POJ 1862 Stripies （优先队列）

Stripies

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

题意：给你n个数，你可以取出a，b两个数进行操作2*sqrt(a*b)后变成c，再将c放入序列，问你最终序列变成一个数的时候的最小值，

思路：感觉是数学题，既然要得到最小值，那每次应该取出两个最大的数，再开方，
那么开方之后得到的数相当于不是两个最大的数开方，变化幅度会大一点，即减少的更多，所以可以用优先队列......

#include<iostream>
#include<cstdio> 
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
using namespace std;
#define ll long long
const int mod=1e9+7;
const int inf=1e9+7;

const int maxn=105;

int main()
{
    //ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    int n;
    cin>>n;
    
    priority_queue<double>q;
    
    double num;
    for(int i=0;i<n;i++)
    {
        cin>>num;
        q.push(num);
    }
    
    double a,b,s;
    while(q.size()>1)
    {
        a=q.top();//取出最大的 
        q.pop();
        b=q.top();//取出另一个最大的 
        q.pop();
        s=2*sqrt(a*b);//操作 
        q.push(s);//放入队列 
    }
    
    printf("%.3f\n",q.top());//注意用%f，用%lf会wa，最好平常都写%f而不是%lf 
    
    return 0;
 }