Best Cow Fences Page29 实数域二分

Best Cow Fences Page29 实数域二分

不得不说真的是道好题

把数列中每个值都减去二分值，转化为判定是否存在子段和非负，这个转化很巧妙

对于限制长度型的最大子段和提供了 $O(n)$解的思路

第一次写实数域二分，区别没啥，形式稍作修改

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<queue>
#include<map>
#define ll long long
#define pb push_back
#define rep(x,a,b) for (int x=a;x<=b;x++)
#define repp(x,a,b) for (int x=a;x<b;x++)
#define W(x) printf("%d\n",x)
#define WW(x) printf("%lld\n",x)
#define pi 3.14159265358979323846
#define mem(a,x) memset(a,x,sizeof a)
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
using namespace std;
const int maxn=2e6+7;
const int INF=1e9;
const ll INFF=1e18;
const double eps=1e-5;
double a[maxn],b[maxn],sum[maxn];
int n,L;
int main()
{
    scanf("%d%d",&n,&L);
    rep(i,1,n)scanf("%lf",&a[i]);
    double l=0,r=3000;
    while(l+eps<r)
    {
        double mid=(l+r)/2;
        rep(i,1,n)b[i]=a[i]-mid;
        rep(i,1,n)sum[i]=sum[i-1]+b[i];
        double minn=1e9;
        double ans=-1e9;
        rep(i,L,n)
        {
            minn=min(minn,sum[i-L]);
            ans=max(ans,sum[i]-minn);
        }
        if (ans>=0)l=mid;
        else r=mid;
    }
    W(int(r*1000));
    return 0;
}