POJ 2833 The Average(nlogn排序)

最新推荐文章于 2022-03-08 09:13:55 发布

steveyg

最新推荐文章于 2022-03-08 09:13:55 发布

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分类专栏： POJ解题报告文章标签： acm poj 解题报告排序

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本文深入探讨了AI音视频处理领域中的视频分割与语义识别技术，介绍了视频分割的基本概念、算法及其应用，并详细阐述了语义识别在不同场景下的实现方法，包括自动驾驶、AR增强现实等领域的应用实例。

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The Average

Time Limit: 6000MS		Memory Limit: 10000K
Total Submissions: 9395		Accepted: 2927
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 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 5
1 2 3 4 5
4 2 10
2121187 902 485 531 843 582 652 926 220 155
0 0 0

Sample Output

3.500000
562.500000

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;
}