查找一个N x N网格，其每行和每列的xor相等 _XOR

本文概述

C ++
Java
Python3
C#

给定一个整数N，它是4的倍数，任务是找到一个N × N的网格，其每一行和每一列的位异或是相同的。
例子：

输入：N = 4
输出：
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
输入：N = 8
输出：
0 1 2 3 16 17 18 19
4 5 6 7 20 21 22 23
8 9 10 11 24 25 26 27
12 13 14 15 28 29 30 31
32 33 34 35 48 49 50 51
36 37 38 39 52 53 54 55
40 41 42 43 56 57 58 59
44 45 46 47 60 61 62 63

方法：要解决此问题, 请修复异或由于从0开始的4个连续数字的异或为0, 因此每一行和每一列的0为0。这是一个4 x 4矩阵的示例：

0 ^ 1 ^ 2 ^ 3 = 0
4 ^ 5 ^ 6 ^ 7 = 0
8 ^ 9 ^ 10 ^ 11 = 0
12 ^ 13 ^ 14 ^ 15 = 0等。

如果你在上面的示例中注意到, 每一行和每一列的xor为0。现在我们需要以这样的方式放置数字：每一行和每一列的xor为0。因此我们可以将N x N矩阵变成较小的4 x 4矩阵N/4行和列, 并以每行和列的xor为0的方式填充单元格。
下面是上述方法的实现：
C ++

//C++ implementation of the approach #include < bits/stdc++.h> using namespace std; //Function to find the n x n matrix //that satisfies the given condition void findGrid( int n) { int arr[n][n]; //Initialize x to 0 int x = 0; //Divide the n x n matrix into n /4 matrices //for each of the n /4 rows where //each matrix is of size 4 x 4 for ( int i = 0; i < n /4; i++) { for ( int j = 0; j < n /4; j++) { for ( int k = 0; k < 4; k++) { for ( int l = 0; l < 4; l++) { arr[i * 4 + k][j * 4 + l] = x; x++; } } } }//Print the generated matrix for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { cout < < arr[i][j] < < " " ; } cout < < "\n" ; } }//Driver code int main() { int n = 4; findGrid(n); return 0; }

Java

//Java implementation of the approach import java.io.*; class GFG {//Function to find the n x n matrix //that satisfies the given condition static void findGrid( int n) { int [][]arr = new int [n][n]; //Initialize x to 0 int x = 0 ; //Divide the n x n matrix into n /4 matrices //for each of the n /4 rows where //each matrix is of size 4 x 4 for ( int i = 0 ; i < n /4 ; i++) { for ( int j = 0 ; j < n /4 ; j++) { for ( int k = 0 ; k < 4 ; k++) { for ( int l = 0 ; l < 4 ; l++) { arr[i * 4 + k][j * 4 + l] = x; x++; } } } }//Print the generated matrix for ( int i = 0 ; i < n; i++) { for ( int j = 0 ; j < n; j++) { System.out.print(arr[i][j] + " " ); } System.out.println( " " ); } }//Driver code public static void main (String[] args) { int n = 4 ; findGrid(n); } }//This code is contributed by ajit.

Python3

# Python3 implementation of the approach # Function to find the n x n matrix # that satisfies the given condition def findGrid(n): arr = [[ 0 for k in range (n)] for l in range (n)] # Initialize x to 0 x = 0# Divide the n x n matrix into n /4 matrices # for each of the n /4 rows where # each matrix is of size 4 x 4 for i in range (n //4 ): for j in range (n //4 ): for k in range ( 4 ): for l in range ( 4 ): arr[i * 4 + k][j * 4 + l] = x x + = 1# Print the generated matrix for i in range (n): for j in range (n): print (arr[i][j], end = " " ) print ()# Driver code n = 4 findGrid(n) # This code is contributed by divyamohan123

//C# implementation of the approach using System; class GFG {//Function to find the n x n matrix //that satisfies the given condition static void findGrid( int n) { int [, ]arr = new int [n, n]; //Initialize x to 0 int x = 0; //Divide the n x n matrix into n /4 matrices //for each of the n /4 rows where //each matrix is of size 4 x 4 for ( int i = 0; i < n /4; i++) { for ( int j = 0; j < n /4; j++) { for ( int k = 0; k < 4; k++) { for ( int l = 0; l < 4; l++) { arr[i * 4 + k, j * 4 + l] = x; x++; } } } }//Print the generated matrix for ( int i = 0; i < n; i++) { for ( int j = 0; j < n; j++) { Console.Write(arr[i, j] + " " ); } Console.WriteLine( " " ); } }//Driver code public static void Main (String[] args) { int n = 4; findGrid(n); } }//This code is contributed by PrinciRaj1992

输出如下：