hdu 1157 (1.3.7) Who's in the Middle

最新推荐文章于 2015-07-04 17:07:22 发布

原创最新推荐文章于 2015-07-04 17:07:22 发布 · 403 阅读

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#ACM HDU #排序

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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。

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<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;
}