给定范围内的值，具有最大XOR

给定正整数N，L和R，我们必须找到N⊕X的最大值，其中X∈[L，R]。
例子：

Input : N = 7 
L = 2 
R = 23 
Output : 23 
Explanation : When X = 16, we get 7 ⊕ 16 = 23 which is the maximum value for all X ∈ [2, 23].
Input : N = 10 
L = 5 
R = 12 
Output : 15 
Explanation : When X = 5, we get 10 ⊕ 5 = 15 which is the maximum value for all X ∈ [5, 12].

蛮力法：我们可以使用蛮力法解决此问题，方法是循环遍历[L，R]范围内的所有整数，并将它们与N进行XOR，同时记录到目前为止所遇到的最大结果。该算法的复杂度为O(R – L)，并且当输入变量接近高值(例如10 ⁹ )时，这是不可行的。
高效的方法：由于仅当两个位互补时，两个位的XOR才为1，因此我们需要X具有与N互补的位，以具有最大值。我们将会从最大的迭代位(日志_2(R)^个位)到最低(^第0位)。每位可能出现以下两种情况：

1. 如果未设置该位(即0)，我们将尝试将其设置为X。如果将该位设置为1，导致X超过R，则我们将不对其进行设置。
2. 如果该位被设置，即1，那么我们将尝试在X中将其取消设置。如果X的当前值已经大于或等于L，则可以安全地将其设置为未设置。在另一种情况下，我们将检查设置所有下一个位是否足以保持X> =L。否则，我们需要设置当前位。请注意，设置所有下一个位等效于加法(1 << b )– 1，其中b是当前位。

此方法的时间复杂度为O(log ₂ (R))。

C++

// CPP program to find the x in range [l, r]
// such that x ^ n is maximum.
#include 
#include 
using namespace std;
 
// Function to calculate the maximum value of
// N ^ X, where X is in the range [L, R]
int maximumXOR(int n, int l, int r)
{
    int x = 0;
    for (int i = log2(r); i >= 0; --i)
    {
        if (n & (1 << i))  // Set bit
        {
            if (x + (1 << i) - 1 < l)
                x ^= (1 << i);
        }
        else // Unset bit
        {
            if ((x ^ (1 << i)) <= r)
                x ^= (1 << i);
        }
    }
    return n ^ x;
}
 
// Driver Code
int main()
{
    int n = 7, l = 2, r = 23;
    cout << "The output is " << maximumXOR(n, l, r);
    return 0;
}

Java

// Java program to find the x in range [l, r]
// such that x ^ n is maximum.
 
import java.util.*;
import java.lang.*;
import java.io.*;
 
class GFG
{
// Function to calculate the maximum value of
// N ^ X, where X is in the range [L, R]
static int maximumXOR(int n, int l, int r)
{
    int x = 0;
    for (int i = (int)(Math.log(r)/Math.log(2)); i >= 0; --i)
    {
        if ((n & (1 << i))>0) // Set bit
        {
            if  (x + (1 << i) - 1 < l)
                x ^= (1 << i);
        }
        else // Unset bit
        {
            if ((x ^ (1 << i)) <= r)
                x ^= (1 << i);
        }
    }
    return n ^ x;
}
 
// Driver function
public static void main(String args[])
{
    int n = 7, l = 2, r = 23;
    System.out.println( "The output is " + maximumXOR(n, l, r));
 
}
}
 
// This code is Contributed by tufan_gupta2000

Python3

# Python program to find the
# x in range [l, r] such that
# x ^ n is maximum.
import math
 
# Function to calculate the
# maximum value of N ^ X,
# where X is in the range [L, R]
def maximumXOR(n, l, r):
    x = 0
    for i in range(int(math.log2(r)), -1, -1):
        if (n & (1 << i)): # Set bit
            if  (x + (1 << i) - 1 < l):
                x ^= (1 << i)
        else: # Unset bit
            if (x ^ (1 << i)) <= r:
                x ^= (1 << i)
    return n ^ x
 
# Driver code
n = 7
l = 2
r = 23
print("The output is",
       maximumXOR(n, l, r))
 
# This code was contributed
# by VishalBachchas

C#

// C# program to find the x in range
// [l, r] such that x ^ n is maximum.
using System;
 
class GFG
{
     
// Function to calculate the
// maximum value of N ^ X,
// where X is in the range [L, R]
public static int maximumXOR(int n,
                             int l, int r)
{
    int x = 0;
    for (int i = (int)(Math.Log(r) /
                       Math.Log(2)); i >= 0; --i)
    {
        if ((n & (1 << i)) > 0) // Set bit
        {
            if (x + (1 << i) - 1 < l)
            {
                x ^= (1 << i);
            }
        }
        else // Unset bit
        {
            if ((x ^ (1 << i)) <= r)
            {
                x ^= (1 << i);
            }
        }
    }
    return n ^ x;
}
 
// Driver Code
public static void Main(string[] args)
{
    int n = 7, l = 2, r = 23;
    Console.WriteLine("The output is " +
                   maximumXOR(n, l, r));
}
}
 
// This code is contributed
// by Shrikant13

PHP

= 0; --$i)
    {
        if ($n & (1 << $i))
        {  
            // Set bit
            if ($x + (1 << $i) - 1 < $l)
                $x ^= (1 << $i);
        }
        else
        {
            // Unset bit
            if (($x ^ (1 << $i)) <= $r)
                $x ^= (1 << $i);
        }
    }
    return $n ^ $x;
}
 
// Driver Code
$n = 7;
$l = 2;
$r = 23;
echo "The output is " ,
      maximumXOR($n, $l, $r);
 
// This code is contributed by ajit
?>

输出

The output is 23