计算生成具有分别等于X和Y的按位XOR和按位AND的对的方法

给定两个整数X和Y中，任务是找到方式的总数量，以产生一对整数A和B的，使得按位XOR和按位与A和B之间分别是X和Y

例子：

Input: X = 2, Y = 5
Output: 2
Explanation:
The two possible pairs are (5, 7) and (7, 5).
Pair 1: (5, 7)
Bitwise AND = 5 & 7 = 2
Bitwise XOR = 5 ^ 7 = 5
Pair 2: (7, 5)
Bitwise AND = 7 & 5 = 2
Bitwise XOR = 7 ^ 5 = 5

Input: X = 7, Y = 5
Output: 0

天真的方法：解决问题的最简单方法是在X和Y中选择最大值，并设置所有位，然后检查从0到该最大数(例如M)的所有可能的对。如果对于任意一对A和B ， A和B和A andB分别等于X和Y ，则增加计数。在检查所有可能的对之后，打印count的最终值。
时间复杂度： O(M ² )
辅助空间： O(1)

高效方法：想法是在每个位置生成所有可能的位组合。 X和Y中的^第i位有4种可能性，如下所示：

1. 如果X _i = 0且Y _i = 1，则A _i = 1且B _i = 1。
2. 如果X _i = 1且Y _i = 1，则不存在可能的位分配。
3. 如果X _i = 0且Y _i = 0，则A _i = 0且B _i = 0。
4. 如果X _i = 1且Y _i = 0，则A _i = 0且B _i = 1或A _i = 1且B _i = 0，其中A _i ， B _i ， X _i和Y _i分别代表^第i^个比特其中。

请按照以下步骤解决问题：

1. 将计数器初始化为1 。
2. 对于^第i^个比特，如果X _I和Y _i为等于1，然后打印0。
3. 如果在^第i位， X _i为1且Y _i为0，则将计数器乘以2，因为有2个选项。
4. 之后，将X和Y分别除以2 。
5. 对^第i^个比特重复上述步骤，直到两个比特都变为0并打印计数器的值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to return the count of
// possible pairs of A and B whose
// Bitwise XOR is X and Y respectively
int countOfPairs(int x, int y)
{
    // Stores the count of pairs
    int counter = 1;
 
    // Iterate till any bit are set
    while (x || y) {
 
        // Extract i-th bit
        // of X and Y
        int bit1 = x % 2;
        int bit2 = y % 2;
 
        // Divide X and Y by 2
        x >>= 1;
        y >>= 1;
 
        // If Xi = 1 and Yi = 2,
        // multiply counter by 2
        if (bit1 == 1 and bit2 == 0) {
 
            // Increase required count
            counter *= 2;
            continue;
        }
 
        // If Xi =1 and Yi = 1
        if (bit1 & bit2) {
 
            // No answer exists
            counter = 0;
            break;
        }
    }
 
    // Return the final count
    return counter;
}
 
// Driver Code
int main()
{
    // Given X and Y
    int X = 2, Y = 5;
 
    // Function Call
    cout << countOfPairs(X, Y);
 
    return 0;
}

Java

// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to return the count of
// possible pairs of A and B whose
// Bitwise XOR is X and Y respectively
static int countOfPairs(int x, int y)
{
  // Stores the count of pairs
  int counter = 1;
 
  // Iterate till any bit are set
  while (x > 0 || y > 0)
  {
    // Extract i-th bit
    // of X and Y
    int bit1 = x % 2;
    int bit2 = y % 2;
 
    // Divide X and Y by 2
    x >>= 1;
    y >>= 1;
 
    // If Xi = 1 and Yi = 2,
    // multiply counter by 2
    if (bit1 == 1 && bit2 == 0)
    {
      // Increase required count
      counter *= 2;
      continue;
    }
 
    // If Xi =1 and Yi = 1
    if ((bit1 & bit2) > 0)
    {
      // No answer exists
      counter = 0;
      break;
    }
  }
 
  // Return the final count
  return counter;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given X and Y
  int X = 2, Y = 5;
 
  // Function Call
  System.out.print(countOfPairs(X, Y));
}
}
 
// This code is contributed by Princi Singh

Python3

# Python3 program for the above approach
 
# Function to return the count of
# possible pairs of A and B whose
# Bitwise XOR is X and Y respectively
def countOfPairs(x, y):
 
    # Stores the count of pairs
    counter = 1
 
    # Iterate till any bit are set
    while(x or y):
 
        # Extract i-th bit
        # of X and Y
        bit1 = x % 2
        bit2 = y % 2
 
        # Divide X and Y by 2
        x >>= 1
        y >>= 1
 
        # If Xi = 1 and Yi = 2,
        # multiply counter by 2
        if (bit1 == 1 and bit2 == 0):
             
            # Increase required count
            counter *= 2
            continue
 
        # If Xi =1 and Yi = 1
        if(bit1 & bit2):
 
            # No answer exists
            counter = 0
            break
 
    # Return the final count
    return counter
 
# Driver Code
 
# Given X and Y
X = 2
Y = 5
 
# Function call
print(countOfPairs(X, Y))
 
# This code is contributed by Shivam Singh

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to return the count of
// possible pairs of A and B whose
// Bitwise XOR is X and Y respectively
static int countOfPairs(int x, int y)
{
  // Stores the count of pairs
  int counter = 1;
 
  // Iterate till any bit are set
  while (x > 0 || y > 0)
  {
    // Extract i-th bit
    // of X and Y
    int bit1 = x % 2;
    int bit2 = y % 2;
 
    // Divide X and Y by 2
    x >>= 1;
    y >>= 1;
 
    // If Xi = 1 and Yi = 2,
    // multiply counter by 2
    if (bit1 == 1 && bit2 == 0)
    {
      // Increase required count
      counter *= 2;
      continue;
    }
 
    // If Xi =1 and Yi = 1
    if ((bit1 & bit2) > 0)
    {
      // No answer exists
      counter = 0;
      break;
    }
  }
 
  // Return the readonly count
  return counter;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given X and Y
  int X = 2, Y = 5;
 
  // Function Call
  Console.Write(countOfPairs(X, Y));
}
}
  
// This code is contributed by Rajput-Ji

Javascript

 
// JavaScript program for
// the above approach
 
// Function to return the count of
// possible pairs of A and B whose
// Bitwise XOR is X and Y respectively
function countOfPairs(x, y)
{
  // Stores the count of pairs
  let counter = 1;
  
  // Iterate till any bit are set
  while (x > 0 || y > 0)
  {
    // Extract i-th bit
    // of X and Y
    let bit1 = x % 2;
    let bit2 = y % 2;
  
    // Divide X and Y by 2
    x >>= 1;
    y >>= 1;
  
    // If Xi = 1 and Yi = 2,
    // multiply counter by 2
    if (bit1 == 1 && bit2 == 0)
    {
      // Increase required count
      counter *= 2;
      continue;
    }
  
    // If Xi =1 and Yi = 1
    if ((bit1 & bit2) > 0)
    {
      // No answer exists
      counter = 0;
      break;
    }
  }
  
  // Return the final count
  return counter;
}
 
// Driver code
 
   // Given X and Y
  let X = 2, Y = 5;
  
  // Function Call
  document.write(countOfPairs(X, Y));
                             

输出：

2

时间复杂度： O(log M)
辅助空间： O(1)