poj 2388 Who's in the Middle

最新推荐文章于 2021-01-17 20:42:53 发布

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本文探讨了在给定奇数数量奶牛及其产奶量的情况下，如何使用快速排序算法找到中位数产奶量。通过实例输入演示了如何识别在产量上位于中间位置的奶牛，即至少一半的奶牛产量不低于该值，另一半不低于或等于该值。此过程展示了在农业数据处理中应用排序算法解决实际问题的方法。

Who's in the Middle
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 25458 Accepted: 14639

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
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.

// 给定n个数，求中间值，所以排序，输出中间值。
// 快速排序做的
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
using namespace std;
int cmp(const void *a,const void *b)
{
    return *(int *)a-*(int *)b;
}
int main()
{
    int n,a[10001];
    int i;
    cin>>n;
    for(i=1;i<=n;i++)
    cin>>a[i];
    qsort(a+1,n,sizeof(a[1]),cmp);
    if(n&1)
    printf("%d\n",a[(n>>1)+1]);
    else
    printf("%d\n",a[n>>1]);
    return 0;
}