Who's in the Middle_c++ who's in the middle-优快云博客

本文链接：https://blog.youkuaiyun.com/liuxinglon/article/details/20242805

本文介绍了一个问题：如何找到一群奶牛中产奶量处于中位数的那一只，即一半以上的奶牛产奶量不低于该数值。文章提供了一个通过输入奶牛数量及各自产奶量来计算中位数产奶量的示例程序。

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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

Sample Output

3

<div style='font-family:Times New Roman;font-size:14px;background-color:F4FBFF;border:#B7CBFF 1px dashed;padding:6px'><div style='font-family:Arial;font-weight:bold;color:#7CA9ED;border-bottom:#B7CBFF 1px dashed'><i>Hint</i></div> 
INPUT DETAILS: 

Five cows with milk outputs of 1..5 

OUTPUT DETAILS: 

1 and 2 are below 3; 4 and 5 are above 3.
</div>

Source

#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{    int i,j,a,b,c[100000],n,m,t;

while(scanf("%d",&n)!=EOF)
{
    

    for(i=0;i<n;i++)
        scanf("%d",&c[i]);

    sort(c,c+n);//默认升序

    printf("%d\n",c[(n-1)/2]);
    
    
    
}

return 0;

}

USACO 2004 November