POJ - 2018 Best Cow Fences （二分 长度>=l的子序列最大和+前缀和）

Best Cow Fences

Description

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

Source

​​USACO 2003 March Green​​

书上讲得完美

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std ;
const int N=100005;
const double eps=1e-5;
double a[N],b[N],sum[N];
int n;
double max_sum_L(double mid,int m)
{
  for(int i=1;i<=n;i++)
    b[i]=a[i]-mid;
  for(int i=1;i<=n;i++)
    sum[i]=sum[i-1]+b[i];//求前缀和，
    
  double ans=-1e10;
  double minn=1e10;
  for(int i=m;i<=n;i++)
  {
    minn=min(minn,sum[i-m]);  ///维护左端最小值
    ans=max(ans,sum[i]-minn); ///维护长度>=的子序列的最大值
  }
  return ans;
}
int main()
{
  int m;
  scanf("%d%d",&n,&m);
  double minn=1e10;
  double maxx=-1e10;
  for(int i=1;i<=n;i++)
  {
        scanf("%lf",&a[i]);
        minn=min(a[i],minn);
        maxx=max(a[i],maxx);
  }

  double l=minn;
  double r=maxx;
  while(r-l>eps)
  {
    double mid=(l+r)/2;
      if(max_sum_L(mid,m)>=0)
        l=mid;
    else
      r=mid;
    
  }
  cout<<int(r*1000)<<endl;
  return 0 ;
}