Best Cow Fences题解

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时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 32768K，其他语言65536K
64bit IO Format: %lld

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.

输入描述:
* 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.
输出描述:
* Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000×ncowsnfields\frac{1000\times ncows}{nfields}nfields1000×ncows.
示例1

输入
10 6
6
4
2
10
3
8
5
9
4
1
输出
6500
二分答案

AC代码：
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-5;
double a[100010],b[100010],sum[100010];
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);int n,L;cin >> n >> L;for(int i = 1;i <= n;i++) cin >> a[i];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++) sum[i] = (sum[i - 1] + 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 >= 0) l = mid;else r = mid;}cout << int(r * 1e3) << endl;return 0;
}

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Best Cow Fences题解

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