Timus 1161. Stripies-优快云博客

化学生物学家发明了一种名为条纹生物的透明、无定形、水母状生物，它们生活在胶状营养介质中并经常移动。当两个条纹生物碰撞时，会产生新的条纹生物，其重量不是两者的简单相加。通过实验观察，科学家们发现新生成的条纹生物重量为两碰撞生物重量的平方根乘以二。本文提供了一个帮助计算特定群体最小总重量的程序，适用于初学者和专业开发者。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Our chemical biologists have invented a new very useful form of life calledstripies(in fact, they were first called in Russian -polosatiki, but the scientists had to invent an English name to apply for an international patent). The stripies are transparent amorphous amebiform creatures that live in flat colonies in a jelly-like nutrient medium. Most of the time the stripies are moving. When two of them collide a new stripie appears instead of them. Long observations made by our scientists enabled them to establish that the weight of the new stripie isn't equal to the sum of weights of two disappeared stripies that collided; nevertheless, they soon learned that when two stripies of weights m1and m2collide the weight of resulting stripie equals to 2·sqrt(m1m2). Our chemical biologists are very anxious to know to what limits can decrease the total weight of a given colony of stripies.

You are to write a program that will help them to answer this question. You may assume that 3 or more stipies never collide together.

Input

The first line contains one integer N (1 ≤ N ≤ 100) - the number of stripies in a colony. Each of next N lines contains one integer ranging from 1 to 10000 - the weight of the corresponding stripie.

Output

The output must contain one line with the minimal possible total weight of colony with the accuracy of two decimal digits after the point.

Sample

input	output
3 72 30 50	120.00

奇怪的数学题：

排序，然后由大到小合并。

这里使用点新东西，使用个multiset<double>容器来解决问题，这样连排序都省了。然后是倒序处理数组。

很好用的容器。

#include <cmath>
#include <algorithm>
#include <iostream>
#include <set>
using namespace std;

void Stripies1161()
{
	multiset<double> s;
	int n = 0;
	double a;
	cin>>n;
	for (int i = 0; i < n; i++)
	{
		cin>>a;
		s.insert(a);
	}
	auto it = s.rbegin();
	a = *it;
	for (it++; it != s.rend(); it++)
	{
		a = 2.0 * sqrt(a * (*it));
	}
	cout<<a;
}