按如下函数原型编程从键盘输入一个m行n列的二维数组，然后根据输入mode值将数组进行水平翻转，垂直翻转或转置操作。其中数组行列数m，n，操作方式mode的值由用户键盘输入。已知m和n的值都不超过10，mode=0时水平翻转数组，mode=1时垂直翻转数组，mode =2时转置数组。向二维数组输入数据,数组输出和操作数组的函数原型如 下： void InputArray(int a[][N],int m,int n); void OutputArray(int a[][N],int m, int n); int ArrayOp(int a[][N],int m,int n,int mode); 程序运行结果示例： Input m,n: 3,3↙ Input 3*3 array: 2 5 7↙ 9 0 2↙ 3 -4 9↙ Input operation mode: 0 After operation: 7 5 2 2 0 9 9 -4 3 Input m,n: 3,3↙ Input 3*3 array: 2 5 7↙ 9 0 2↙ 3 -4 9↙ Input operation mode: 1 After operation: 3 -4 9 9 0 2 2 5 7 Input m,n: 3,3↙ Input 3*3 array: 2 5 7↙ 9 0 2↙ 3 -4 9↙ Input operation mode: 2 After operation: 2 9 3 5 0 -4 7 2 9 输入格式: 输入数组大小："%d,%d" 输入数组元素："%d" 输入操作模式："%d" 输出格式： 数组大小输入提示信息："Input m,n:\n" 数组元素输入提示信息："Input %d*%d array:\n" 操作模式输入提示信息 "Input operation mode:\n" 操作后数组输出提示信息"After operation:\n" 数组输出格式： "%4d"

#include <stdio.h>

#define N 10

void InputArray(int a[][N], int m, int n);
void OutputArray(int a[][N], int m, int n);
int ArrayOp(int a[][N], int m, int n, int mode);

int main()
{
    int a[N][N];
    int m, n, mode;

    printf("Input m,n:\n");
    scanf("%d,%d", &m, &n);

    InputArray(a, m, n);

    printf("Input operation mode:\n");
    scanf("%d", &mode);

    printf("After operation:\n");
    ArrayOp(a, m, n, mode);
    OutputArray(a, m, n);

    return 0;
}

void InputArray(int a[][N], int m, int n)
{
    printf("Input %d*%d array:\n", m, n);
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            scanf("%d", &a[i][j]);
        }
    }
}

void OutputArray(int a[][N], int m, int n)
{
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            printf("%4d", a[i][j]);
        }
        printf("\n");
    }
}

int ArrayOp(int a[][N], int m, int n, int mode)
{
    int tmp[N][N];
    switch (mode)
    {
    case 0: //水平翻转
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                tmp[i][n - j - 1] = a[i][j];
            }
        }
        break;

    case 1: //垂直翻转
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                tmp[m - i - 1][j] = a[i][j];
            }
        }
        break;

    case 2: //转置
        for (int i = 0; i < m; i++)
        {
            for (int j = 0; j < n; j++)
            {
                tmp[j][i] = a[i][j];
            }
        }
        int t = m;
        m = n;
        n = t;
        break;
    }
    //将tmp中的数据拷贝回a
    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            a[i][j] = tmp[i][j];
        }
    }
    return 0;
}