7-6 Exchange minimum and maximum scores 15

This question requires writing a program to first exchange the minimum value of a series of input integers with the first number, then exchange the maximum value with the last number, and finally output the exchanged sequence.

Note: The question ensures that the maximum and minimum values ​​are unique.

Input format:

The input gives a positive integer N (≤10) in the first line, and N integers in the second line, separated by spaces.

Output format:

Print the swapped sequence sequentially on a line, with each integer followed by a space.

Input example:

5
8 2 5 1 4

Output sample:

1 2 5 4 8 

Code length limit

16 KB

time limit

400 ms

memory limit

64 MB

#include <stdio.h>

int arrmin(int *arr, int length)
{

    int m = arr[0];
    for (size_t i = 0; i < length; i++)
    {
        if (arr[i] < m)
        {
            m = arr[i];
        }
    }

    return m;
}

int arrmax(int *arr, int length)
{

    int m = arr[0];
    for (size_t i = 0; i < length; i++)
    {
        if (arr[i] > m)
        {
            m = arr[i];
        }
    }

    return m;
}

void inputarr(int *arr)
{
    int num, i;
    char ch;
    i = 0;
    do
    {

        scanf("%d", &num);
        ch = getchar();
        arr[i] = num;
        i++;

    } while (ch != '\n');
}

int printarr(int *arr, int length)
{
    if (sizeof(arr) == 0)
    {
        return 0;
    }
    size_t i;
    for (i = 0; i < length; i++)
    {
        printf("%d%c", arr[i], 32);
    }
    putchar(' \n');

    return 1;
}

int main()
{
    // 输入样例                8 2 5 1 4
    // 最小值与第一个数交换     1 2 5 8 4
    // 最大值与最后一个数交换   1 2 5 4 8
    // 预期结果                1 2 5 4 8

    int n, min, max;
    scanf("%d", &n);
    int arr[n];
    inputarr(arr);

    min = arrmin(arr, n);
    max = arrmax(arr, n);

    for (size_t i = 0; i < n; i++)
    {
        if (arr[i] == min)
        {
            arr[i] = arr[0]; // 最小值与第一个数值交换
            arr[0] = min;    // 第一个数值与最小值交换
        }
    }

    for (size_t i = 0; i < n; i++)
    {
        if (arr[i] == max)
        {
            arr[i] = arr[n - 1]; // 最大值与最后一个数交换
            arr[n - 1] = max;    // 最后一个值与最大值交换
        }
    }

    printarr(arr, n);

    return 0;
}