最大化表达式(A AND X)*(B AND X)|位操作

给定两个正整数A和B使得A！= B ，任务是找到一个使表达式(A AND X)*(B AND X)最大化的正整数X。

例子：

Input: A = 9 B = 8
Output: 8
(9 AND 8) * (8 AND 8) = 8 * 8 = 64 (maximum possible)

Input: A = 11 and B = 13
Output: 9

天真的方法：可以运行从1到max(A，B)的循环，并且可以轻松地找到使给定表达式最大化的X。

高效的方法：众所周知，

(a – b)² ≥ 0
which implies (a + b)² – 4*a*b ≥ 0
which implies a * b ≤ (a + b)² / 4

Hence, it concludes that a * b will be maximum when a * b = (a + b)² / 4
which implies a = b
From the above result, (A AND X) * (B AND X) will be maximum when (A AND X) = (B AND X)

现在X可以找到为：

A = 11 = 1011
B = 13 = 1101
X = ? = abcd

At 0th place: (1 AND d) = (1 AND d) implies d = 0, 1 but to maximize (A AND X) * (B AND X) d = 1
At 1st place: (1 AND d) = (0 AND d) implies c = 0
At 2nd place: (0 AND d) = (1 AND d) implies b = 0
At 3rd place: (1 AND d) = (1 AND d) implies a = 0, 1 but to maximize (A AND X) * (B AND X) a = 1

Hence, X = 1001 = 9

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 find X according
// to the given conditions
int findX(int A, int B)
{
    int X = 0;
  
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++) {
  
        // Temporary ith bit
        int tempBit = 1 << bit;
  
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
  
        // Add the ith bit of X to X
        X += bitOfX;
    }
  
    return X;
}
  
// Driver code
int main()
{
    int A = 11, B = 13;
  
    cout << findX(A, B);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
static int MAX = 32;
  
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
    int X = 0;
  
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++)
    {
  
        // Temporary ith bit
        int tempBit = 1 << bit;
  
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
  
        // Add the ith bit of X to X
        X += bitOfX;
    }
    return X;
}
  
// Driver code
public static void main(String []args) 
{
    int A = 11, B = 13;
  
    System.out.println(findX(A, B));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
MAX = 32
  
# Function to find X according 
# to the given conditions 
def findX(A, B) :
  
    X = 0; 
  
    # int can have 32 bits 
    for bit in range(MAX) :
  
        # Temporary ith bit 
        tempBit = 1 << bit; 
  
        # Compute ith bit of X according to 
        # given conditions 
        # Expression below is the direct 
        # conclusion from the illustration 
        # we had taken earlier 
        bitOfX = A & B & tempBit; 
  
        # Add the ith bit of X to X 
        X += bitOfX; 
  
    return X; 
  
# Driver code 
if __name__ == "__main__" :
      
    A = 11; B = 13; 
    print(findX(A, B)); 
  
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
      
class GFG
{
static int MAX = 32;
  
// Function to find X according
// to the given conditions
static int findX(int A, int B)
{
    int X = 0;
  
    // int can have 32 bits
    for (int bit = 0; bit < MAX; bit++)
    {
  
        // Temporary ith bit
        int tempBit = 1 << bit;
  
        // Compute ith bit of X according to
        // given conditions
        // Expression below is the direct
        // conclusion from the illustration
        // we had taken earlier
        int bitOfX = A & B & tempBit;
  
        // Add the ith bit of X to X
        X += bitOfX;
    }
    return X;
}
  
// Driver code
public static void Main(String []args) 
{
    int A = 11, B = 13;
  
    Console.WriteLine(findX(A, B));
}
}
  
// This code is contributed by 29AjayKumar

输出：

9

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