小于N的最大数M，使得M和N的XOR为偶数

给定正整数N ，任务是找到最大整数M ，以使0 <= M 且XOR(M，N)为偶数。如果对于给定的N无法获得M的值，则打印-1 。

例子：

Input: N = 10 
Output: 8 
Explanation: 
(10 XOR 9) = 3, so M = 9 is not possible because 3 is not an even number. 
(10 XOR 8) = 2, so M = 8 is possible as it is the largest possible value of M satisfying the necessary conditions.

Input: N = 5 
Output: 3 
(5 XOR 4) = 1, so M = 4 is not possible because 1 is not an even number. 
(5 XOR 3) = 6, so M = 3 is possible as 6 is an even number and 3 is the largest possible value of M (which is less than N). 
 

方法：
解决问题时需要进行以下观察：

• 以二进制形式表示的奇数以1作为其最低有效位(LSB)，而以二进制形式表示的偶数以0作为其最低有效位(LSB)。
• 在执行XOR操作时，如果设置的位数为奇数，则XOR值为1。如果设置的位数为偶数，则XOR值为0。
• 因此，我们需要对两个奇数或两个偶数执行XOR运算，以得到一个偶数。

根据以上说明，可以通过以下步骤解决问题：

1. 如果N为奇数，则小于N的最大奇数将为N – 2 。
2. 如果N为偶数，则小于N的最大偶数将为N – 2 。
3. 因此，如果N = 1，则在这种情况下无法获得作为合适的M的打印-1。对于所有其他情况，请打印N – 2作为请求者。

下面是上述方法的实现：

C++

// C++ program for the above problem.
#include 
using namespace std;
 
// Function to find the maximum
// possible value of M
int getM(int n)
{
    // Edge case
    if (n == 1)
        return -1;
 
    // M = N - 2 is maximum
    // possible value
    else
        return n - 2;
}
 
// Driver Code
int main()
{
    int n = 10;
 
    int ans = getM(n);
    cout << ans;
}

Java

// Java program for the above problem.
import java.util.*;
 
class GFG{
     
// Function to find the maximum
// possible value of M
static int getM(int n)
{
     
    // Edge case
    if (n == 1)
        return -1;
             
    // M = N - 2 is maximum
    // possible value
    else
        return n - 2;
}
     
// Driver code
public static void main(String[] args)
{
    int n = 10;
         
    System.out.print(getM(n));
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above problem
 
# Function to find the maximum
# possible value of M
def getM(n):
     
    # Edge case
    if (n == 1):
        return -1;
         
    # M = N - 2 is maximum
    # possible value
    else:
        return n - 2;                
 
# Driver code
n = 10
 
print(getM(n))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above problem.
using System;
class GFG{
     
// Function to find the maximum
// possible value of M
static int getM(int n)
{
     
    // Edge case
    if (n == 1)
        return -1;
                 
    // M = N - 2 is maximum
    // possible value
    else
        return n - 2;
}
 
// Driver code
static void Main()
{
    int n = 10;
         
    Console.Write(getM(n));
}
}
 
// This code is contributed by divyeshrabadiya07

Javascript

 
// Javascript program for the above problem.
 
// Function to find the maximum
// possible value of M
function getM(n)
{
        // Edge case
    if (n == 1)
        return -1;
         
    // M = N - 2 is maximum
    // possible value
    else
        return n - 2;
}
     
// Driver code
var n = 10;
 
document.write(getM(n));
 
// This code is contributed by Khushboogoyal499
    

输出：

8

时间复杂度： O(1)
辅助空间： O(1)