hdu 1157 (1.3.7) Who's in the Middle

本文介绍了一种算法挑战,即找到一群奶牛中产量处于中位数的那头奶牛的方法。通过输入奇数个奶牛的产奶量数据,利用排序算法如快速排序或STL sort函数来确定中位数奶量。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

<div class="panel_title" align="left">Problem Description</div><div class="panel_content">FJ is surveying his herd to find the most average cow. He wants to know how much milk this 'median' cow gives: half of the cows give as much or more than the median; half give as much or less. 

Given an odd number of cows N (1 <= N < 10,000) and their milk output (1..1,000,000), find the median amount of milk given such that at least half the cows give the same amount of milk or more and at least half give the same or less.
</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Input</div><div class="panel_content">* Line 1: A single integer N 

* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.
</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Output</div><div class="panel_content">* Line 1: A single integer that is the median milk output.
</div><div class="panel_bottom"> </div>
<div class="panel_title" align="left">Sample Input</div><div class="panel_content"><pre><div style="FONT-FAMILY: Courier New,Courier,monospace">5
2
4
1
3
5</div>
 

Sample Output
3
题目实际上是排序,但是用c的输入输出貌似就过不去啊,cin和cout才可以
Hint
INPUT DETAILS: Five cows with milk outputs of 1..5 OUTPUT DETAILS: 1 and 2 are below 3; 4 and 5 are above 3.


// 解法1:使用qsort快排 
//#include <iostream>
//#include <stdlib.h>
//using namespace std;
//int cmp(const void *a, const void *b)
//{
//	return (*(int*)a - *(int*)b);
//}
//
//int main(void)
//{
//	int n, i;
//    int number[100000] = {0};	
//	while(cin >> n)
//	{
//		for(i = 0; i < n; i++)
//			cin >> number[i];
//		qsort(number, n, sizeof(int), cmp);
//		cout << number[(n-1)/2] << endl;
//	}
//	
//	return 0;
//}
// 解法2:使用sort排序 
#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(int a, int b)
{
	return a > b;
}
int main(void)
{
	int n, i;
	int num[10000];
	while(cin >> n)
	{
		for(i = 0; i < n; i++)
			cin >> num[i];
		sort(num, num + n, cmp);
		cout << num[(n-1)/2] << endl; 
	}
	
	return 0;
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值