用二维数组作为函数参数，实现矩阵转置。按如下函数原型编程计算并输出m×n阶矩阵的转置矩阵。其中，m和n的值由用户从键盘输入。已知m和n的值都不超过10。 void Transpose(int a[][N], int at[][M], int m, int n); void InputMatrix(int a[][N], int m, int n); void PrintMatrix(int at[][M], int n, int m); 输入提示信息："Input m, n:" "Input %d*%d matrix:\n" 输入格式："%d,%d" 输出提示信息和格式："The transposed matrix is:\n" 输出格式（输完矩阵的一行要换行）："%d\t"

好的，我明白了。以下是代码实现：

#include <iostream>
using namespace std;

const int N = 10, M = 10;

void InputMatrix(int a[][N], int m, int n);
void Transpose(int a[][N], int at[][M], int m, int n);
void PrintMatrix(int at[][M], int n, int m);

int main()
{
    int m, n;
    cout << "Input m, n:";
    cin >> m >> n;

    int a[N][N], at[M][M];
    InputMatrix(a, m, n);
    Transpose(a, at, m, n);
    PrintMatrix(at, n, m);

    return 0;
}

void InputMatrix(int a[][N], int m, int n)
{
    cout << "Input " << m << "*" << n << " matrix:" << endl;
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            cin >> a[i][j];
}

void Transpose(int a[][N], int at[][M], int m, int n)
{
    for (int i = 0; i < m; i++)
        for (int j = 0; j < n; j++)
            at[j][i] = a[i][j];
}

void PrintMatrix(int at[][M], int n, int m)
{
    cout << "The transposed matrix is:" << endl;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
            cout << at[i][j] << "\t";
        cout << endl;
    }
}

希望能帮到你！