x <= n的值计数，其中(n XOR x)=(n – x)

给定整数n ，任务是找到满足n XOR x = n – x的可能值0≤x≤n的数量。
例子：

Input: n = 5 
Output: 4 
Following values of x satisfy the equation 
5 XOR 0 = 5 – 0 = 5 
5 XOR 1 = 5 – 1 = 4 
5 XOR 4 = 5 – 4 = 1 
5 XOR 5 = 5 – 5 = 0
Input: n = 2 
Output: 2 
 

天真的方法：一种简单的方法是检查从0到n(包括两个端点)的所有值，并确定它们是否满足方程式。下面的代码实现了这种方法：
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to return the count of
// valid values of x
static int countX(int n)
{
    int count = 0;
 
    for (int i = 0; i <= n; i++)
    {
 
        // If n - x = n XOR x
        if (n - i == (n ^ i))
                count++;
    }
 
        // Return the required count;
        return count;
}
 
// Driver code
int main()
{
    int n = 5;
    int answer = countX(n);
    cout << answer;
}
 
// This code is contributed by
// Shivi_Aggarwal

Java

// Java implementation of the approach
public class GFG {
 
    // Function to return the count of
    // valid values of x
    static int countX(int n)
    {
        int count = 0;
 
        for (int i = 0; i <= n; i++) {
 
            // If n - x = n XOR x
            if (n - i == (n ^ i))
                count++;
        }
 
        // Return the required count;
        return count;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        int answer = countX(n);
        System.out.println(answer);
    }
}

Python3

# Python3 implementation of the approach
import math as mt
 
# Function to return the count of
# valid values of x
def countX(n):
    count = 0
 
    for i in range(n + 1):
 
        if n - i == (n ^ i):
            count += 1
 
    return count
 
# Driver Code
if __name__ == '__main__':
    n = 5
    answer = countX(n)
    print(answer)
 
# This code is contributed by
# Mohit kumar 29

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
    // Function to return the count of
    // valid values of x
    static int countX(int n)
    {
        int count = 0;
 
        for (int i = 0; i <= n; i++)
        {
 
            // If n - x = n XOR x
            if (n - i == (n ^ i))
                count++;
        }
 
        // Return the required count;
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 5;
        int answer = countX(n);
        Console.WriteLine(answer);
    }
}
 
// This code is contributed by Ryuga

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count of
// valid values of x
function countX(n)
{
    let count = 0;
 
    for (let i = 0; i <= n; i++)
    {
 
        // If n - x = n XOR x
        if (n - i == (n ^ i))
                count++;
    }
 
        // Return the required count;
        return count;
}
 
// Driver code
    let n = 5;
    let answer = countX(n);
    document.write(answer);
 

输出：

4

时间复杂度： O(N)
高效方法：将n转换为其二进制表示形式。现在，对于二进制字符串中的每1 ，无论我们减去1还是0 ，都将等于1与0或1的XOR，即
(1-1)=(1 XOR 1)= 0
(1 – 0)=(1 XOR 0)= 1
但是0不满足此条件。因此，我们只需要考虑n的二进制表示形式中的所有那些。现在，每1有两种可能性，即0或1 。因此，如果我们在n中有m个1的数目，那么我们的解将是2 ^m 。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Function to return the count of
// valid values of x
int countX(int n)
{
    // Convert n into binary String
    string binary = bitset<8>(n).to_string();
     
    // To store the count of 1s
    int count = 0;
    for (int i = 0; i < binary.length(); i++)
    {
        // If current bit is 1
        if (binary.at(i) == '1')
            count++;
    }
     
    // Calculating answer
    int answer = (int)pow(2, count);
    return answer;
}
 
// Driver code
int main()
{
    int n = 5;
    int answer = countX(n);
    cout << (answer);
}
 
// This code is contributed by Rajput-Ji

Java

// Java implementation of the approach
public class GFG {
 
    // Function to return the count of
    // valid values of x
    static int countX(int n)
    {
        // Convert n into binary String
        String binary = Integer.toBinaryString(n);
 
        // To store the count of 1s
        int count = 0;
 
        for (int i = 0; i < binary.length(); i++) {
 
            // If current bit is 1
            if (binary.charAt(i) == '1')
                count++;
        }
 
        // Calculating answer
        int answer = (int)Math.pow(2, count);
        return answer;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 5;
        int answer = countX(n);
        System.out.println(answer);
    }
}

的Python 3

# Python3 implementation of the approach
 
# Function to return the count of
# valid values of x
def countX(n):
 
    # Convert n into binary String
    binary = "{0:b}".format(n)
 
    # To store the count of 1s
    count = 0
 
    for i in range(len(binary)):
 
        # If current bit is 1
        if (binary[i] == '1'):
            count += 1
 
    # Calculating answer
    answer = int(pow(2, count))
    return answer
 
# Driver code
if __name__ == "__main__":
     
    n = 5
    answer = countX(n)
    print(answer)
 
# This code is contributed by ita_c

C＃

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count of
// valid values of x
static int countX(int n)
{
    // Convert n into binary String
    string binary = Convert.ToString(n, 2);
 
    // To store the count of 1s
    int count = 0;
 
    for (int i = 0; i < binary.Length; i++)
    {
 
        // If current bit is 1
        if (binary[i] == '1')
            count++;
    }
 
    // Calculating answer
    int answer = (int)Math.Pow(2, count);
    return answer;
}
 
// Driver code
public static void Main()
{
    int n = 5;
    int answer = countX(n);
    Console.WriteLine(answer);
}
}
 
// This code is contributed
// by Akanksha Rai

输出：

4

时间复杂度： $$O(\log{}n)$$