原文:
给定一个正方形mat【】【】和一个整数 n ，任务是将矩阵乘以后。
示例:
输入: mat[][] = {{1，2，3}，{3，4，5}，{6，7，9}}，n = 2 t3】输出:t5】25 31 40 45 57 74 81 103 134
输入: mat[][] = {{1，2}，{3，4}}，n = 3 t3】输出:t5】37 54 81 118
进场:思路是使用 。即 a = ia 、 a = ai ，其中 a 为 n * m 阶次尺寸的矩阵， i 为尺寸 m * n 的恒等式矩阵，其中 n 为总行数， m 为矩阵中的总列数。
其思路是迭代范围【1，n】，用 a.i 更新身份矩阵，这样在计算出a2t9】的值后= a.a ，a3t15】就可以计算为a . a2t19】等等，直到 a 【t21**
下面是上述方法的实现:
c
// c program for the above approach
#include
using namespace std;
// function for matrix multiplication
void power(vector>& i,
vector>& a,
int rows, int cols)
{
// stores the resultant matrix
// after multiplying a[][] by i[][]
vector> res(rows, vector(cols));
// matrix multiplication
for(int i = 0; i < rows; i)
{
for(int j = 0; j < cols; j)
{
for(int k = 0; k < rows; k)
{
res[i][j] = a[i][k] * i[k][j];
}
}
}
// updating identity element
// of a matrix
for(int i = 0; i < rows; i)
{
for(int j = 0; j < cols; j)
{
i[i][j] = res[i][j];
}
}
}
// function to print the given matrix
void print(vector >& a)
{
// traverse the row
for(int i = 0; i < a.size(); i)
{
// traverse the column
for(int j = 0; j < a[0].size(); j)
{
cout << a[i][j] << " ";
}
cout << "\n";
}
}
// function to multiply the given
// matrix n times
void multiply(vector >& arr, int n)
{
// identity element of matrix
vector> i(arr.size(),
vector(arr[0].size()));
// update the identity matrix
for(int i = 0; i < arr.size(); i)
{
for(int j = 0; j < arr[0].size(); j)
{
// for the diagonal element
if (i == j)
{
i[i][j] = 1;
}
else
{
i[i][j] = 0;
}
}
}
// multiply the matrix n times
for(int i = 1; i <= n; i)
{
power(i, arr, arr.size(), arr[0].size());
}
// update the matrix arr[i] to
// to identity matrix
for(int i = 0; i < arr.size(); i)
{
for(int j = 0; j < arr[0].size(); j)
{
arr[i][j] = i[i][j];
}
}
// print the matrix
print(arr);
}
// driver code
int main()
{
// given 2d array
vector> arr = { { 1, 2, 3 },
{ 3, 4, 5 },
{ 6, 7, 9 } };
// given n
int n = 2;
// function call
multiply(arr, n);
return 0;
}
// this code is contributed by akhilsaini
java 语言(一种计算机语言，尤用于创建网站)
// java program for the above approach
import java.io.*;
class gfg {
// function for matrix multiplication
static void power(int i[][], int a[][],
int rows, int cols)
{
// stores the resultant matrix
// after multiplying a[][] by i[][]
int res[][] = new int[rows][cols];
// matrix multiplication
for (int i = 0; i < rows; i) {
for (int j = 0;
j < cols; j) {
for (int k = 0;
k < rows; k) {
res[i][j] = a[i][k]
* i[k][j];
}
}
}
// updating identity element
// of a matrix
for (int i = 0; i < rows; i) {
for (int j = 0; j < cols; j) {
i[i][j] = res[i][j];
}
}
}
// function to print the given matrix
static void print(int a[][])
{
// traverse the row
for (int i = 0;
i < a.length; i) {
// traverse the column
for (int j = 0;
j < a[0].length; j) {
system.out.print(a[i][j]
" ");
}
system.out.println();
}
}
// function to multiply the given
// matrix n times
public static void multiply(
int arr[][], int n)
{
// identity element of matrix
int i[][]
= new int[arr.length][arr[0].length];
// update the identity matrix
for (int i = 0; i < arr.length; i) {
for (int j = 0;
j < arr[0].length; j) {
// for the diagonal element
if (i == j) {
i[i][j] = 1;
}
else {
i[i][j] = 0;
}
}
}
// multiply the matrix n times
for (int i = 1; i <= n; i) {
power(i, arr, arr.length,
arr[0].length);
}
// update the matrix arr[i] to
// to identity matrix
for (int i = 0;
i < arr.length; i) {
for (int j = 0;
j < arr[0].length; j) {
arr[i][j] = i[i][j];
}
}
// print the matrix
print(arr);
}
// driver code
public static void main(string[] args)
{
// given 2d array
int arr[][]
= { { 1, 2, 3 },
{ 3, 4, 5 },
{ 6, 7, 9 } };
// given n
int n = 2;
// function call
multiply(arr, n);
}
}
python 3
# python3 program for the above approach
# function for matrix multiplication
def power(i, a, rows, cols):
# stores the resultant matrix
# after multiplying a[][] by i[][]
res = [[0 for i in range(cols)]
for j in range(rows)]
# matrix multiplication
for i in range(0, rows):
for j in range(0, cols):
for k in range(0, rows):
res[i][j] = a[i][k] * i[k][j]
# updating identity element
# of a matrix
for i in range(0, rows):
for j in range(0, cols):
i[i][j] = res[i][j]
# function to print the given matrix
def prints(a):
# traverse the row
for i in range(0, len(a)):
# traverse the column
for j in range(0, len(a[0])):
print(a[i][j], end = ' ')
print()
# function to multiply the given
# matrix n times
def multiply(arr, n):
# identity element of matrix
i = [[1 if i == j else 0 for i in range(
len(arr))] for j in range(
len(arr[0]))]
# multiply the matrix n times
for i in range(1, n 1):
power(i, arr, len(arr), len(arr[0]))
# update the matrix arr[i] to
# to identity matrix
for i in range(0, len(arr)):
for j in range(0, len(arr[0])):
arr[i][j] = i[i][j]
# print the matrix
prints(arr)
# driver code
if __name__ == '__main__':
# given 2d array
arr = [ [ 1, 2, 3 ],
[ 3, 4, 5 ],
[ 6, 7, 9 ] ]
# given n
n = 2
# function call
multiply(arr, n)
# this code is contributed by akhilsaini
c
// c# program for the above approach
using system;
class gfg{
// function for matrix multiplication
static void power(int[,] i, int[,] a,
int rows, int cols)
{
// stores the resultant matrix
// after multiplying a[][] by i[][]
int[,] res = new int[rows, cols];
// matrix multiplication
for(int i = 0; i < rows; i)
{
for(int j = 0; j < cols; j)
{
for(int k = 0; k < rows; k)
{
res[i, j] = a[i, k] * i[k, j];
}
}
}
// updating identity element
// of a matrix
for(int i = 0; i < rows; i)
{
for(int j = 0; j < cols; j)
{
i[i, j] = res[i, j];
}
}
}
// function to print the given matrix
static void print(int[, ] a)
{
// traverse the row
for(int i = 0; i < a.getlength(0); i)
{
// traverse the column
for(int j = 0; j < a.getlength(1); j)
{
console.write(a[i, j] " ");
}
console.writeline();
}
}
// function to multiply the given
// matrix n times
public static void multiply(int[, ] arr, int n)
{
// identity element of matrix
int[, ] i = new int[arr.getlength(0),
arr.getlength(1)];
// update the identity matrix
for(int i = 0; i < arr.getlength(0); i)
{
for(int j = 0; j < arr.getlength(1); j)
{
// for the diagonal element
if (i == j)
{
i[i, j] = 1;
}
else
{
i[i, j] = 0;
}
}
}
// multiply the matrix n times
for(int i = 1; i <= n; i)
{
power(i, arr, arr.getlength(0),
arr.getlength(1));
}
// update the matrix arr[i] to
// to identity matrix
for(int i = 0; i < arr.getlength(0); i)
{
for(int j = 0; j < arr.getlength(1); j)
{
arr[i, j] = i[i, j];
}
}
// print the matrix
print(arr);
}
// driver code
public static void main()
{
// given 2d array
int[, ] arr = { { 1, 2, 3 },
{ 3, 4, 5 },
{ 6, 7, 9 } };
// given n
int n = 2;
// function call
multiply(arr, n);
}
}
// this code is contributed by akhilsaini
output:
25 31 40
45 57 74
81 103 134
时间复杂度:o(n3) 辅助空间: o(n)
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