创建一个数组，使长度为K的子数组的XOR为X

给定三个整数N ， K和X ，任务是构造一个长度为N的数组，其中每个长度为K的连续子数组的所有元素的XOR为X。
例子：

Input: N = 5, K = 1, X = 4 
Output: 4 4 4 4 4 
Explanation: 
Each subarray of length 1 has Xor value equal to 4.
Input: N = 5, K = 2, X = 4 
Output: 4 0 4 0 4 
Explanation: 
Each subarray of length 2 has Xor value equal to 4. 
 

方法：
要解决上述问题，我们需要执行以下步骤：

• 任意数字的按位XOR， X与0等于数字本身。因此，如果将数组A的第一个元素设置为X，并将接下来的K – 1个元素设置为0 ，则将使长度为K的第一个子数组的元素的XOR等于X。
• 如果将A [K]设置为A [0] ，则将具有XOR(A [1]，…，A [K])=X。类似地，如果将A [K + 1]设置为A [1] ，那么我们将得到XOR(A [2]，…，A [K + 1])= X
• 以这种方式继续，我们可以观察到通式可以描述如下：

下面是上述方法的实现：

C++

// C++ implementation to Create an array
// in which the XOR of all elements of
// each contiguous sub-array of
// length K is X
 
#include 
using namespace std;
 
// Function to construct the array
void constructArray(int N, int K, int X)
{
 
    // Creating a vector of size K,
    // initialised with 0
    vector ans(K, 0);
 
    // Initialising the first element
    // with the given XOR
    ans[0] = X;
 
    for (int i = 0; i < N; ++i) {
        cout << ans[i % K] << " ";
    }
 
    cout << endl;
}
 
// Driver code
int main()
{
    int N = 5, K = 2, X = 4;
 
    constructArray(N, K, X);
 
    return 0;
}

Java

// Java implementation to create an array
// in which the XOR of all elements of
// each contiguous sub-array of
// length K is X
class GFG{
     
// Function to construct the array
public static void constructArray(int N, int K,
                                         int X)
{
     
    // Creating an array of size K,
    // initialised with 0
    int[] ans = new int[K];
      
    // Initialising the first element
    // with the given XOR
    ans[0] = X;
     
    for(int i = 0; i < N; ++i)
    {
       System.out.print(ans[i % K] + " ");
    }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 5, K = 2, X = 4;
     
    constructArray(N, K, X);
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation to create an array
# in which the XOR of all elements of
# each contiguous sub-array of
# length K is X
 
# Function to construct the array
def constructArray(N, K, X):
     
    # Creating a list of size K,
    # initialised with 0
    ans = []
     
    for i in range(0, K):
        ans.append(0)
         
    # Initialising the first element
    # with the given XOR
    ans[0] = X
         
    for i in range(0, N):
        print(ans[i % K], end = " ")
 
# Driver code
N = 5
K = 2
X = 4
 
# Function call
constructArray(N, K, X)
 
# This code is contributed by ishayadav181

C#

// C# implementation to create an array
// in which the XOR of all elements of
// each contiguous sub-array of
// length K is X
using System;
 
class GFG{
     
// Function to construct the array
public static void constructArray(int N, int K,
                                         int X)
{
     
    // Creating an array of size K,
    // initialised with 0
    int[] ans = new int[K];
     
    // Initialising the first element
    // with the given XOR
    ans[0] = X;
     
    for(int i = 0; i < N; ++i)
    {
        Console.Write(ans[i % K] + " ");
    }
}
 
// Driver code
public static void Main(string[] args)
{
    int N = 5, K = 2, X = 4;
     
    constructArray(N, K, X);
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript implementation to Create an array
// in which the XOR of all elements of
// each contiguous sub-array of
// length K is X
 
// Function to construct the array
function constructArray(N, K, X)
{
 
    // Creating a vector of size K,
    // initialised with 0
    let ans = new Array(K).fill(0);
 
    // Initialising the first element
    // with the given XOR
    ans[0] = X;
 
    for (let i = 0; i < N; ++i) {
        document.write(ans[i % K] + " ");
    }
 
    document.write("<br>");
}
 
// Driver code
    let N = 5, K = 2, X = 4;
 
    constructArray(N, K, X);
 

输出：

4 0 4 0 4

时间复杂度： O(N)

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