poj 2388 Quick_sort 求中间值

Who's in the Middle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 42058 Accepted: 24335

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

524135

Sample Output

3

//下面是代码，题意就是排好序后求中间的那个数
#include<stdio.h>#define MAX 10005int a[MAX];void quick_sort(int s[], int l, int r){if (l < r){int i = l, j = r, x = s[l];while (i < j){while(i < j && s[j] >= x) // 从右向左找第一个小于x的数j--;  if(i < j) s[i++] = s[j];while(i < j && s[i] < x) // 从左向右找第一个大于等于x的数i++;  if(i < j) s[j--] = s[i];}s[i] = x;quick_sort(s, l, i - 1); // 递归调用 quick_sort(s, i + 1, r);}}int main(){int T;scanf("%d",&T);for(int i=0;i<T;i++){scanf("%d",&a[i]); }quick_sort(a,0,T-1); printf("%d\n",a[T/2]);return 0;}