HDU 1157 快速排序

      快速排序的水题，，用sort应该也可以，用来练习快速排序了。。题目：

Who's in the Middle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4266    Accepted Submission(s): 2148

Problem 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

5
2
4
1
3
5

 

Sample Output

3

ac代码：

#include <iostream>
#include <cstdio>
using namespace std;
const int N=100005;
int num[N];
int partition(int low,int high){
  int i=low,j=high,key=num[low];
  while(i<j){
    while(i<j&&num[j]>=key) --j;
	int t=num[i];num[i]=num[j];num[j]=t;
	while(i<j&&num[i]<=key) ++i;
	t=num[i];num[i]=num[j];num[j]=t;
  }
  return i;
}
void quick_sort(int low,int high){
	if(low<high){
	  int x=partition(low,high);
	  quick_sort(low,x-1);
	  quick_sort(x+1,high);
	}
}
int main(){
 // freopen("11.txt","r",stdin);
  int n;
  while(~scanf("%d",&n)){
    for(int i=0;i<n;++i)
		scanf("%d",&num[i]);
	quick_sort(0,n-1);
	printf("%d\n",num[n/2]);
  }
  return 0;
}