poj 2388 Who's in the Middle

Who's in the Middle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 39218 Accepted: 22785

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

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

提示

题意：

不废话了，叫你排序输出最中间的那个值。

思路：

看楼上。

示例程序

Source CodeProblem: 2388Code Length: 558BMemory: 428KTime: 32MSLanguage: GCCResult: Accepted#include <stdio.h>int a[10000];void quicksort(int l,int r){    int i=l,j=r,key=a[i];    if(i>=j)    {        return;    }    while(i<j)    {        while(i<j&&key<=a[j])        {            j--;        }        a[i]=a[j];        while(i<j&&key>=a[i])        {            i++;        }        a[j]=a[i];    }    a[i]=key;    quicksort(l,i-1);    quicksort(i+1,r);}int main(){    int n,i;    scanf("%d",&n);    for(i=0;n>i;i++)    {        scanf("%d",&a[i]);    }    quicksort(0,n-1);    printf("%d",a[n/2]);    return 0;}