（优先队列）The Average

In a speech contest, when a contestant finishes his speech, the judges will then grade his performance. The staff remove the highest grade and the lowest grade and compute the average of the rest as the contestant’s final grade. This is an easy problem because usually there are only several judges.

Let’s consider a generalized form of the problem above. Given n positive integers, remove the greatestn₁ ones and the leastn₂ ones, and compute the average of the rest.

Input

The input consists of several test cases. Each test case consists two lines. The first line contains three integersn₁,n₂ and n (1 ≤ n₁,n₂ ≤ 10,n₁ + n₂ < n ≤ 5,000,000) separate by a single space. The second line containsn positive integersa_i (1 ≤ a_i ≤ 10⁸ for alli s.t. 1 ≤i ≤ n) separated by a single space. The last test case is followed by three zeroes.

Output

For each test case, output the average rounded to six digits after decimal point in a separate line.

Sample Input

1 2 51 2 3 4 54 2 102121187 902 485 531 843 582 652 926 220 1550 0 0

Sample Output

3.500000562.500000

Hint

This problem has very large input data. scanf and printf are recommended for C++ I/O.

The memory limit might not allow you to store everything in the memory.

这道题用了优先队列，分别把a个最大值和b个最小值存到有限队列里，然后用sum减去求平均值i就好了。

#include<iostream>#include<queue>#include<cstdio>#include<vector>using namespace std;priority_queue<int,vector<int>,greater<int> >qmin;priority_queue<int,vector<int>,less<int> >qmax;int main(){    int a,b,n;    while(scanf("%d%d%d",&a,&b,&n)&&n)    {        int c;        int maxsize=0,minsize=0;        long long sum=0;        for(int i=0; i<n; ++i)        {            scanf("%d",&c);            sum+=(long long )c;            if(maxsize<a)            {                qmin.push(c);                ++maxsize;            }            else            {                if(c>qmin.top())                {                    qmin.pop();                    qmin.push(c);                }            }            if(minsize<b)            {                qmax.push(c);                ++minsize;            }            else            {                if(c<qmax.top())                {                    qmax.pop();                    qmax.push(c);                }            }        }        int del=0;        while(!qmin.empty())        {            del+=qmin.top();            qmin.pop();        }        while(!qmax.empty())        {            del+=qmax.top();            qmax.pop();        }        printf("%f\n",(1.0 * sum - (long long)del) / (1.0 * (n- a - b)));    }}