Stripies (贪心)

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原文链接：http://poj.org/problem?id=1862

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#算法 #贪心

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本文介绍了一种名为Stripies的新型生物形态，其在碰撞时遵循特定的重量合并规则。通过数学运算，我们的化学生物学家希望找到一种方法来最小化整个群体的总重量。文章提供了一个算法解决方案，首先对群体中个体的重量进行排序，然后应用计算公式以达到最小化总重量的目的。

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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

思路

这道题是求输入的数值里面，如何进行计算才能算出最后权值的最小值（计算公式：2*根号（m1*m2））。也就是说，最大的数值开方的次数越多，则最后所得到的权值就会越小。所以，要先对输入的数值进行排序，再从大到小进行计算。

#include <iostream>
#include <math.h>
#include <algorithm>
#include <stdio.h>

using namespace std;

double s(const double &i,const double &j)
{
    return 2*sqrt(i*j);
}
bool cmp(const int &i,const int &j)
{
    return i>j;
}
int main()
{
    int n,i;
    int a[100];
    cin>>n;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    sort(a,a+n,cmp);
    double b=a[0];
    for(i=0;i<n-1;i++)
    {
        b=s(b,a[i+1]);
    }
    printf("%.3f\n",b);
    return 0;
}