POJ 2833 The Average(nlogn排序)

The Average

Time Limit: 6000MS Memory Limit: 10000KTotal Submissions: 9395 Accepted: 2927Case 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 a_i (1 ≤ a_i ≤ 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 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.

Source

POJ Monthly--2006.05.28, zby03

已知几个数，去掉几个最大值、几个最小值，求平均值。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <string>#include <sstream>#include <vector>#include <queue>#include <set>#include <map>#include <ctime>using namespace std;typedef long long LL;int cmp(int a, int b)  {      return a>b;  }  int main(){LL n1,n2,n;while(scanf("%I64d%I64d%I64d",&n1,&n2,&n)!=EOF){if(n1 == 0 && n2 == 0 && n == 0)break;int h[11],l[11];int i = 0;for(i = 0;i<11;i++){h[i] = 0;l[i] = 1e8+1;}LL sum = 0;for(i = 0;i<n;i++){LL num;scanf("%I64d",&num);sum+=num;if(num > h[n1]){h[n1] = num;sort(h,h+n1+1,cmp);}if(num < l[n2]){l[n2] = num;sort(l,l+n2+1);}}LL n3 = n - n1 - n2;//for(i = 0;i<n;i++){//cout<<v[i]<<endl;//}for(i = 0;i<n1;i++){//cout<<h[i]<<endl;sum = sum - h[i];}for(i = 0;i<n2;i++){//cout<<l[i]<<endl;sum = sum - l[i];}//cout<<sum<<endl;printf("%.6lf\n",1.0*sum/n3);   }return 0;}