最大均值【二分】

>Link

ybtoj最大均值

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>解题思路

考虑二分答案（最大平均数），将$a$序列减去当前二分值存到$b$序列中，这是问题就转换为“长度不小于L且区间和不小于0的序列”
定义$su{m}_i$为前$i$位的和，$ma{x}_i$为前$i$位中长度不小于L且区间和最大的那个值，
运用前缀和的思想，$ma{x}_i=su{m}_i-mi{n}_{j=1}^{i-L}(su{m}_j)$
因为$i$每累加1，$j$的范围就累加1，所以我们只需要每次记录下当前最小的$su{m}_j$，就可以$O(n)$计算出$ma{x}_i$

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>代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define N 100010
#define inf 1 << 30
#define db double
using namespace std;

const db eps = 0.00001;
int n, m;
db a[N], sum[N], l, r, mid;

bool check (db x)
{
    
    
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + a[i] - x;
	db minn = inf, maxn = -inf;
	for (int i = m; i <= n; i++)
	{
    
    
		minn = min (minn, sum[i - m]);
		maxn = max (maxn, sum[i] - minn);
	}
	return maxn >= 0;
}

int main()
{
    
    
	scanf ("%d%d", &n, &m);
	l = inf;
	for (int i = 1; i <= n; i++)
	{
    
    
		scanf ("%lf", &a[i]);
		l = min (l, a[i]), r = max (r, a[i]);
	}
	while (r - l > eps)
	{
    
    
		mid = (l + r) / 2;
 		if (check (mid)) l = mid;
		else r = mid;
	}
	printf ("%d", (int)floor(r * 1000));
	return 0;
}