给定范围内所有奇数的按位XOR

给定一个整数N ，任务是找到范围[1，N]中所有奇数的按位XOR。

例子：

Input: 11
Output: 2
Explanation: Bitwise XOR of all odd numbers up to 11 = 1 ^ 3 ^ 5 ^ 7 ^ 9 ^ 11 = 2.

Input: 10
Output: 9
Explanation: Bitwise XOR of all odd numbers up to 10 = 1 ^ 3 ^ 5 ^ 7 ^ 9 = 9.

天真的方法：解决问题的最简单方法是在[1，N]范围内进行迭代，并针对每个值检查其是否为奇数。计算找到的每个奇数元素的按位XOR，并将其打印为所需结果。
时间复杂度： O(N)
辅助空间： O(1)

高效方法：为了优化上述方法，我们的想法是使用以下公式计算所有小于或等于N的奇数的按位XOR ：

Let f(N) = 2 ^ 4 ^ 6 ^ … ^ (N − 2) ^ n and g(n) = 1 ^ 2 ^ 3 ^ … ^ (N − 2) / 2 ^ (N / 2) 
=> f(N) = 2 * g(N)

Now, let k(N) = 1 ^ 2 ^ 3 ^ … ^ (N − 2) ^ N

XOR of all odd numbers less than or equal to N = k(N) ^ f(N) [Since all even numbers cancel their own bits leaving only odd numbers]. 
Substituting the value of f(N) = 2 * g(N), XOR of all odd numbers up to N = k(N) ^ (2 * g(N))

请按照以下步骤解决问题：

• 将范围为[1，N]的数字的XOR存储在变量中，例如K。
• 如果N为奇数，则将[X，(N – 1)/ 2]范围内的数字的XOR存储在变量g中。否则，以g为单位存储数字[X，N / 2]的XOR。
• 使用导出的公式，将结果更新为K ^(2＆g) 。
• 完成上述步骤后，打印结果值作为答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to calculate Bitwise
// XOR of odd numbers in the range [1, N]
int findXOR(int n)
{
    // N & 3 is equivalent to n % 4
    switch (n & 3) {
 
    // If n is multiple of 4
    case 0:
        return n;
 
    // If n % 4 gives remainder 1
    case 1:
        return 1;
 
    // If n % 4 gives remainder 2
    case 2:
        return n + 1;
 
    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}
 
// Function to find the XOR of odd
// numbers less than or equal to N
void findOddXOR(int n)
{
 
    // If number is even
    if (n % 2 == 0)
 
        // Print the answer
        cout << ((findXOR(n))
                 ^ (2 * findXOR(n / 2)));
 
    // If number is odd
    else
 
        // Print the answer
        cout << ((findXOR(n))
                 ^ (2 * findXOR((n - 1) / 2)));
}
 
// Driver Code
int main()
{
    int N = 11;
 
    // Function Call
    findOddXOR(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
  // Function to calculate Bitwise
  // XOR of odd numbers in the range [1, N]
  static int findXOR(int n)
  {
 
    // N & 3 is equivalent to n % 4
    switch (n & 3)
    {
 
        // If n is multiple of 4
      case 0:
        return n;
 
        // If n % 4 gives remainder 1
      case 1:
        return 1;
 
        // If n % 4 gives remainder 2
      case 2:
        return n + 1;
    }
    // If n % 4 gives remainder 3
    return 0;
 
  }
 
  // Function to find the XOR of odd
  // numbers less than or equal to N
  static void findOddXOR(int n)
  {
 
    // If number is even
    if (n % 2 == 0)
 
      // Print the answer
      System.out.print(((findXOR(n))
                        ^ (2 * findXOR(n / 2))));
 
    // If number is odd
    else
 
      // Print the answer
      System.out.print(((findXOR(n))
                        ^ (2 * findXOR((n - 1) / 2))));
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 11;
 
    // Function Call
    findOddXOR(N);
  }
}
 
// This code is contributed by shikhasingrajput

Python3

# Python program for the above approach
 
# Function to calculate Bitwise
# XOR of odd numbers in the range [1, N]
def findXOR(n):
   
    # N & 3 is equivalent to n % 4
    if (n % 4 == 0):
       
        # If n is multiple of 4
        return n;
    elif (n % 4 == 1):
       
        # If n % 4 gives remainder 1
        return 1;
 
    # If n % 4 gives remainder 2
    elif (n % 4 == 2):
        return n + 1;
 
    # If n % 4 gives remainder 3
    elif (n % 4 == 3):
        return 0;
 
# Function to find the XOR of odd
# numbers less than or equal to N
def findOddXOR(n):
   
    # If number is even
    if (n % 2 == 0):
 
        # Prthe answer
        print(((findXOR(n)) ^ (2 * findXOR(n // 2))));
 
    # If number is odd
    else:
 
        # Prthe answer
        print(((findXOR(n)) ^ (2 * findXOR((n - 1) // 2))));
 
# Driver Code
if __name__ == '__main__':
    N = 11;
 
    # Function Call
    findOddXOR(N);
 
# This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
public class GFG
{
 
  // Function to calculate Bitwise
  // XOR of odd numbers in the range [1, N]
  static int findXOR(int n)
  {
 
    // N & 3 is equivalent to n % 4
    switch (n & 3)
    {
 
        // If n is multiple of 4
      case 0:
        return n;
 
        // If n % 4 gives remainder 1
      case 1:
        return 1;
 
        // If n % 4 gives remainder 2
      case 2:
        return n + 1;
    }
    // If n % 4 gives remainder 3
    return 0;
 
  }
 
  // Function to find the XOR of odd
  // numbers less than or equal to N
  static void findOddXOR(int n)
  {
 
    // If number is even
    if (n % 2 == 0)
 
      // Print the answer
      Console.Write(((findXOR(n))
                     ^ (2 * findXOR(n / 2))));
 
    // If number is odd
    else
 
      // Print the answer
      Console.Write(((findXOR(n))
                     ^ (2 * findXOR((n - 1) / 2))));
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int N = 11;
 
    // Function Call
    findOddXOR(N);
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

// JavaScript program for the above approach
 
// Function to calculate Bitwise
// XOR of odd numbers in the range [1, N]
function findXOR(n)
{
    // N & 3 is equivalent to n % 4
    switch (n & 3) {
 
    // If n is multiple of 4
    case 0:
        return n;
 
    // If n % 4 gives remainder 1
    case 1:
        return 1;
 
    // If n % 4 gives remainder 2
    case 2:
        return n + 1;
 
    // If n % 4 gives remainder 3
    case 3:
        return 0;
    }
}
 
// Function to find the XOR of odd
// numbers less than or equal to N
function findOddXOR(n)
{
 
    // If number is even
    if (n % 2 == 0)
 
        // Print the answer
        document.write((findXOR(n))
                ^ (2 * findXOR(n / 2)));
 
    // If number is odd
    else
 
        // Print the answer
        document.write((findXOR(n))
                ^ (2 * findXOR((n - 1) / 2)));
}
 
// Driver Code
    let N = 11;
 
    // Function Call
    findOddXOR(N);
 
// This code is contributed by Surbhi Tyagi.
 

输出：

2

时间复杂度： O(1)
辅助空间： O(1)