Best Cow Fences POJ - 2018（经典二分）

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本文介绍了一种解决最大化平均值围栏问题的方法，通过枚举可能的平均值并使用二分查找法来确定最优解。该问题要求在一个包含不同数量奶牛的田地中找出连续的一段田地，使得这段田地的平均奶牛数最大，同时这段田地的数量至少为给定值。

Farmer John’s farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000.

FJ wants to build a fence around a contiguous group of these fields in order to maximize the average number of cows per field within that block. The block must contain at least F (1 <= F <= N) fields, where F given as input.

Calculate the fence placement that maximizes the average, given the constraint.
Input
* Line 1: Two space-separated integers, N and F.

Lines 2..N+1: Each line contains a single integer, the number of cows in a field. Line 2 gives the number of cows in field 1,line 3 gives the number in field 2, and so on.
Output
Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000*ncows/nfields.
Sample Input
10 6
6
4
2
10
3
8
5
9
4
1
Sample Output
6500
评价：题意是在一个数组里，寻找一段连续和，使其平均和最大，但是长度不能小于F,
首先可以看出是满足单调性的，但是怎么二分呢，
我们先枚举一个可能的数。
然后数组里的值全部减去这个值（结果会有正有负），那么我们就看是否存一段长度大于等于F,且和为正。对于此的判断，可谓经典，见代码。
AC

#include<iostream>
#include<cstdio>
#include<vector>
#define maxx 400005
using namespace std;
double a[100005];
double b[100005];
double c[100005];
int main()
{

        int n;
        int f;
        cin>>n>>f;
        for(int i=1;i<=n;i++)
            scanf("%lf",a+i);
        double eps=1e-5;
        double l=-1e6,r=1e6;
        while(r-l>eps)
        {
            double mid=(l+r)/2;
            for(int i=1;i<=n;i++)b[i]=a[i]-mid;
            for(int i=1;i<=n;i++)
                c[i]=c[i-1]+b[i];//求前缀和，
            double ans=-1e10;
            double minn=1e10;
            for(int i=f;i<=n;i++)
            {
                minn=min(minn, c[i-f]);不断维护左端的最小值
                ans=max(ans, c[i]-minn);//相见得到区间和，不断忘正值靠近
            }//核心
            if(ans>=0)
                l=mid;
            else
                r=mid;
        }
        cout<<(int)(r*1000);
    return 0;
}