题目链接：
http://poj.org/problem?id=2833
The Average
Time Limit: 6000MS Memory Limit: 10000K Total Submissions: 13247 Accepted: 3947 Case Time Limit: 4000MS Description
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 greatest n1 ones and the least n2 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 integers n1, n2 and n (1 ≤ n1, n2 ≤ 10, n1 + n2 < n ≤ 5,000,000) separate by a single space. The second line contains n positive integers ai (1 ≤ ai ≤ 108 for all i 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 5 1 2 3 4 5 4 2 10 2121187 902 485 531 843 582 652 926 220 155 0 0 0Sample Output
3.500000 562.500000Hint
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.
Source
POJ Monthly--2006.05.28, zby03
题目大意：
给你n个值，去掉n1个最大值，去掉n2个最小值，求剩下的平均值
排序的话会超内存，需要用优先队列储存最小值与最大值
This is the code
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define pppp cout<<endl;
#define EPS 1e-8
#define LL long long
#define ULL unsigned long long //1844674407370955161
#define INT_INF 0x3f3f3f3f //1061109567
#define LL_INF 0x3f3f3f3f3f3f3f3f //4557430888798830399
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int dr[]={0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]={-1, 1, 0, 0, -1, 1, -1, 1};
inline int read()//输入外挂
{int ret=0, flag=0;char ch;if((ch=getchar())=='-')flag=1;else if(ch>='0'&&ch<='9')ret = ch - '0';while((ch=getchar())>='0'&&ch<='9')ret=ret*10+(ch-'0');return flag ? -ret : ret;
}
int main()
{long long n1,n2,n;//n1最大值，n2最小值while(scanf("%lld%lld%lld",&n1,&n2,&n)){if(!n1&&!n2&&!n)break;priority_queue< int, vector<int>, greater<int> > big;//小根堆储存最大值priority_queue< int, vector<int>, less<int> > small;//大根堆储存最小值long long sum=0;for(int i=0;i<n;++i){long long tem;scanf("%lld",&tem);sum+=tem;big.push(tem);if(big.size()>n1)//储存最大值big.pop();small.push(tem);if(small.size()>n2)//储存最小值small.pop();}while(!big.empty()){sum-=big.top();big.pop();}while(!small.empty()){sum-=small.top();small.pop();}n=n-n1-n2;double ans=1.0*sum/n;printf("%.6f\n",ans);}return 0;
}