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📜  创建一个数组,使得长度为 K 的子数组的 XOR 为 X

📅  最后修改于: 2021-09-02 06:39:57             🧑  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,我们需要按照以下步骤操作:

  • 任何数字的按位异或, X0等于数字本身。因此,如果我们将数组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 ,那么我们将得到长度为 K 的第一个子数组的元素的异或,等于 X。
  • 如果我们将A[K]设置为A[0] ,那么我们将有 XOR(A[1], …, A[K]) = X。同样,如果我们将A[K + 1] 设置A[1] ,那么我们将有 XOR(A[2], …, A[K+1]) = X
  • 以这种方式继续,我们可以观察到通用公式可以描述如下:

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


输出:
4 0 4 0 4

时间复杂度: O(N)

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