给定一个范围[L，R]，我们需要找到此范围内的两个整数，以使它们的XOR在两个整数的所有可能选择中最大。更正式地说，
给定[L，R]，求最大值(A ^ B)其中L <= A，B
例子 ：
Input : L = 8
R = 20
Output : 31
31 is XOR of 15 and 16.
Input : L = 1
R = 3
Output : 3一个简单的解决方案是生成所有对，找到它们的XOR值，最后返回最大XOR值。
一个有效的解决方案是考虑从L到R的二进制值的模式。我们可以看到，从L到R的第一位从0变为1或保持为1，即如果我们将任意两个数字的XOR作为最大值，则为第一个该位将被固定，与L和R本身的XOR的第一位相同。
观察获得第一位的技术后，我们可以看到，如果对L和R进行XOR，则此XOR的最高有效位将告诉我们可以实现的最大值，即让L和R的XOR为1xxx，其中x可以为0或1，那么我们可以得到的最大XOR值是1111，因为从L到R，我们具有xxx的所有可能组合，并且总是可以从两个数字中选择这些位，以使它们的XOR变为全1。下面有一些例子，
Examples 1:
L = 8 R = 20
L ^ R = (01000) ^ (10100) = (11100)
Now as L ^ R is of form (1xxxx) we
can get maximum XOR as (11111) by
choosing A and B as 15 and 16 (01111
and 10000)
Examples 2:
L = 16 R = 20
L ^ R = (10000) ^ (10100) = (00100)
Now as L ^ R is of form (1xx) we can
get maximum xor as (111) by choosing
A and B as 19 and 20 (10011 and 10100)因此，此问题的解决方案仅取决于(L ^ R)的值。我们将首先计算L ^ R值，然后从该值的最高有效位开始，将所有1加1以获得最终结果。
C++
// C/C++ program to get maximum xor value
// of two numbers in a range
#include
using namespace std;
// method to get maximum xor value in range [L, R]
int maxXORInRange(int L, int R)
{
// get xor of limits
int LXR = L ^ R;
// loop to get msb position of L^R
int msbPos = 0;
while (LXR)
{
msbPos++;
LXR >>= 1;
}
// construct result by adding 1,
// msbPos times
int maxXOR = 0;
int two = 1;
while (msbPos--)
{
maxXOR += two;
two <<= 1;
}
return maxXOR;
}
// Driver code to test above methods
int main()
{
int L = 8;
int R = 20;
cout << maxXORInRange(L, R) << endl;
return 0;
} Java
// Java program to get maximum xor value
// of two numbers in a range
class Xor
{
// method to get maximum xor value in range [L, R]
static int maxXORInRange(int L, int R)
{
// get xor of limits
int LXR = L ^ R;
// loop to get msb position of L^R
int msbPos = 0;
while (LXR > 0)
{
msbPos++;
LXR >>= 1;
}
// construct result by adding 1,
// msbPos times
int maxXOR = 0;
int two = 1;
while (msbPos-- >0)
{
maxXOR += two;
two <<= 1;
}
return maxXOR;
}
// main function
public static void main (String[] args)
{
int L = 8;
int R = 20;
System.out.println(maxXORInRange(L, R));
}
}Python3
# Python3 program to get maximum xor
# value of two numbers in a range
# Method to get maximum xor
# value in range [L, R]
def maxXORInRange(L, R):
# get xor of limits
LXR = L ^ R
# loop to get msb position of L^R
msbPos = 0
while(LXR):
msbPos += 1
LXR >>= 1
# construct result by adding 1,
# msbPos times
maxXOR, two = 0, 1
while (msbPos):
maxXOR += two
two <<= 1
msbPos -= 1
return maxXOR
# Driver code
L, R = 8, 20
print(maxXORInRange(L, R))
# This code is contributed by Anant Agarwal.C#
// C# program to get maximum xor
// value of two numbers in a range
using System;
class Xor
{
// method to get maximum xor
// value in range [L, R]
static int maxXORInRange(int L, int R)
{
// get xor of limits
int LXR = L ^ R;
// loop to get msb position of L^R
int msbPos = 0;
while (LXR > 0)
{
msbPos++;
LXR >>= 1;
}
// construct result by
// adding 1, msbPos times
int maxXOR = 0;
int two = 1;
while (msbPos-- >0)
{
maxXOR += two;
two <<= 1;
}
return maxXOR;
}
// Driver code
public static void Main()
{
int L = 8;
int R = 20;
Console.WriteLine(maxXORInRange(L, R));
}
}
// This code is contributed by Anant Agarwal.PHP
>= 1;
}
// construct result by
// adding 1, msbPos times
$maxXOR = 0;
$two = 1;
while ($msbPos--)
{
$maxXOR += $two;
$two <<= 1;
}
return $maxXOR;
}
// Driver Code
$L = 8;
$R = 20;
echo maxXORInRange($L, $R), "\n";
// This code is contributed by aj_36
?>Javascript
输出 ：
31