找出一组数最多为 N 且按位异或为 N 的数

给定一个正整数N ，任务是找到整数对(X, Y)使得X和Y的按位异或为N并且X * Y最大，其中X和Y中的位数小于或等于N 。

例子：

Input: N = 10
Output: 13 7
Explanation: The case where X = 13 and Y = 7 is the most optimal choice as 13 xor 7 = 10 and 13 * 7 = 91 which is maximum possible.

Input: N = 45
Output: 50 31

方法：给定的问题可以通过以下观察来解决：

• 如果N的^{第 i}位是0 ，那么X和Y的^第i位必须是0或1 。为了最大化产品，将这些位设置为1 。
• 如果N的^{第 i}位为1 ，则X或Y的第ⁱ位中的一个必须为1 ，另一个必须为0 。由于N必须具有恒定数量的设置位，因此 X 和 Y 的总和必须是恒定的。
• 如果两个数之和为常数，则当两个数之差最小时，它们的乘积最大。

根据上述观察，将两个整数X和Y初始化为 0。为了最小化X和Y之间的差异， X必须等于MSB _N并且Y必须等于N – MSB _N其中MSB表示最重要少量。此外，对于N中的所有0位，将X和Y中的相应位设置为1 。

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 pair (X, Y) such
// that X xor Y = N and the count of set
// bits in X and Y is less than count of
// set bit in N
void maximizeProduct(int N)
{
    // Stores MSB (Most Significant Bit)
    int MSB = (int)log2(N);
 
    // Stores the value of X
    int X = 1 << MSB;
 
    // Stores the value of Y
    int Y = N - (1 << MSB);
 
    // Traversing over all bits of N
    for (int i = 0; i < MSB; i++) {
 
        // If ith bit of N is 0
        if (!(N & (1 << i))) {
 
            // Set ith bit of X to 1
            X += 1 << i;
 
            // Set ith bit of Y to 1
            Y += 1 << i;
        }
    }
 
    // Print Answer
    cout << X << " " << Y;
}
 
// Driver Code
int main()
{
    int N = 45;
    maximizeProduct(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
 
// Function to find the pair (X, Y) such
// that X xor Y = N and the count of set
// bits in X and Y is less than count of
// set bit in N
static void maximizeProduct(int N)
{
    // Stores MSB (Most Significant Bit)
    int MSB = (int)(Math.log(N) / Math.log(2));
   
    // Stores the value of X
    int X = 1 << MSB;
   
    // Stores the value of Y
    int Y = N - (1 << MSB);
   
    // Traversing over all bits of N
    for (int i = 0; i < MSB; i++) {
   
        // If ith bit of N is 0
        if ((N & (1 << i))==0) {
   
            // Set ith bit of X to 1
            X += 1 << i;
   
            // Set ith bit of Y to 1
            Y += 1 << i;
        }
    }
   
    // Print Answer
    System.out.println(X+" "+Y);
}
// Driver Code
public static void main(String[] args)
{
    int N = 45;
    maximizeProduct(N);
}
}
 
// This code is contributed by dwivediyash

Python3

# python 3 program for the above approach
import math
 
# Function to find the pair (X, Y) such
# that X xor Y = N and the count of set
# bits in X and Y is less than count of
# set bit in N
def maximizeProduct(N):
 
    # Stores MSB (Most Significant Bit)
    MSB = (int)(math.log2(N))
 
    # Stores the value of X
    X = 1 << MSB
 
    # / Stores the value of Y
    Y = N - (1 << MSB)
 
    # Traversing over all bits of N
    for i in range(MSB):
 
        # If ith bit of N is 0
        if (not (N & (1 << i))):
 
            # / Set ith bit of X to 1
            X += 1 << i
 
            # Set ith bit of Y to 1
            Y += 1 << i
 
    # Print Answer
    print(X, Y)
 
# Driver Code
if __name__ == "__main__":
    N = 45
    maximizeProduct(N)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
 
class GFG {
 
// Function to find the pair (X, Y) such
// that X xor Y = N and the count of set
// bits in X and Y is less than count of
// set bit in N
static void maximizeProduct(int N)
{
   
    // Stores MSB (Most Significant Bit)
    int MSB = (int)(Math.Log(N) / Math.Log(2));
   
    // Stores the value of X
    int X = 1 << MSB;
   
    // Stores the value of Y
    int Y = N - (1 << MSB);
   
    // Traversing over all bits of N
    for (int i = 0; i < MSB; i++) {
   
        // If ith bit of N is 0
        if ((N & (1 << i))==0) {
   
            // Set ith bit of X to 1
            X += 1 << i;
   
            // Set ith bit of Y to 1
            Y += 1 << i;
        }
    }
   
    // Print Answer
    Console.Write(X+" "+Y);
}
 
 
    // Driver Code
    public static void Main()
    {
        int N = 45;
        maximizeProduct(N);
    }
}
 
// This code is contributed by code_hunt.

Javascript

        // JavaScript Program to implement
        // the above approach
 
        // Function to find the pair (X, Y) such
        // that X xor Y = N and the count of set
        // bits in X and Y is less than count of
        // set bit in N
        function maximizeProduct(N)
        {
         
            // Stores MSB (Most Significant Bit)
            let MSB = Math.log2(N);
 
            // Stores the value of X
            let X = 1 << MSB;
 
            // Stores the value of Y
            let Y = N - (1 << MSB);
 
            // Traversing over all bits of N
            for (let i = 0; i < MSB; i++) {
 
                // If ith bit of N is 0
                if (!(N & (1 << i))) {
 
                    // Set ith bit of X to 1
                    X += 1 << i;
 
                    // Set ith bit of Y to 1
                    Y += 1 << i;
                }
            }
 
            // Print Answer
            document.write(X + " " + Y);
        }
 
        // Driver Code
 
        let N = 45;
        maximizeProduct(N);
 
 
// This code is contributed by Potta Lokesh
 
    

输出：

50 31

时间复杂度： O(log N)
辅助空间： O(N)