将矩阵元素相乘 N 次后打印矩阵
原文:https://www . geesforgeks . org/print-matrix-乘法后-matrix-elements-n-times/
给定一个正方形矩阵mat【】【】和一个整数 N ，任务是将矩阵乘以后打印矩阵。
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:思路是使用矩阵乘法 身份矩阵。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 <bits/stdc++.h>
using namespace std;
// Function for matrix multiplication
void power(vector<vector<int>>& I,
vector<vector<int>>& a,
int rows, int cols)
{
// Stores the resultant matrix
// after multiplying a[][] by I[][]
vector<vector<int>> res(rows, vector<int>(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<vector<int> >& 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<vector<int> >& arr, int N)
{
// Identity element of matrix
vector<vector<int>> I(arr.size(),
vector<int>(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<vector<int>> 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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