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Best Cow Fences

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16945		Accepted: 5425

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

OJ-ID：
poj-2018

author：
Caution_X

date of submission：
20191118

tags：
二分

description modelling：
给定正整数数列A[]，求一个平均值最大的，长度不小于L的（连续的）子段，输出该平均值*1000

major steps to solve it：
（1）二分枚举所有的平均值
（2）每一次二分，将A[]-mid，之后求前缀和。判断所取子段平均值是否大于等于mid只要判断选取区间累加是否大于等于0即可

AC code:

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
double a[],b[];
double sum[];
double eps=1e-;
int main()
{
    //freopen("input.txt","r",stdin);
    int N,L;
    scanf("%d%d",&N,&L);
    for(int i=;i<=N;i++)
    {
        scanf("%lf",&a[i]);
    }
    double l=-1e6,r=1e6;
    while(r-l>eps)
    {
        double mid=(r+l)/;
        for(int i=;i<=N;i++)    b[i]=a[i]-mid;
        for(int i=;i<=N;i++)
        {
            sum[i]=sum[i-]+b[i];
        }
        double ans=-1e10;
        double min_val=1e10;
        for(int i=L;i<=N;i++)
        {
            min_val=min(min_val,sum[i-L]);
            ans=max(ans,sum[i]-min_val);
        }
        if(ans>=)    l=mid;
        else r=mid;
    }
    printf("%d\n",int(r*));
}