【ICPC】The 2023 ICPC Asia Hangzhou Regional Contest (The 2nd Universal Cup. Stage 22 Hangzhou) M

V-Diagram

#贪心 #枚举

题目描述

A $1$-indexed integer sequence $a$ of length $n$ is a V-diagram if $n\ge 3$ and there exists an index $i$ ($1\le i\le n$) satisfying the following:

• $a_j\ge a_{j+1}$ for $1\le j\le i$;
• $a_j\ge a_{j-1}$ for $i\le j\le n$.

Given a V-diagram $a$, find a V-diagram $b$ with the maximum possible average such that $b$ is a consecutive subsequence of $a$.

A consecutive subsequence of a sequence can be obtained by removing some (possibly zero) elements from the beginning and end of the sequence.

输入格式

Each test contains multiple test cases. The first line contains a single integer $t$ ($1\le t\le 10^5$) denoting the number of test cases. For each test case:

The first line contains one integer $n$ ($3\le n\le 3\cdot 10^5$) denoting the length of the integer sequence $a$.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 10^9$) denoting the sequence $a$ itself.

It is guaranteed that $a$ is a V-diagram, and the sum of $n$ over all test cases does not exceed $3\cdot 10^5$.

输出格式

For each test case, output a real number denoting the maximum possible average.

Your answer is considered correct if its absolute or relative error does not exceed $10^{-9}$.

Formally, let your answer be $x$, and the jury’s answer be $y$. Your answer will be considered correct if and only if $\frac{|x-y|}{\max (1,|y|)}\le 10^{-9}$.

样例 #1

样例输入 #1

2
4
8 2 7 10
6
9 6 5 3 4 8

样例输出 #1

6.75000000000000000000
5.83333333333333303727

解题思路

首先我们满足$V$形图的前提下（取中间三个数），分别枚举右侧连续取的段数和左边连续取的段数。

具体而言，先枚举右边连续取的段数，如果当前取的大于平均值，则更新平均值，如果不大于，保持右边段取的状态，直到后面大于平均值才更新，然后对左边一段进行同样的操作。

最后，我们今天同样的操作，只不过这次先从左侧开始取，然后才是右侧。

最终答案为二者的最大值。

代码

void solve() {
	int n;
	cin >> n;
	vector<int>a(n + 1);
	for (int i = 1; i <= n; ++i) {
		cin >> a[i];
	}
	auto idx = min_element(a.begin() + 1, a.end()) - a.begin();
	int s = a[idx] + a[idx - 1] + a[idx+1];
	int sum = s, siz = 3, now = 3;
	for (int i = idx+2; i <= n; ++i) {
		sum += a[i];
		siz++;
		if (sum * now > s * siz) {
			now = siz;
			s = sum;
		}
	}
	siz = now, sum = s;
	for (int i = idx - 2; i >= 1; --i) {
		sum += a[i];
		siz++;
		if (sum * now > s * siz) {
			now = siz;
			s = sum;
		}
	}

	double res = (double)s / now;
	
	s = a[idx] + a[idx - 1] + a[idx + 1];
	sum = s, siz = 3, now = 3;
	for (int i = idx - 2; i >= 1; --i) {
		sum += a[i];
		siz++;
		if (sum * now > s * siz) {
			now = siz;
			s = sum;
		}
	}
	siz = now, sum = s;
	for (int i = idx + 2; i <= n; ++i) {
		sum += a[i];
		siz++;
		if (sum * now > s * siz) {
			now = siz;
			s = sum;
		}
	}

	res = std::max(res, (double)s / now);

	std::cout << fixed << std::setprecision(20) << res<<"\n";
}

signed main() {
	ios::sync_with_stdio(0);
	std::cin.tie(0);
	std::cout.tie(0);
	int t = 1;
	cin >> t;
	while (t--) {
		solve();
	}
};