使得a = b +(a ^ b)的b的数量

给定一个整数$a$ 。查找解决方案的数量$b$满足等式：

a = b + (a^b)

例子：

Input: a = 4 
Output: 2
The only values of b are 0 and 4 itself.

Input: a = 3 
Output: 4 

一个幼稚的解决方案是从0迭代到$a$并计算满足给定方程的值的数量。我们只需要走到$a$因为任何大于$a$将给出XOR值> $a$ ，因此不能满足方程式。

下面是上述方法的实现：

C++

// C++ program to find the number of values
// of b such that a = b + (a^b)
#include 
using namespace std;
 
// function to return the number of solutions
int countSolutions(int a)
{
    int count = 0;
 
    // check for every possible value
    for (int i = 0; i <= a; i++) {
        if (a == (i + (a ^ i)))
            count++;
    }
 
    return count;
}
 
// Driver Code
int main()
{
    int a = 3;
    cout << countSolutions(a);
}

Java

// Java program to find the number of values
// of b such that a = b + (a^b)
 
import java.io.*;
 
class GFG {
 
 
// function to return the number of solutions
 static int countSolutions(int a)
{
    int count = 0;
 
    // check for every possible value
    for (int i = 0; i <= a; i++) {
        if (a == (i + (a ^ i)))
            count++;
    }
 
    return count;
}
 
// Driver Code
 
 
    public static void main (String[] args) {
        int a = 3;
    System.out.println( countSolutions(a));
    }
}
// This code is contributed by inder_verma

Python3

# Python 3 program to find
# the number of values of b
# such that a = b + (a^b)
 
# function to return the
# number of solutions
def countSolutions(a):
 
    count = 0
 
    # check for every possible value
    for i in range(a + 1):
        if (a == (i + (a ^ i))):
            count += 1
 
    return count
 
# Driver Code
if __name__ == "__main__":
    a = 3
    print(countSolutions(a))
 
# This code is contributed
# by ChitraNayal

C#

// C# program to find the number of
// values of b such that a = b + (a^b)
using System;
 
class GFG
{
 
// function to return the
// number of solutions
static int countSolutions(int a)
{
    int count = 0;
 
    // check for every possible value
    for (int i = 0; i <= a; i++)
    {
        if (a == (i + (a ^ i)))
            count++;
    }
 
    return count;
}
 
// Driver Code
public static void Main ()
{
    int a = 3;
    Console.WriteLine(countSolutions(a));
}
}
 
// This code is contributed by inder_verma

Javascript

 
// Javascript program to find the number of values
// of b such that a = b + (a^b)
 
// function to return the number of solutions
function countSolutions(a)
{
    let count = 0;
 
    // check for every possible value
    for (let i = 0; i <= a; i++) {
        if (a == (i + (a ^ i)))
            count++;
    }
 
    return count;
}
 
// Driver Code
let a = 3;
document.write(countSolutions(a));
 

输出：

4

时间复杂度：O(a)

一种有效的方法是，当我们针对不同的值编写可能的解决方案时，观察答案的模式。 $a$ 。仅将设置的位用于确定可能的答案的数量。问题的答案将始终是2 ^(设置位数)，可以通过观察确定。

下面是上述方法的实现：

C++

// C++ program to find the number of values
// of b such that a = b + (a^b)
#include 
using namespace std;
 
// function to return the number of solutions
int countSolutions(int a)
{
    int count = __builtin_popcount(a);
 
    count = pow(2, count);
    return count;
}
 
// Driver Code
int main()
{
    int a = 3;
    cout << countSolutions(a);
}

Java

// Java program to find the number of values
// of b such that a = b + (a^b)
import java.io.*;
class GFG
{
  
    // function to return the number of solutions
    static int countSolutions(int a)
    {
        int count = Integer.bitCount(a);
      
        count =(int) Math.pow(2, count);
        return count;
    }
      
    // Driver Code
        public static void main (String[] args)
    {
        int a = 3;
        System.out.println(countSolutions(a));
    }
}
// This code is contributed by Raj

Python3

# Python3 program to find the number
# of values of b such that a = b + (a^b)
 
# function to return the number
# of solutions
def countSolutions(a):
 
    count = bin(a).count('1')
    return 2 ** count
 
# Driver Code
if __name__ == "__main__":
 
    a = 3
    print(countSolutions(a))
 
# This code is contributed by
# Rituraj Jain

C＃

// C# program to find the number of
// values of b such that a = b + (a^b)
class GFG
{
 
// function to return the number
// of solutions
static int countSolutions(int a)
{
    int count = bitCount(a);
 
    count =(int) System.Math.Pow(2, count);
    return count;
}
 
static int bitCount(int n)
{
    int count = 0;
    while (n != 0)
    {
        count++;
        n &= (n - 1);
    }
    return count;
}
 
// Driver Code
public static void Main()
{
    int a = 3;
    System.Console.WriteLine(countSolutions(a));
}
}
 
// This code is contributed by mits

Java脚本

 
// Javascript program to find the number
// of values of b such that a = b + (a^b)
function bitCount(n)
{
    let count = 0;
    while (n != 0)
    {
        count++;
        n &= (n - 1);
    }
    return count;
}
 
// Function to return the number of solutions
function countSolutions(a)
{
    let count = bitCount(a);
 
    count = Math.pow(2, count);
    return count;
}
 
// Driver Code
let a = 3;
 
document.write(countSolutions(a));
 
// This code is contributed by subhammahato348
 

输出：

4

时间复杂度： O(log N)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。