Stripies POJ - 1862 ---优先队列

目录

题目：

 证明：​编辑

 解题思路：

代码：

---

题目：

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

Inputcopy	Outputcopy
3 72 30 50	120.000

 证明：

 解题思路：

由数学知识可知最大的两个数先进行2*sqrt(min1*min2)的数学运算得出的值最小，然后继续放在优先对列里面继续计算最后得出我们想要的值

代码：

#include<iostream>
#include<cmath>
#include<queue>
#include<iomanip>
using namespace std;
priority_queue <double> q;
int main()
{
    double n,min1,min2,x,sum=0;
    cin>>n;
    while(n--)
    {
        cin>>x;
        q.push(x);
    }
    while(q.size()>1)
    {
        min1=q.top();
        q.pop();
        min2=q.top();
        q.pop();
        min1=2*sqrt(min1*min2);
        q.push(min1);
    }
    //sum=q.top();
    cout << fixed << setprecision(3) << q.top() << endl;

    return 0;
}