位大小最大为X并且按位或等于X的对的计数

给定数字X ，计算可能的对(a，b)的数量，以使a和b的按位或等于X并且a和b的位数都小于等于X的位数。

例子：

Input: X = 6 
Output: 9 
Explanation: 
The possible pairs of (a, b) are (4, 6), (6, 4), (6, 6), (6, 2), (4, 2), (6, 0), (2, 6), (2, 4), (0, 6).
Input: X = 21 
Output: 27 
Explanation: 
In total there are 27 pairs possible.

方法：要解决上述问题，请执行以下步骤：

• 遍历给定数字X的每一位。
• 如果该位为1，则根据按位或的真值表，我们知道数字a和b中的给定位可能有3种组合，即(0，1)，(1，0)，(1，1)有3种可能的方式。
• 如果该位为0，则从按位或的真值表中，我们知道数字a和b中的给定位只有(0，0)的1种组合。
• 因此，我们的答案将是3 ^(X中的接通位数)。

下面是上述方法的实现：

C++

// C++ implementation to Count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
 
#include 
using namespace std;
 
// Function to count the pairs
int count_pairs(int x)
{
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0) {
 
        // check if bit is 1
        if (x % 2 == 1)
 
            // multiplying ans by 3
            // if bit is 1
            ans = ans * 3;
 
        x = x / 2;
    }
    return ans;
}
 
// Driver code
int main()
{
    int X = 6;
 
    cout << count_pairs(X)
         << endl;
 
    return 0;
}

Java

// Java implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
class GFG{
 
// Function to count the pairs
static int count_pairs(int x)
{
     
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0)
    {
         
        // Check if bit is 1
        if (x % 2 == 1)
 
            // Multiplying ans by 3
            // if bit is 1
            ans = ans * 3;
 
        x = x / 2;
    }
    return ans;
}
 
// Driver code
public static void main(String[] args)
{
    int X = 6;
 
    System.out.print(count_pairs(X) + "\n");
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to count number of
# possible pairs of (a, b) such that
# their Bitwise OR gives the value X
 
# Function to count the pairs
def count_pairs(x):
 
    # Initializing answer with 1
    ans = 1;
 
    # Iterating through bits of x
    while (x > 0):
 
        # Check if bit is 1
        if (x % 2 == 1):
 
            # Multiplying ans by 3
            # if bit is 1
            ans = ans * 3;
 
        x = x // 2;
     
    return ans;
 
# Driver code
if __name__ == '__main__':
     
    X = 6;
 
    print(count_pairs(X));
 
# This code is contributed by amal kumar choubey

C#

// C# implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X
using System;
class GFG{
 
  // Function to count the pairs
  static int count_pairs(int x)
  {
 
    // Initializing answer with 1
    int ans = 1;
 
    // Iterating through bits of x
    while (x > 0)
    {
 
      // Check if bit is 1
      if (x % 2 == 1)
 
        // Multiplying ans by 3
        // if bit is 1
        ans = ans * 3;
 
      x = x / 2;
    }
    return ans;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int X = 6;
 
    Console.Write(count_pairs(X) + "\n");
  }
}
 
// This code is contributed by sapnasingh4991

Javascript

 
// javascript implementation to count number of
// possible pairs of (a, b) such that
// their Bitwise OR gives the value X   
// Function to count the pairs
    function count_pairs(x) {
 
        // Initializing answer with 1
        var ans = 1;
 
        // Iterating through bits of x
        while (x > 0) {
 
            // Check if bit is 1
            if (x % 2 == 1)
 
                // Multiplying ans by 3
                // if bit is 1
                ans = ans * 3;
 
            x = parseInt(x / 2);
        }
        return ans;
    }
 
    // Driver code
     
        var X = 6;
 
        document.write(count_pairs(X) + "\n");
 
// This code contributed by Rajput-Ji
 

输出：

9

时间复杂度： O(log(X))