hdu 1157 Who's in the Middle

最新推荐文章于 2024-07-21 14:30:00 发布

原创最新推荐文章于 2024-07-21 14:30:00 发布 · 385 阅读

0 ·

CC 4.0 BY-SA版权

排序专栏收录该内容

3 篇文章

订阅专栏

本文详细解释并提供了解决HDU平台上的HDU 1157问题的算法解决方案。该问题涉及找到一组奶牛产出奶量的中位数，即至少一半奶牛产奶量不低于此值，另一半不高于此值。通过使用排序算法和数组操作，可以有效地找出中位数奶牛的产奶量。

http://acm.hdu.edu.cn/showproblem.php?pid=1157

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6012 Accepted Submission(s): 3003

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

求中位数

Sample Output

  
   3


   #include <stdio.h>
#include<algorithm>
using namespace std;
int a[10001];
int main()
{
 int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=0;i<n;i++)
   scanf("%d",&a[i]);
sort(a,a+n);
printf("%d\n",a[n/2]);
}
}