Median Value

Problem A: Median Value

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 874 Solved: 307

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Description

Figure out the median of a floating point number array. If the size of an array is an odd number, the median is the middle element after arranging all the array elements from lowest value to highest value; If there is an even number of elements, then there is no single middle value, so the median is the mean of the two middle values.

Input

The input may contain several test cases.

The first line of each test case is an integer n (n >= 1), representing the size of the array.

The second line of each test case contains all the elements in the array.

Input is terminated by EOF.

Output

For each test case, output the median , which must be formatted as a floating point number with exactly two digit after the decimal point.

Sample Input

6

5.0 4.0 3.0 2.0 11.0 3.0

11

5.0 6.0 222.0 23.0 23.0 4.0 2.0 5.0 99.0 1.0 8.0

Sample Output

3.50

6.00

问题描述：

该题为水题，只需注意数组中长度。数组长度为奇数时，输出中间值即可；若数组长度为偶数，输出中间两数和的一半。

另外注意输出保留小数点位两位，采用cout<

#include <iostream>
#include <iomanip>
using namespace std;
void sort(float* shu, int length) {
    int i, j;
    float temp;
    for (i = 0; i<length - 1; ++i) {
        for (j = 0; j<length - i - 1; j++) {
            if (shu[j + 1]<shu[j]) {
                temp = shu[j + 1];
                shu[j + 1] = shu[j];
                shu[j] = temp;
            }
        }
    }
}
int main() {
    int n;
    float a[101];
    while (cin >> n)
    {
        for (int i = 0; i < n; i++)
            cin >> a[i];
        sort(a, n);
        if (n % 2 == 0)
            cout << fixed << setprecision(2) << ((a[n / 2] + a[n / 2 - 1]) / 2) << endl;
        else
            cout << fixed << setprecision(2) << a[n / 2] << endl;
    }
    return 0;
}
/**************************************************************
    Problem: 1009
    User: 20151000332
    Language: C++
    Result: Accepted
    Time:12 ms
    Memory:1268 kb