POJ 1862 Stripies

Stripies

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10373		Accepted: 5017

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

3
72
30
50

Sample Output

120.000

有一些单细胞生物，每两个可以相碰撞形成一个新的，新的生物重量为2*sqrt(m1*m2)

给定一些单细胞生物，求一种最佳的方式，使它们经过碰撞后最终剩下的一个单细胞生物的重量最小。

现假设重量为b和c的两个单细胞生物都想与a碰撞，那么我们应该让哪个先与a碰撞呢

若b先，则最终重量为2*sqrt(c*2*sqrt(a*b))

若c先，则最终重量为2*sqrt(b*2*sqrt(a*c))

假设要让b先，则就2*sqrt(c*2*sqrt(a*b))<=2*sqrt(b*2*sqrt(a*c))，化简得到c<=b。

由此我们知道要从重量大的先碰撞，重量小的后碰撞。

 

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>

using namespace std;

int n;
int num[110];

bool cmp(int a,int b)
{
    return a>=b;
}

int main()
{
    while(cin>>n)
    {
        for(int i=0;i<n;i++)
            scanf("%d",&num[i]);
        if(n==1)
        {
            printf("%.3lf
",(double)num[0]);
            continue;
        }
        sort(num,num+n,cmp);
        double ans=2*sqrt((double)num[0]*num[1]);
        for(int i=2;i<n;i++)
            ans=2*sqrt(ans*num[i]);
        printf("%.3lf
",ans);
    }

    return 0;
}