将两个矩阵相乘的程序
原文： https://www.geeksforgeeks.org/c-program-multiply-two-matrices/
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。
示例：
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}}
方阵的乘积：
下面的程序将两个大小为4 * 4的方阵相乘，我们可以为不同的维数更改N。
C++
// C++ program to multiply
// two square matrices.
#include <iostream>
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 \n";
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
cout << res[i][j] << " ";
cout << "\n";
}
return 0;
}
// This code is contributed
// by Soumik Mondal
C
// C program to multiply two square matrices.
#include <stdio.h>
#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];
}
}
}
int main()
{
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}};
int res[N][N]; // To store result
int i, j;
multiply(mat1, mat2, res);
printf("Result matrix is \n");
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
printf("%d ", res[i][j]);
printf("\n");
}
return 0;
}
Java
// Java program to multiply two square
// matrices.
import java.io.*;
class GFG {
static int N = 4;
// This function multiplies mat1[][]
// and mat2[][], and stores the result
// in res[][]
static void multiply(int mat1[][],
int mat2[][], int res[][])
{
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
public static void main (String[] args)
{
int mat1[][] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
int mat2[][] = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
// To store result
int res[][] = new int[N][N] ;
int i, j;
multiply(mat1, mat2, res);
System.out.println("Result matrix"
+ " is ");
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
System.out.print( res[i][j]
+ " ");
System.out.println();
}
}
}
// This code is contributed by anuj_67\.
Python 3
# 4x4 matrix multiplication using Python3
# Function definition
def matrix_multiplication(M,N):
# List to store matrix multiplication result
R = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
for i in range(0, 4):
for j in range(0, 4):
for k in range(0, 4):
R[i][j] += M[i][k] * N[k][j]
for i in range(0,4):
for j in range(0,4):
#if we use print(), by default cursor moves to next line each time,
#Now we can explicitly define ending character or sequence passing
#second parameter as end="<character or string>"
#syntax: print(<variable or value to print>, end="<ending with>")
#Here space (" ") is used to print a gape after printing
#each element of R
print(R[i][j],end=" ")
print("\n",end="")
# First matrix. M is a list
M = [[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3],
[4, 4, 4, 4]]
# Second matrix. N is a list
N = [[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3],
[4, 4, 4, 4]]
# Call matrix_multiplication function
matrix_multiplication(M,N)
# This code is contributed by Santanu
C#
// C# program to multiply two square
// matrices.
using System;
class GFG {
static int N = 4;
// This function multiplies mat1[][]
// and mat2[][], and stores the result
// in res[][]
static void multiply(int[,] mat1,
int [,]mat2, int [,]res)
{
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
public static void Main ()
{
int [,]mat1 = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
int [,]mat2 = { {1, 1, 1, 1},
{2, 2, 2, 2},
{3, 3, 3, 3},
{4, 4, 4, 4}};
// To store result
int [,]res = new int[N,N] ;
int i, j;
multiply(mat1, mat2, res);
Console.WriteLine("Result matrix"
+ " is ");
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
Console.Write( res[i,j]
+ " ");
Console.WriteLine();
}
}
}
// This code is contributed by anuj_67\.
PHP
<?php
// PHP program to multiply two
// square matrices.
// This function multiplies mat1[][] and
// mat2[][], and stores the result in res[][]
function multiply(&$mat1, &$mat2, &$res)
{
$N = 4;
for ($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
$mat1 = array(array(1, 1, 1, 1),
array(2, 2, 2, 2),
array(3, 3, 3, 3),
array(4, 4, 4, 4));
$mat2 = array(array(1, 1, 1, 1),
array(2, 2, 2, 2),
array(3, 3, 3, 3),
array(4, 4, 4, 4));
multiply($mat1, $mat2, $res);
$N = 4;
echo ("Result matrix is \n");
for ($i = 0; $i < $N; $i++)
{
for ($j = 0; $j < $N; $j++)
{
echo ($res[$i][$j]);
echo(" ");
}
echo ("\n");
}
// This code is contributed
// by Shivi_Aggarwal
?>
输出：
Result matrix is
10 10 10 10
20 20 20 20
30 30 30 30
40 40 40 40
矩形矩阵的乘法：
我们在 C 中使用指针来乘法到矩阵。 请参考以下帖子作为代码的前提条件。
C++
// C++ program to multiply two
// rectangular matrices
#include<bits/stdc++.h>
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 << "\n";
}
}
// 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;
multiply(m1, m2, mat1, n1, n2, mat2);
return 0;
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
C
// C program to multiply two square matrices.
#include <stdio.h>
#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];
}
}
}
int main()
{
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 } };
int res[N][N]; // To store result
int i, j;
multiply(mat1, mat2, res);
printf("Result matrix is \n");
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++)
printf("%d ", res[i][j]);
printf("\n");
}
return 0;
}
Java
// Java program to multiply two square
// matrices.
import java.io.*;
class GFG {
static int N = 4;
// This function multiplies mat1[][]
// and mat2[][], and stores the result
// in res[][]
static void multiply(int mat1[][],
int mat2[][], int res[][])
{
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
public static void main(String[] args)
{
int mat1[][] = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
int mat2[][] = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
// To store result
int res[][] = new int[N][N];
int i, j;
multiply(mat1, mat2, res);
System.out.println("Result matrix"
+ " is ");
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++)
System.out.print(res[i][j]
+ " ");
System.out.println();
}
}
}
// This code is contributed by anuj_67.
Python 3
# 4x4 matrix multiplication using Python3
# Function definition
def matrix_multiplication(M, N):
# List to store matrix multiplication result
R = [[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0],
[0, 0, 0, 0]]
for i in range(0, 4):
for j in range(0, 4):
for k in range(0, 4):
R[i][j] += M[i][k] * N[k][j]
for i in range(0, 4):
for j in range(0, 4):
# if we use print(), by default cursor moves to next line each time,
# Now we can explicitly define ending character or sequence passing
# second parameter as end ="<character or string>"
# syntax: print(<variable or value to print>, end ="<ending with>")
# Here space (" ") is used to print a gape after printing
# each element of R
print(R[i][j], end =" ")
print("\n", end ="")
# First matrix. M is a list
M = [[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3],
[4, 4, 4, 4]]
# Second matrix. N is a list
N = [[1, 1, 1, 1],
[2, 2, 2, 2],
[3, 3, 3, 3],
[4, 4, 4, 4]]
# Call matrix_multiplication function
matrix_multiplication(M, N)
# This code is contributed by Santanu
C
// C# program to multiply two square
// matrices.
using System;
class GFG {
static int N = 4;
// This function multiplies mat1[][]
// and mat2[][], and stores the result
// in res[][]
static void multiply(int[, ] mat1,
int[, ] mat2, int[, ] res)
{
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
public static void Main()
{
int[, ] mat1 = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
int[, ] mat2 = { { 1, 1, 1, 1 },
{ 2, 2, 2, 2 },
{ 3, 3, 3, 3 },
{ 4, 4, 4, 4 } };
// To store result
int[, ] res = new int[N, N];
int i, j;
multiply(mat1, mat2, res);
Console.WriteLine("Result matrix"
+ " is ");
for (i = 0; i < N; i++) {
for (j = 0; j < N; j++)
Console.Write(res[i, j]
+ " ");
Console.WriteLine();
}
}
}
// This code is contributed by anuj_67.
PHP
<?php
// PHP program to multiply two
// square matrices.
// This function multiplies mat1[][] and
// mat2[][], and stores the result in res[][]
function multiply(&$mat1, &$mat2, &$res)
{
$N = 4;
for ($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
$mat1 = array(array(1, 1, 1, 1),
array(2, 2, 2, 2),
array(3, 3, 3, 3),
array(4, 4, 4, 4));
$mat2 = array(array(1, 1, 1, 1),
array(2, 2, 2, 2),
array(3, 3, 3, 3),
array(4, 4, 4, 4));
multiply($mat1, $mat2, $res);
$N = 4;
echo ("Result matrix is \n");
for ($i = 0; $i < $N; $i++)
{
for ($j = 0; $j < $N; $j++)
{
echo ($res[$i][$j]);
echo(" ");
}
echo ("\n");
}
// This code is contributed
// by Shivi_Aggarwal
?>
输出：
Result matrix is
10 10 10 10
20 20 20 20
30 30 30 30
40 40 40 40
矩形矩阵的乘法：
我们在 C 中使用指针来乘以矩阵。 请参考以下帖子作为代码的前提条件。
C++
// C++ program to multiply two
// rectangular matrices
#include <bits/stdc++.h>
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 << "\n";
}
}
// 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)
C
// C program to multiply two rectangular matrices
#include <stdio.h>
// 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[][m2], int n1,
int n2, int mat2[][n2])
{
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++)
{
printf("%d ", *(*(res + i) + j));
}
printf("\n");
}
}
// 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;
}
JAVA
// Java program to multiply two matrices.
public class GFG
{
/**
* to find out matrix multiplication
*
* @param matrix1 First matrix
* @param rows1 Number of rows in matrix 1
* @param cols1 Number of columns in matrix 1
* @param matrix2 Second matrix
* @param rows2 Number of rows in matrix 2
* @param cols2 Number of columns in matrix 2
* @return the result matrix (matrix 1 and matrix 2
* multiplication)
*/
public static int[][] matrixMultiplication(
int[][] matrix1, int rows1, int cols1,
int[][] matrix2, int rows2, int cols2)
throws Exception
{
// Required condition for matrix multiplication
if (cols1 != rows2) {
throw new Exception("Invalid matrix given.");
}
// create a result matrix
int resultMatrix[][] = new int[rows1][cols2];
// Core logic for 2 matrices multiplication
for (int i = 0; i < resultMatrix.length; i++)
{
for (int j = 0;
j < resultMatrix[i].length;
j++)
{
for (int k = 0; k < cols1; k++)
{
resultMatrix[i][j]
+= matrix1[i][k] * matrix2[k][j];
}
}
}
return resultMatrix;
}
// Driver code
public static void main(String[] args) throws Exception
{
// Initial matrix 1 and matrix 2
int matrix1[][] = { { 2, 4 }, { 3, 4 } };
int matrix2[][] = { { 1, 2 }, { 1, 3 } };
// Function call to get a matrix multiplication
int resultMatrix[][] = matrixMultiplication(
matrix1, 2, 2, matrix2, 2, 2);
// Display result matrix
System.out.println("Result Matrix is:");
for (int i = 0; i < resultMatrix.length; i++)
{
for (int j = 0;
j < resultMatrix[i].length;
j++)
{
System.out.print(resultMatrix[i][j] + "\t");
}
System.out.println();
}
}
// This code is contributed by darshatandel1998 (Darshan
// Tandel)
}
PHP
<?php
// PHP program to multiply two
// rectangular matrices
// Multiplies two matrices mat1[][]
// and mat2[][] and prints result.
// (m1) x (m2) and (n1) x (n2) are
// dimensions of given matrices.
function multiply($m1, $m2, $mat1,
$n1, $n2, $mat2)
{
for ($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++)
{
echo $res[$i][$j] . " ";
}
echo "\n";
}
}
// Driver code
$mat1 = array( array( 2, 4 ), array( 3, 4 ));
$mat2 = array( array( 1, 2 ), array( 1, 3 ));
$m1 = 2;
$m2 = 2;
$n1 = 2;
$n2 = 2;
//Function call
multiply($m1, $m2, $mat1, $n1, $n2, $mat2);
// This code is contributed by rathbhupendra
?>
输出：
6 16
7 18
时间复杂度：O(n3)。 可以使用 Strassen 的矩阵乘法对其进行优化。
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