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📜  创建一个数组,使长度为K的子数组的XOR为X

📅  最后修改于: 2021-05-07 01:37:47             🧑  作者: Mango

给定三个整数NKX ,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XORX。

例子:

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, X0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


输出:
4 0 4 0 4

时间复杂度: O(N)