给定两个正整数A和B。让我们定义D ，使B AND D = D。任务是使表达式A XOR D最大化。
例子：
Input: A = 11 B = 4
Output: 15
Take D = 4 as (B AND D) = (4 AND 4) = 4.
Also, (A XOR D) = (11 XOR 4) = 15 which is the
maximum according to the given condition.
Input: A = 9 and B = 13
Output: 13
天真的方法：由于B AND D = D ，所以D总是小于或等于B。因此，可以从1到B循环运行，并检查是否满足给定条件。
高效的方法：无需运行循环并检查每个D ，可以使用位操作技术轻松计算表达式的最大值(A XOR D) 。
让我们举一个例子来了解解决问题的方法：
A = 11 = 1011, B = 14 = 1110
Let's assume D = abcd in base 2 notation
B AND D: 1110 A XOR D: 1011
abcd abcd
------ ------
abcd ????
At 0th place: (0 AND d) = d implies d = 0
At 1st place: (1 AND c) = c implies c = 0, 1 but to maximize (A XOR D), take c = 0
At 2nd place: (1 AND b) = b implies b = 0, 1 but to maximize (A XOR D), take b = 1
At 3rd place: (1 AND a) = a implies a = 0, 1 but to maximize (A XOR D), take a = 0
Hence, D = 0100 = 4 and maximum value of (A XOR D) = (11 XOR 4) = 15.
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
#define MAX 32
// Function to return the value of
// the maximized expression
int maximizeExpression(int a, int b)
{
int result = a;
// int can have 32 bits
for (int bit = MAX - 1; bit >= 0; bit--) {
// Consider the ith bit of D to be 1
int bitOfD = 1 << bit;
// Calculate the value of (B AND bitOfD)
int x = b & bitOfD;
// Check if bitOfD satisfies (B AND D = D)
if (x == bitOfD) {
// Check if bitOfD can maximize (A XOR D)
int y = result & bitOfD;
if (y == 0) {
result = result ^ bitOfD;
}
}
// Note that we do not need to consider ith bit of D
// to be 0 because if above condition are not satisfied
// then value of result will not change
// which is similar to considering bitOfD = 0
// as result XOR 0 = result
}
return result;
}
// Driver code
int main()
{
int a = 11, b = 14;
cout << maximizeExpression(a, b);
return 0;
} Java
// Java implementation of the approach
class GFG
{
final static int MAX = 32;
// Function to return the value of
// the maximized expression
static int maximizeExpression(int a, int b)
{
int result = a;
// int can have 32 bits
for (int bit = MAX - 1; bit >= 0; bit--)
{
// Consider the ith bit of D to be 1
int bitOfD = 1 << bit;
// Calculate the value of (B AND bitOfD)
int x = b & bitOfD;
// Check if bitOfD satisfies (B AND D = D)
if (x == bitOfD) {
// Check if bitOfD can maximize (A XOR D)
int y = result & bitOfD;
if (y == 0)
{
result = result ^ bitOfD;
}
}
// Note that we do not need to consider ith bit of D
// to be 0 because if above condition are not satisfied
// then value of result will not change
// which is similar to considering bitOfD = 0
// as result XOR 0 = result
}
return result;
}
// Driver code
public static void main (String[] args)
{
int a = 11, b = 14;
System.out.println(maximizeExpression(a, b));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the approach
MAX = 32
# Function to return the value of
# the maximized expression
def maximizeExpression(a, b) :
result = a
# int can have 32 bits
for bit in range(MAX - 1, -1, -1) :
# Consider the ith bit of D to be 1
bitOfD = 1 << bit
# Calculate the value of (B AND bitOfD)
x = b & bitOfD
# Check if bitOfD satisfies (B AND D = D)
if (x == bitOfD) :
# Check if bitOfD can maximize (A XOR D)
y = result & bitOfD
if (y == 0) :
result = result ^ bitOfD
# Note that we do not need to consider ith bit of D
# to be 0 because if above condition are not satisfied
# then value of result will not change
# which is similar to considering bitOfD = 0
# as result XOR 0 = result
return result
# Driver code
a = 11
b = 14
print(maximizeExpression(a, b))
# This code is contributed by divyamohan123C#
// C# implementation of the above approach
using System;
class GFG
{
static int MAX = 32;
// Function to return the value of
// the maximized expression
static int maximizeExpression(int a, int b)
{
int result = a;
// int can have 32 bits
for (int bit = MAX - 1; bit >= 0; bit--)
{
// Consider the ith bit of D to be 1
int bitOfD = 1 << bit;
// Calculate the value of (B AND bitOfD)
int x = b & bitOfD;
// Check if bitOfD satisfies (B AND D = D)
if (x == bitOfD)
{
// Check if bitOfD can maximize (A XOR D)
int y = result & bitOfD;
if (y == 0)
{
result = result ^ bitOfD;
}
}
// Note that we do not need to consider
// ith bit of D to be 0 because if
// above condition are not satisfied then
// value of result will not change which is
// similar to considering bitOfD = 0 as
// result XOR 0 = result
}
return result;
}
// Driver code
public static void Main (String []args)
{
int a = 11, b = 14;
Console.WriteLine(maximizeExpression(a, b));
}
}
// This code is contributed by Arnab Kundu输出：
15