根据给定的按位与和按位异或值计算两个整数的按位或

给定两个整数X和Y ，代表两个正整数的按位异或和按位与，任务是计算这两个正整数的按位或值。

例子：

Input: X = 5, Y = 2 
Output: 7 
Explanation: 
If A and B are two positive integers such that A ^ B = 5, A & B = 2, then the possible value of A and B is 3 and 6 respectively. 
Therefore, (A | B) = (3 | 6) = 7.

Input: X = 14, Y = 1 
Output: 15 
Explanation: 
If A and B are two positive integers such that A ^ B = 14, A & B = 1, then the possible value of A and B is 7 and 9 respectively. 
Therefore, (A | B) = (7 | 9) = 15.

朴素的方法：解决这个问题的最简单的方法是迭代到X和Y的最大值，比如N ，并生成前N 个自然数的所有可能对。对于每一对，检查该对的按位异或和按位与是否分别为X和Y 。如果发现为真，则打印该对的按位或。

时间复杂度： O(N ² )，其中 N = max(X, Y)
辅助空间： O(1)

高效的方法：为了优化上述方法，该想法基于以下观察：

(A ^ B) = (A | B) – (A & B) 
=> (A | B) = (A ^ B) + (A & B) = X + Y 
 

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
int findBitwiseORGivenXORAND(int X, int Y)
{
    return X + Y;
}
 
// Driver Code
int main()
{
    int X = 5, Y = 2;
    cout << findBitwiseORGivenXORAND(X, Y);
}

Java

// Java program to implement
// the above approach
class GFG {
 
    // Function to calculate Bitwise OR from given
    // bitwise XOR and bitwise AND values
    static int findBitwiseORGivenXORAND(int X, int Y)
    {
        return X + Y;
    }
     
    // Driver Code
    public static void main (String[] args)
    {
        int X = 5, Y = 2;
        System.out.print(findBitwiseORGivenXORAND(X, Y));
    }
}
 
// This code is contributed by AnkitRai01

Python3

# Python3 program to implement
# the above approach
 
# Function to calculate Bitwise OR from
# given bitwise XOR and bitwise AND values
def findBitwiseORGivenXORAND(X, Y):
 
    return X + Y
 
# Driver Code
if __name__ == "__main__" :
     
    X = 5
    Y = 2
     
    print(findBitwiseORGivenXORAND(X, Y))
 
# This code is contributed by AnkitRai01

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
static int findBitwiseORGivenXORAND(int X, int Y)
{
    return X + Y;
}
 
// Driver Code
public static void Main(string []args)
{
    int X = 5, Y = 2;
     
    Console.Write(findBitwiseORGivenXORAND(X, Y));
}
}
 
// This code is contributed by ipg2016107

Javascript

// JavaScript program to implement
// the above approach
 
// Function to calculate Bitwise OR from given
// bitwise XOR and bitwise AND values
function findBitwiseORGivenXORAND(X, Y)
{
    return X + Y;
}
 
// Driver Code
 
    let X = 5, Y = 2;
    document.write(findBitwiseORGivenXORAND(X, Y));
 
// This code is contributed by Surbhi Tyagi.
 

输出：

7

时间复杂度： O(1)
辅助空间： O(1)

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