POJ2388-Who's in the Middle

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此编程挑战要求从一组奶牛的牛奶产量中找出中位数产量，即一半的奶牛产奶量高于该值，另一半低于该值。通过排序算法解决此问题。

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Who's in the Middle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 42039		Accepted: 24320

Description

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.

Input

* Line 1: A single integer N

* Lines 2..N+1: Each line contains a single integer that is the milk output of one cow.

Output

* Line 1: A single integer that is the median milk output.

Sample Input

Sample Output

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.

Source

USACO 2004 November

题意：给定n个数，输出中间值

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const LL INF = 0x3f3f3f3f3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;

int a[1000005];

int main()
{
	int n;
	while (~scanf("%d", &n))
	{
		for (int i = 0; i<n; i++)
			scanf("%d", &a[i]);
		sort(a, a + n);
		printf("%d\n", a[n / 2]);
	}
	return 0;
}