Best Cow Fences（二分）

问题 K: Best Cow Fences

时间限制: 1 Sec  内存限制: 128 MB
提交: 61  解决: 27
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题目描述

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. 

 

输入

* 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. 

 

输出

* 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. 

 

样例输入

10 6
6 
4
2
10
3
8
5
9
4
1

 

样例输出

6500

题意：给你n个牛的自身价值，让你找出连续的且数量大于等于F的一段区间，使这段区间内的牛的平均价值最大。

分析：一个简单的二分，再加一个经典的处理。

1、首先二分区间的平均值为多少，让数组中的所有数都减去平均值。

2、再从数组中找出一段长度大于等于F的区间，看区间和是否会大于等于0。

3、要找这段区间需要维护左端点最小值，枚举右端点。先把减去平均值的数组求一个前缀和，再设K，对于当前枚举到的 i 位置，我们想让a[i]-a[k]，尽量大，那么a[k]，就必须尽量小，所以一直维护最小值就可以了minn=（minn，sum[i-m]）。