# 查找一个N x N网格，其每行和每列的xor相等

2021年4月28日19:56:54 发表评论 645 次浏览

## 本文概述

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

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

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

## 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#

``````//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``````

``````0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15``````