构造一个数组，其从X开始的Prefix XOR数组是一个长度为N的递增序列

给定两个整数N和X ，任务是生成一个大小为 N的数组，使得 X 的前缀xor 数组与生成的数组将是 第一个 N 个自然数的排列。

例子：

Input: N = 4, X = 3
Output: [2, 3, 1, 7]
Explanation: Prefix XOR array for the array {2, 3, 1, 7} is as follows:

1. X ⊕ 2 = 1. Now, X = 1
2. X ⊕ 3 = 2. Now, X = 2
3. X ⊕ 1 = 3. Now, X = 3
4. X ⊕ 7 = 4. Now, X = 4

The array [1, 2, 3, 4] is a permutation of first 4 natural numbers.

Input: N = 7, X = 52
Output: [53, 3, 1, 7, 1, 3, 1]

方法：这个问题可以使用XOR 的性质来解决（如果x ⊕ a = b ，则x ⊕ b = a ）。假设生成的数组中的元素与X直到第 i^个索引为X进行异或，并且生成的数组的第(i + 1)^个元素为B ，则可以使用以下步骤计算B ：

X ⊕ B = i + 1 
According to problem statement, using the property of XOR (if x⊕ a = b then, x⊕ b = a)… 
X ⊕ i + 1 = B
Or
B = X ⊕ i + 1, which is the required value of B.

请按照以下步骤解决问题：

• 初始化一个变量，比如prev_xor为X 。
• 使用变量i迭代一个循环，从1到N并打印prev_xor ⊕ i 。
• 将 prev_xor 更新为 i 。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print the required array
void GenerateArray(int N, int X)
{
    int prev_xor = X;
 
    // Iterative from 1 to N
    for (int i = 1; i <= N; i++) {
 
        // Print the i-th element
        cout << (i ^ prev_xor);
 
        if (i != N) {
            cout << " ";
        }
 
        // Update prev_xor to i
        prev_xor = i;
    }
}
 
// Driver Code
int main()
{
    // Given Input
    int N = 4, X = 3;
 
    // Function Call
    cout << "The generated array is ";
    GenerateArray(N, X);
    return 0;
}

Java

// Java program for the above approach
class GFG {
 
  // Function to print the required array
  static void GenerateArray(int N, int X)
  {
    int prev_xor = X;
 
    // Iterative from 1 to N
    for (int i = 1; i <= N; i++) {
 
      // Print the i-th element
      System.out.print(i ^ prev_xor);
 
      if (i != N) {
        System.out.print(" ");
      }
 
      // Update prev_xor to i
      prev_xor = i;
    }
  }
 
  // Driver Code
  public static void main(String args[]) {
 
    // Given Input
    int N = 4, X = 3;
 
    // Function Call
    System.out.print("The generated array is ");
    GenerateArray(N, X);
  }
}
 
// This code is contributed by splevel62.

Python3

# Python 3 program for the above approach
 
# Function to print the required array
def GenerateArray(N, X):
    prev_xor = X
 
    # Iterative from 1 to N
    for i in range(1,N+1,1):
        # Print the i-th element
        print(i ^ prev_xor,end="")
 
        if (i != N):
            print(" ",end="")
 
        # Update prev_xor to i
        prev_xor = i
 
# Driver Code
if __name__ == '__main__':
    # Given Input
    N = 4
    X = 3
 
    # Function Call
    print("The generated array is ",end="")
    GenerateArray(N, X)
     
    # This code is contributed by ipg2016107

C#

// C# program for above approach
using System;
class GFG
{
 
// Function to print the required array
static void GenerateArray(int N, int X)
{
    int prev_xor = X;
 
    // Iterative from 1 to N
    for (int i = 1; i <= N; i++) {
 
        // Print the i-th element
        Console.Write(i ^ prev_xor);
 
        if (i != N) {
            Console.Write(" ");
        }
 
        // Update prev_xor to i
        prev_xor = i;
    }
}
 
 
// Driver Code
static void Main()
{
   
    // Given Input
    int N = 4, X = 3;
 
    // Function Call
    Console.Write("The generated array is ");
    GenerateArray(N, X);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

 
// JavaScript program for the above approach;
 
// Function to print the required array
function GenerateArray(N, X)
{
    let prev_xor = X;
     
    // Iterative from 1 to N
    for(let i = 1; i <= N; i++)
    {
         
        // Print the i-th element
        document.write((i ^ prev_xor));
 
        if (i != N)
        {
            document.write(" ");
        }
 
        // Update prev_xor to i
        prev_xor = i;
    }
}
 
// Driver Code
 
// Given Input
let N = 4, X = 3;
 
// Function Call
document.write("The generated array is ");
GenerateArray(N, X);
 
// This code is contributed by Potta Lokesh
 

输出：

The generated array is 2 3 1 7

时间复杂度： O(N)
辅助空间：O(1)