POJ 2623 Sequence Median_不见不散的结局是曲终人散_新浪博客

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最新推荐文章于 2022-01-23 21:02:48 发布

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本文介绍了一种求解给定非负整数序列中位数的方法，包括处理偶数和奇数长度序列的情况，以及如何通过排序找到中位数。

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Sequence Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 16366		Accepted: 4560

Description

Given a sequence of N nonnegative integers. Let's define the median of such sequence. If N is odd the median is the element with stands in the middle of the sequence after it is sorted. One may notice that in this case the median has position (N+1)/2 in sorted sequence if sequence elements are numbered starting with 1. If N is even then the median is the semi-sum of the two "middle" elements of sorted sequence. I.e. semi-sum of the elements in positions N/2 and (N/2)+1 of sorted sequence. But original sequence might be unsorted.

Your task is to write program to find the median of given sequence.

Input

The first line of input contains the only integer number N - the length of the sequence. Sequence itself follows in subsequent lines, one number in a line. The length of the sequence lies in the range from 1 to 250000. Each element of the sequence is a positive integer not greater than 2^32 - 1 inclusive.

Output

You should print the value of the median with exactly one digit after decimal point.

Sample Input

Sample Output

4.5

Hint

Huge input,scanf is recommended.

题意：求中位数，如果中位数有两个，求两个和的平均；

代码如下：

#include
#include
#include
using namespace std;
int main()
{
    int n,m[250000];
    while(~scanf("%d",&n))
    {
        for(int i=0;i
        {
            scanf("%d",&m[i]);
        }
        sort(m,m+n);
        if(!(n%2))
        {
             printf("%.1f\n",(m[n/2-1]/2.0+m[n/2]/2.0));//注意坐标要减一；
        }
        else
        {
             printf("%.1f\n",m[(n+1)/2-1]/(1.0));
        }
    }
}