通过将每个元素转换为其数字的异或来修改矩阵
给定维度为 M*N 的矩阵arr【】【】【】，任务是将每个矩阵元素转换为元素中数字的位异或。
示例:
输入: arr[][] = {{27，173}，{5，21}} 输出: 5 5 5 3 t8】解释: arr[0]0的数字按位异或是 5 (2^7). arr[0]1的数字按位异或值为 5 (1 ^ 7 ^ 3)。 arr[1]0位数的按位异或值为 5。 arr[1]1的数字按位异或值为 3(1 ^ 2)。
输入: arr[][] = {{11，12，33}，{64，57，61}，{74，88，39}} 输出: 0 3 0 2 2 7 3 0 10
逼近:为了解决这个问题，逼近思想是遍历给定的矩阵，对于每个矩阵元素，打印其数字的按位异或。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int M = 3;
const int N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
int findXOR(int X)
{
// Stores the Bitwise XOR
int ans = 0;
// While X is true
while (X) {
// Update Bitwise
// XOR of its digits
ans ^= (X % 10);
X /= 10;
}
// Return the result
return ans;
}
// Function to print matrix after
// converting each matrix element
// to XOR of its digits
void printXORmatrix(int arr[M][N])
{
// Traverse each row of arr[][]
for (int i = 0; i < M; i++) {
// Traverse each column of arr[][]
for (int j = 0; j < N; j++) {
cout << arr[i][j] << " ";
}
cout << "\n";
}
}
// Function to convert the given
// matrix to required XOR matrix
void convertXOR(int arr[M][N])
{
// Traverse each row of arr[][]
for (int i = 0; i < M; i++) {
// Traverse each column of arr[][]
for (int j = 0; j < N; j++) {
// Store the current
// matrix element
int X = arr[i][j];
// Find the xor of
// digits present in X
int temp = findXOR(X);
// Stores the XOR value
arr[i][j] = temp;
}
}
// Print resultant matrix
printXORmatrix(arr);
}
// Driver Code
int main()
{
int arr[][3] = { { 27, 173, 5 },
{ 21, 6, 624 },
{ 5, 321, 49 } };
convertXOR(arr);
return 0;
}
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{
static int M = 3;
static int N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
static int findXOR(int X)
{
// Stores the Bitwise XOR
int ans = 0;
// While X is true
while (X != 0)
{
// Update Bitwise
// XOR of its digits
ans ^= (X % 10);
X /= 10;
}
// Return the result
return ans;
}
// Function to print matrix after
// converting each matrix element
// to XOR of its digits
static void printXORmatrix(int arr[][])
{
// Traverse each row of arr[][]
for(int i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(int j = 0; j < N; j++)
{
System.out.print(arr[i][j] + " ");
}
System.out.println();
}
}
// Function to convert the given
// matrix to required XOR matrix
static void convertXOR(int arr[][])
{
// Traverse each row of arr[][]
for(int i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(int j = 0; j < N; j++)
{
// Store the current
// matrix element
int X = arr[i][j];
// Find the xor of
// digits present in X
int temp = findXOR(X);
// Stores the XOR value
arr[i][j] = temp;
}
}
// Print resultant matrix
printXORmatrix(arr);
}
// Driver Code
public static void main (String[] args)
{
int arr[][] = { { 27, 173, 5 },
{ 21, 6, 624 },
{ 5, 321, 49 } };
convertXOR(arr);
}
}
// This code is contributed by sanjoy_62
Python 3
# Python3 program for the above approach
M = 3
N = 3
# Function to calculate Bitwise
# XOR of digits present in X
def findXOR(X):
# Stores the Bitwise XOR
ans = 0
# While X is true
while (X):
# Update Bitwise
# XOR of its digits
ans ^= (X % 10)
X //= 10
# Return the result
return ans
# Function to print matrix after
# converting each matrix element
# to XOR of its digits
def printXORmatrix(arr):
# Traverse each row of arr[][]
for i in range(3):
# Traverse each column of arr[][]
for j in range(3):
print(arr[i][j], end = " ")
print()
# Function to convert the given
# matrix to required XOR matrix
def convertXOR(arr):
# Traverse each row of arr[][]
for i in range(3):
# Traverse each column of arr[][]
for j in range(3):
# Store the current
# matrix element
X = arr[i][j]
# Find the xor of
# digits present in X
temp = findXOR(X)
# Stores the XOR value
arr[i][j] = temp
# Print resultant matrix
printXORmatrix(arr)
# Driver Code
if __name__ == '__main__':
arr=[[27, 173, 5],
[ 21, 6, 624 ],
[ 5, 321, 49 ]]
convertXOR(arr)
# This code is contributed by mohit kumar 29.
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
static int M = 3;
static int N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
static int findXOR(int X)
{
// Stores the Bitwise XOR
int ans = 0;
// While X is true
while (X != 0)
{
// Update Bitwise
// XOR of its digits
ans ^= (X % 10);
X /= 10;
}
// Return the result
return ans;
}
// Function to print matrix after
// converting each matrix element
// to XOR of its digits
static void printXORmatrix(int[,] arr)
{
// Traverse each row of arr[][]
for(int i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(int j = 0; j < N; j++)
{
Console.Write(arr[i, j] + " ");
}
Console.WriteLine();
}
}
// Function to convert the given
// matrix to required XOR matrix
static void convertXOR(int[,] arr)
{
// Traverse each row of arr[][]
for(int i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(int j = 0; j < N; j++)
{
// Store the current
// matrix element
int X = arr[i, j];
// Find the xor of
// digits present in X
int temp = findXOR(X);
// Stores the XOR value
arr[i, j] = temp;
}
}
// Print resultant matrix
printXORmatrix(arr);
}
// Driver Code
static public void Main()
{
int[,] arr = { { 27, 173, 5 },
{ 21, 6, 624 },
{ 5, 321, 49 } };
convertXOR(arr);
}
}
// This code is contributed by splevel62
java 描述语言
<script>
// JavaScript program to implement
// the above approach
let M = 3;
let N = 3;
// Function to calculate Bitwise
// XOR of digits present in X
function findXOR(X)
{
// Stores the Bitwise XOR
let ans = 0;
// While X is true
while (X != 0)
{
// Update Bitwise
// XOR of its digits
ans ^= (X % 10);
X /= 10;
}
// Return the result
return ans;
}
// Function to print matrix after
// converting each matrix element
// to XOR of its digits
function printXORmatrix(arr)
{
// Traverse each row of arr[][]
for(let i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(let j = 0; j < N; j++)
{
document.write(arr[i][j] + " ");
}
document.write("<br/>");
}
}
// Function to convert the given
// matrix to required XOR matrix
function convertXOR(arr)
{
// Traverse each row of arr[][]
for(let i = 0; i < M; i++)
{
// Traverse each column of arr[][]
for(let j = 0; j < N; j++)
{
// Store the current
// matrix element
let X = arr[i][j];
// Find the xor of
// digits present in X
let temp = findXOR(X);
// Stores the XOR value
arr[i][j] = temp;
}
}
// Print resultant matrix
prletXORmatrix(arr);
}
// Driver code
let arr = [[ 27, 173, 5 ],
[ 21, 6, 624 ],
[ 5, 321, 49 ]];
convertXOR(arr);;
// This code is contributed by susmitakundugoaldanga.
</script>
Output:
5 5 5
3 6 0
5 0 13
时间复杂度: O(MNlog 10 K)其中 K 是 矩阵中存在的最大元素 。 辅助空间: O(1)
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