具有最小差异的最小整数对，其按位异或为 N

给定一个正整数N ，任务是找到两个最小的整数A和B ，使得 A 和B的按位异或为N ，并且A和B之间的差最小。

例子：

Input: N = 26
Output: 10 16
Explanation:
The Bitwise XOR of 10 and 16 is 26 and the difference between them is minimum.

Input: N = 1
Output: 0 1

朴素方法：解决给定问题的最简单方法是生成范围[0, N]内所有可能的数字对，并打印这对数字，其按位异或是给定数字N并且两个数字都是最小的。

时间复杂度： O(N ² )
辅助空间： O(1)

有效的方法：上述方法也可以基于以下观察进行优化：

• 考虑任何数字的二进制表示是“1100011” ，然后该数字可以围绕它们的最高有效位（MSB）拆分为“1000000”和“100011” ，并且这些数字的按位异或是给定的数字。
• 从上面的拆分可以看出， “1000000” （比如A ）和“100011” （比如B ）组成的数字最小，它们之间的差异最小，因为B形成的值总是更小和最接近到A 。

根据以上观察，满足给定标准的A和B的最小值是将给定数N围绕其最高有效位 (MSB)拆分。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the numbers A and
// B whose Bitwise XOR is N and the
// difference between them is minimum
void findAandB(int N)
{
 
    // Find the MSB of the N
    int K = log2(N);
 
    // Find the value of B
    int B = (1 << K);
 
    // Find the value of A
    int A = B ^ N;
 
    // Print the result
    cout << A << ' ' << B;
}
 
// Driver Code
int main()
{
 
    int N = 26;
    findAandB(N);
 
    return 0;
}

Java

// Java program for the above approach
public class MyClass
{
   
// Function to find the numbers A and
// B whose Bitwise XOR is N and the
// difference between them is minimum
static void findAandB(int N)
{
 
    // Find the MSB of the N
    int K = (int)(Math.log(N) / Math.log(2));
 
    // Find the value of B
    int B = (1 << K);
 
    // Find the value of A
    int A = B ^ N;
 
    // Print the result
    System.out.println(A + " " + B);
}
 
    public static void main(String args[]) {
      int N = 26;
      findAandB(N);
 
    }
}
 
// This code is contributed by SoumikMondal

Python3

# Python3 program for the above approach
from math import log2
 
# Function to find the numbers A and
# B whose Bitwise XOR is N and the
# difference between them is minimum
def findAandB(N):
     
    # Find the MSB of the N
    K = int(log2(N))
 
    # Find the value of B
    B = (1 << K)
 
    # Find the value of A
    A = B ^ N
 
    # Print the result
    print(A, B)
 
# Driver Code
if __name__ == '__main__':
     
    N = 26
     
    findAandB(N)
 
# This code is contributed by SURENDRA_GANGWAR

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the numbers A and
// B whose Bitwise XOR is N and the
// difference between them is minimum
static void findAandB(int N)
{
     
    // Find the MSB of the N
    int K = (int)(Math.Log(N) /
                  Math.Log(2));
 
    // Find the value of B
    int B = (1 << K);
 
    // Find the value of A
    int A = B ^ N;
 
    // Print the result
    Console.Write(A + " " + B);
}
 
// Driver Code
public static void Main(String[] args)
{
    int N = 26;
     
    findAandB(N);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
        // JavaScript program for the above approach
 
        // Function to find the numbers A and
        // B whose Bitwise XOR is N and the
        // difference between them is minimum
        function findAandB(N) {
 
            // Find the MSB of the N
            let K = Math.log2(N);
 
            // Find the value of B
            let B = (1 << K);
 
            // Find the value of A
            let A = B ^ N;
 
            // Print the result
            document.write(A + ' ' + B);
        }
 
        // Driver Code
 
        let N = 26;
        findAandB(N);
 
 
  // This code is contributed by Potta Lokesh
   
 

输出：

10 16

时间复杂度： O(1)
辅助空间： O(1)