【二分+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

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毒瘤题,用long double会比double慢一倍时间

枚举平均值，用原数列减去当前平均值，找到一个长度大于f的非负子段就可以了

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#include <iostream>
#include <cstring>
#include <iomanip>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const ll maxn = 1e5+100;
const ll mod = 1e9+7;
const ld pi = acos(-1.0);
const ll inf = 1e18;
const ld eps = 1e-5;

ll n,f;
ld arr[maxn],sum[maxn],sub[maxn];
ld l = inf,r = -inf;

/*bool check(ld x)
{
	ld ans = -1e10,minval = 1e10;
	
	for(ll i = 1; i <= n; i++)
	{
		sub[i] = arr[i]-x;
		sum[i] = sum[i-1] + sub[i];
	}
	
	for(ll i = f; i <= n; i++)
	{
		minval = min(minval,sum[i-f]);
		
		ans = max(ans,sum[i]-minval);	
	}
	
	if(ans >= 0)
		return true;
	
	return false;

}*/

int main()
{	
    ios::sync_with_stdio(false);
    //cin.tie(0),cout.tie(0);
	
	//cout << eps << endl;
	
	cin >> n >> f;
	
	for(ll i = 1; i <= n; i++)
	{
		//scanf("%llf",&arr[i]);
		cin >> arr[i];
		//sum[i] = sum[i-1] + arr[i];
		l = min(arr[i],l);
		r = max(arr[i],r);
	}
	
	while(r-l > eps)
	{
		ld mid = (l+r)/2;
		
		ld ans = -inf,minval = inf;
	
		for(ll i = 1; i <= n; i++)
		{
			sub[i] = arr[i]-mid;
			sum[i] = sum[i-1] + sub[i];
		}
		
		for(ll i = f; i <= n; i++)  //一定大于f段
		{
			minval = min(minval,sum[i-f]);
			
			ans = max(ans,sum[i]-minval);	
		}
		
		if(ans >= 0)
			l = mid;
		else
			r = mid;
	}
	
	cout << ll(r*1000) << endl;
	
	return 0;
}