POJ ACM习题【No.1862】

最新推荐文章于 2025-08-12 21:03:53 发布

最新推荐文章于 2025-08-12 21:03:53 发布 · 131 阅读

文章标签：

#java

ACM 专栏收录该内容

42 篇文章

订阅专栏

本文介绍了一种关于Stripies生物的重量优化问题及其解决方法。Stripies为一种新生命形式，当两个不同重量的Stripies碰撞后会形成一个新的Stripie，其重量遵循特定的数学规则。文章提供了一个算法，用于计算通过最优碰撞方式后剩余Stripies的总重量，并给出具体实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Stripies

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 5363		Accepted: 2746

Description

Our chemical biologists have invented a new very useful form of life called stripies (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 m1 and m2 collide the weight of resulting stripie equals to 2*sqrt(m1*m2). 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 of the input 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 three decimal digits after the point.

Sample Input

Sample Output

120.000

解题思路的关键在于：

最小的值是从大向小进行计算，可以使用不等式进行证明。

import java.util.*;
import java.text.*;

public class Main {

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		
		int num = Integer.valueOf(cin.nextLine()).intValue();
		List list = new ArrayList();
		
		for(int i = 0; i < num; i++)
			list.add(Double.valueOf(cin.nextLine()));
		
		Collections.sort(list);
		
		double result, temp = 0;
		double a, b = 0;
		
		if(list.size() == 1)
			result = Double.valueOf((Double)list.get(0)).doubleValue();
		else
		{
			int index = list.size() - 1;
			a = Double.valueOf((Double)list.get(index)).doubleValue();
			b = Double.valueOf((Double)list.get(index-1)).doubleValue();
			
			result = 2 * Math.sqrt(a * b);
			
			for(int i = index - 2; i>=0; i--)
			{
				a = Double.valueOf((Double)list.get(i)).doubleValue();
				result = 2 * Math.sqrt(a * result);
			}
		}
		
		DecimalFormat df = new DecimalFormat("#.000");
		System.out.println(df.format(result));
	}

}