具有 N 位计数的最大数 M 使得它们的 OR 和 XOR 值之间的差异最大化

给定一个自然数N ，任务是找到在二进制表示中与N具有相同长度的最大数M ，使得N | M和N^M是最大值。

例子：

Input: N = 6
Output: 7
Explanation:  
All number numbers having same length in binary representation as N are 4, 5, 6, 7.
(6 | 4) – (6 ^ 4) = 4
(6 | 5) – (6 ^ 5) = 4
(6 | 6) – (6 ^ 6) = 6
(6 | 7) – (6 ^ 7) = 6
Hence, largest M for which (N | M) – (N ^ M) is maximum is 7

Input: N = 10
Output: 15
Explanation:  
The largest number M = 15 which has the same length in binary representation as 10 and the difference between N | M and N ^ M is maximum.

朴素的方法：这个想法是简单地找到所有与N具有相同长度的二进制表示的数字，然后对每个数字进行迭代并找到最大的 整数 具有(N | i) – (N ^ i)最大值。

时间复杂度： O(N)
辅助空间： O(1)

有效的方法：这个想法是初始化M = 0并在 N 中逐位迭代（比如i ）并根据以下 2 个观察设置或取消设置M 的^第i位：

• 当N的^第i位设置：在这种情况下，如果我们取消设置M的^第i位，我都^第N位| M和N^M将被设置，而在设置 M 的这一位时， N|M 的^第i 位将被设置并且N^M将被取消设置，这将增加(N | M) – (N ^ M) 。因此，设置M 的这一位是最佳的。
• 当 N 的^第i 位未设置时：在这种情况下，如果我们设置了该位 M， N|M和 N^M 都将设置此位，或者保持 M 的此位未设置， N|M和N^M都将未设置此位。因此，在这种情况下，我们不能增加它们之间的差异，但由于要求输出最大可能的M ，因此设置M 的这一位。
• 根据以上观察，很明显M将设置所有位。

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 largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
int maxORminusXOR(int N)
{
    // Find the most significant
    // bit of N
    int MSB = log2(N);
 
    // Initialize M
    int M = 0;
 
    // Set all the bits of M
    for (int i = 0; i <= MSB; i++)
        M += (1 << i);
 
    // Return the answer
    return M;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 10;
 
    // Function Call
    cout << maxORminusXOR(N);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
static int maxORminusXOR(int N)
{
     
    // Find the most significant
    // bit of N
    int MSB = (int)Math.ceil(Math.log(N));
 
    // Initialize M
    int M = 0;
 
    // Set all the bits of M
    for(int i = 0; i <= MSB; i++)
        M += (1 << i);
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given number N
    int N = 10;
 
    // Function call
    System.out.print(maxORminusXOR(N));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
import math
 
# Function to find the largest number
# M having the same length in binary
# form as N such that the difference
# between N | M and N ^ M is maximum
def maxORminusXOR(N):
 
    # Find the most significant
    # bit of N
    MSB = int(math.log2(N));
 
    # Initialize M
    M = 0
 
    # Set all the bits of M
    for i in range(MSB + 1):
        M += (1 << i)
 
    # Return the answer
    return M
 
# Driver code
if __name__ == '__main__':
     
    # Given Number N
    N = 10
 
    # Function call
    print(maxORminusXOR(N))
 
# This code is contributed by jana_sayantan

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
static int maxORminusXOR(int N)
{
     
    // Find the most significant
    // bit of N
    int MSB = (int)Math.Ceiling(Math.Log(N));
 
    // Initialize M
    int M = 0;
 
    // Set all the bits of M
    for(int i = 0; i <= MSB; i++)
        M += (1 << i);
 
    // Return the answer
    return M;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given number N
    int N = 10;
 
    // Function call
    Console.Write(maxORminusXOR(N));
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// JavaScript implementation of the above approach
  
// Function to find the largest number
// M having the same length in binary
// form as N such that the difference
// between N | M and N ^ M is maximum
function maxORminusXOR(N)
{
       
    // Find the most significant
    // bit of N
    let MSB = Math.ceil(Math.log(N));
   
    // Initialize M
    let M = 0;
   
    // Set all the bits of M
    for(let i = 0; i <= MSB; i++)
        M += (1 << i);
   
    // Return the answer
    return M;
}
 
// Driver code
         
    // Given number N
    let N = 10;
   
    // Function call
    document.write(maxORminusXOR(N));
   
  // This code is contributed by code_hunt.

输出：

15

时间复杂度： O(log N)
辅助空间： O(1)

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