📜  C++程序将两个矩阵相乘

📅  最后修改于: 2022-05-13 01:54:31.532000             🧑  作者: Mango

C++程序将两个矩阵相乘

给定两个矩阵,将它们相乘的任务。矩阵可以是正方形或矩形。

例子:

Input : mat1[][] = {{1, 2}, 
                   {3, 4}}
        mat2[][] = {{1, 1}, 
                    {1, 1}}
Output : {{3, 3}, 
          {7, 7}}
Input : mat1[][] = {{2, 4}, 
                    {3, 4}}
        mat2[][] = {{1, 2}, 
                    {1, 3}}       
Output : {{6, 16}, 
          {7, 18}}

平方矩阵的乘法:
下面的程序将两个大小为 4*4 的方阵相乘,我们可以将 N 更改为不同的维度。

C++
// C++ program to multiply
// two square matrices.
#include 
  
using namespace std;
  
#define N 4
  
// This function multiplies
// mat1[][] and mat2[][], and
// stores the result in res[][]
void multiply(int mat1[][N],
              int mat2[][N],
              int res[][N])
{
    int i, j, k;
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++) {
            res[i][j] = 0;
            for (k = 0; k < N; k++)
                res[i][j] += mat1[i][k] * mat2[k][j];
        }
    }
}
  
// Driver Code
int main()
{
    int i, j;
    int res[N][N]; // To store result
    int mat1[N][N] = { { 1, 1, 1, 1 },
                       { 2, 2, 2, 2 },
                       { 3, 3, 3, 3 },
                       { 4, 4, 4, 4 } };
  
    int mat2[N][N] = { { 1, 1, 1, 1 },
                       { 2, 2, 2, 2 },
                       { 3, 3, 3, 3 },
                       { 4, 4, 4, 4 } };
  
    multiply(mat1, mat2, res);
  
    cout << "Result matrix is 
";
    for (i = 0; i < N; i++) {
        for (j = 0; j < N; j++)
            cout << res[i][j] << " ";
        cout << "
";
    }
  
    return 0;
}
  
// This code is contributed
// by Soumik Mondal


C++
// C++ program to multiply two
// rectangular matrices
#include 
using namespace std;
  
// Multiplies two matrices mat1[][]
// and mat2[][] and prints result.
// (m1) x (m2) and (n1) x (n2) are
// dimensions of given matrices.
void multiply(int m1, int m2, int mat1[][2], int n1, int n2,
              int mat2[][2])
{
    int x, i, j;
    int res[m1][n2];
    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            res[i][j] = 0;
            for (x = 0; x < m2; x++) 
            {
                *(*(res + i) + j) += *(*(mat1 + i) + x)
                                     * *(*(mat2 + x) + j);
            }
        }
    }
    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            cout << *(*(res + i) + j) << " ";
        }
        cout << "
";
    }
}
  
// Driver code
int main()
{
    int mat1[][2] = { { 2, 4 }, { 3, 4 } };
    int mat2[][2] = { { 1, 2 }, { 1, 3 } };
    int m1 = 2, m2 = 2, n1 = 2, n2 = 2;
    
    // Function call
    multiply(m1, m2, mat1, n1, n2, mat2);
    return 0;
}
  
// This code is contributed
// by Akanksha Rai(Abby_akku)


输出
Result matrix is 
10 10 10 10 
20 20 20 20 
30 30 30 30 
40 40 40 40

时间复杂度: O(n 3 )。可以使用 Strassen 的矩阵乘法进行优化

辅助空间: O(n 2 )

矩形矩阵的乘法:
我们在 C 中使用指针来乘以矩阵。请参阅以下帖子作为代码的先决条件。
如何在C中将二维数组作为参数传递?

C++

// C++ program to multiply two
// rectangular matrices
#include 
using namespace std;
  
// Multiplies two matrices mat1[][]
// and mat2[][] and prints result.
// (m1) x (m2) and (n1) x (n2) are
// dimensions of given matrices.
void multiply(int m1, int m2, int mat1[][2], int n1, int n2,
              int mat2[][2])
{
    int x, i, j;
    int res[m1][n2];
    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            res[i][j] = 0;
            for (x = 0; x < m2; x++) 
            {
                *(*(res + i) + j) += *(*(mat1 + i) + x)
                                     * *(*(mat2 + x) + j);
            }
        }
    }
    for (i = 0; i < m1; i++) 
    {
        for (j = 0; j < n2; j++) 
        {
            cout << *(*(res + i) + j) << " ";
        }
        cout << "
";
    }
}
  
// Driver code
int main()
{
    int mat1[][2] = { { 2, 4 }, { 3, 4 } };
    int mat2[][2] = { { 1, 2 }, { 1, 3 } };
    int m1 = 2, m2 = 2, n1 = 2, n2 = 2;
    
    // Function call
    multiply(m1, m2, mat1, n1, n2, mat2);
    return 0;
}
  
// This code is contributed
// by Akanksha Rai(Abby_akku)
输出
6 16 
7 18

时间复杂度: O(n 3 )。可以使用 Strassen 的矩阵乘法进行优化

辅助空间: O(m1 * n2)

有关更多详细信息,请参阅有关将两个矩阵相乘的完整文章!