最大均值

• • Description--
  • 解题思路--
  • 代码--

Description–

高效进阶「基础算法」第3章 二分算法课堂过关 例题3

---

解题思路–

二分答案
把所选子串每个数减去mid，若子串和仍为正数，则该子串的平均值大于mid

---

代码–

#include <iostream>
#include <cstdio>
#include <cmath>
#define db double

using namespace std;

const db eps = 1e-5;

db l, r, a[100005], s[100005];
int n, ll;

bool check(db k)
{
	for (int i = 1; i <= n; ++i) s[i] = s[i - 1] + a[i] - k;
	db minn = 1e10, maxx = -1e10;
	for (int i = ll; i <= n; ++i)
	{
		minn = min(minn, s[i - ll]);
		maxx = max(maxx, s[i] - minn);
	}
	return maxx >= 0;
}

int main()
{
	scanf("%d%d", &n, &ll);
	l = 1e10;
	for (int i = 1; i <= n; ++i)
	{
		scanf("%lf", &a[i]);
		l = min(l, a[i]), r = max(r, a[i]);
	}
	while (l + eps < r)
	{
		db mid = (l + r) / 2;
		if (check(mid)) l = mid;
		else r = mid;
	}
	printf("%d", (int)floor(r * 1000));
	
	return 0;
}