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
题意：给出一个数组，寻找一段连续和，使得其平均和最大，但是长度不可以小于F
思路：平均和肯定是单调的，所以我们二分平均和，然后将每个数都减去这二分的数，然后统计序列的和的是不是正数，长度是否大于F。
#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) //二分平均值
{
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);//相减得到区间和，不断往正值靠近
}
if(ans>=0) //实数域上的二分
l=mid;
else
r=mid;
}
cout<<(int)(r*1000);
return 0;
}