<=n; i++) {
            dp[i]=maxx[i-l+1]+sum[i]-sum[i-l+1];
            if (ans<dp[i]) {
                ans=dp[i];
            }
        }
 
        //修正边界
        if (ans>=0) {
            left = mid;
        } else {
            right = mid;
        }
    }
 
    //输出
    printf("%d\n", (int)(right*1000));
 
    return 0;
}