解题报告——POJ 2623

Sequence Median

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 15792		Accepted: 4409

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

4
3
6
4
5

Sample Output

4.5

Hint

Huge input,scanf is recommended.

 

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这道题可以直接用快排进行计算，然后直接计算中位数。

需要注意的小地方如下：

1. 数组类型：采用long long的原因是为了计算sum，这个值超过了unsinged的范围

2. 浮点数输出：对于大数字（2^32 - 1），不能用浮点数表示，而应该将浮点数拆分为整数和小数进行拼装。

-----------------------------------------------------------------------------------------------

#include <cstdio>
#include <algorithm>

using namespace std;

int main()
{
    long long a[250005];
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        scanf("%lld", &a[i]);
    }
    sort(a, a + n);
    if(n & 1)
    {
        printf("%lld.0
", a[n / 2]);
    }
    else
    {
        long long sum = a[n / 2] + a[n / 2 - 1];
        printf("%lld", sum / 2);
        if(sum & 1)
        {
            printf(".5
");
        }
        else
        {
            printf(".0
");
        }
    }

    return 0;
}