快排 谁在中间

 原题

Who's in the Middle

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.

FJ正在调查他的牛群，以找到最普通的奶牛。他想知道这头“中位数”奶牛产奶量：一半的奶牛产奶量等于或超过中位数奶牛;一半的人给予或多或少。

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.

给定奇数奶牛 N （1 <= N < 10,000） 和它们的产奶量 （1..1,000,000），求出给定的牛奶中位数，即至少一半的奶牛产奶量相同或更多，至少一半的奶牛产奶量相同或更少。

Input

* Line 1: A single integer N

* 第 1 行：单个整数 N

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

* 第 2..N+1 行：每行包含一个整数，即一头奶牛的产奶量。

Output

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

* 第 1 行：一个整数，即牛奶输出量的中位数。

Sample

Input	Output
5 2 4 1 3 5	3

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.

AC代码

#include <iostream>
#include <algorithm>
using namespace std;
int a[10005];
int n;
void quick_sort(int l,int r){if(l>=r) return ; int mid=a[(l+r)/2];int i=l,j=r;while(i<=j){while(a[i]<mid) i++;while(a[j]>mid) j--;if(i<=j) {swap(a[i],a[j]);i++,j--; }} quick_sort(l,j);//quick_sort((l+r)/2+1,r);错误 不一定就把mid大的放右，mid小的放右 quick_sort(i,r);//quick_sort((l+r)/2+1,r);
}
int main(){scanf("%d",&n);for(int i=0;i<n;i++) scanf("%d",a+i);quick_sort(0,n-1);printf("%d",a[n/2]);return 0;
}