三个数的总和与XOR

给定整数N。任务是计算整数A和B的对数，以使A + B + N = A ^ B ^ N且A和B小于N。

例子：

Input: N = 2
Output: 3
Explanation:-
For N = 2
2 XOR 0 XOR 0 = 2+0+0
2 XOR 0 XOR 1 = 2+0+1
2 XOR 0 XOR 2 != 2+0+2
2 XOR 1 XOR 0 = 2+1+0
2 XOR 1 XOR 1 != 2+1+1
2 XOR 1 XOR 2 != 2+1+2
2 XOR 2 XOR 0 != 2+2+0
2 XOR 2 XOR 1 != 2+2+1
2 XOR 2 XOR 2 != 2+2+2

So (0, 0), (0, 1) and (1, 0) are the required pairs. So the output is 3.

Input: N = 4
Output: 9

方法：
为了使三个数字的和等于三个数字的xor与给定数字之一的异或，我们可以执行以下操作：

1. 以二进制形式表示固定数字。
2. 遍历固定数字的二进制扩展。
   • 如果找到1，则只有一个条件，即将其他两个数字的二进制位分别设为0和0。
   • 如果找到0，则将存在三个条件，即，您可以将二进制位设置为(0，0)，(1，0)
     或(0，1)。
3. 以下上面的三元组将永远不会进位，因此它们是有效的。
4. 因此答案将是3 ^(二进制表示中的零数) 。

下面是上述方法的实现：

C++

// C++ implementation of the above approach
#include 
using namespace std;
  
// Defining ull to unsigned long long int
typedef unsigned long long int ull;
  
// Function to calculate power of 3
ull calculate(int bit_cnt)
{
    ull res = 1;
    while (bit_cnt--) {
        res = res * 3;
    }
  
    return res;
}
  
// Function to return the count of the
// unset bit ( zeros )
int unset_bit_count(ull n)
{
    int count = 0;
    while (n) {
  
        // Check the bit is 0 or not
        if ((n & 1) == 0)
            count++;
        // Right shifting ( dividing by 2 )
        n = n >> 1;
    }
    return count;
}
  
// Driver Code
int main()
{
    ull n;
    n = 2;
  
    int count = unset_bit_count(n);
  
    ull ans = calculate(count);
  
    cout << ans << endl;
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to calculate power of 3
static long calculate(int bit_cnt)
{
    long res = 1;
    while (bit_cnt-- > 0) 
    {
        res = res * 3;
    }
  
    return res;
}
  
// Function to return the count of the
// unset bit ( zeros )
static int unset_bit_count(long n)
{
    int count = 0;
    while (n > 0) 
    {
  
        // Check the bit is 0 or not
        if ((n & 1) == 0)
            count++;
              
        // Right shifting ( dividing by 2 )
        n = n >> 1;
    }
    return count;
}
  
// Driver Code
public static void main(String[] args)
{
    long n;
    n = 2;
  
    int count = unset_bit_count(n);
  
    long ans = calculate(count);
  
    System.out.println(ans);
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
  
# Function to calculate power of 3
def calculate(bit_cnt):
  
    res = 1;
    while (bit_cnt > 0): 
        bit_cnt -= 1;
        res = res * 3;
    return res;
  
# Function to return the count of the
# unset bit ( zeros )
def unset_bit_count(n):
  
    count = 0;
    while (n > 0):
          
        # Check the bit is 0 or not
        if ((n & 1) == 0):
            count += 1;
              
        # Right shifting ( dividing by 2 )
        n = n >> 1;
      
    return count;
  
# Driver Code
if __name__ == '__main__':
  
    n = 2;
  
    count = unset_bit_count(n);
  
    ans = calculate(count);
  
    print(ans);
  
# This code contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;
  
class GFG
{
  
// Function to calculate power of 3
static long calculate(int bit_cnt)
{
    long res = 1;
    while (bit_cnt-- > 0) 
    {
        res = res * 3;
    }
  
    return res;
}
  
// Function to return the count of the
// unset bit (zeros)
static int unset_bit_count(long n)
{
    int count = 0;
    while (n > 0) 
    {
  
        // Check the bit is 0 or not
        if ((n & 1) == 0)
            count++;
              
        // Right shifting ( dividing by 2 )
        n = n >> 1;
    }
    return count;
}
  
// Driver Code
public static void Main(String[] args)
{
    long n;
    n = 2;
  
    int count = unset_bit_count(n);
  
    long ans = calculate(count);
  
    Console.WriteLine(ans);
}
}
  
// This code is contributed by 29AjayKumar

输出：

3

时间复杂度：O(未设置位数)