位与或位与之和等于N的非负对

给定一个正整数N，任务是找到这样所有非负对(A，B)，其按位或和按位的AND A的总和，B等于N，即，(A | B)+(A＆ B)= N。

例子：

Input: N = 5
Output: (0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0)
Explanation: All possible pairs satisfying the necessary conditions: 

1. (0 | 5) + (0 & 5) = 5 + 0 = 5
2. (1 | 4) + (1 & 4) = 5 + 0 = 5
3. (2 | 3) + (2 & 3) = 3 + 2 = 5
4. (3 | 2) + (3 & 2) = 3 + 2 = 5
5. (4 | 1) + (4 & 1) = 5 + 0 = 5
6. (5 | 0) + (5 & 0) = 5 + 0 = 5

Input: N = 7
Output: (0, 7), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (7, 0)
Explanation: All possible pairs satisfying the necessary conditions:  

1. (0 | 7) + (0 & 7) = 7 + 0 =7
2. (1 | 6) + (1 & 6) = 7 + 0 =7
3. (2 | 5) + (2 & 5) = 7 + 0 =7
4. (3 | 4) + (3 & 4) = 7 + 0 =7
5. (4 | 3) + (4 & 3) = 7 + 0 =7
6. (5 | 2) + (5 & 2) = 7 + 0 =7
7. (6 | 1) + (6 & 1) = 7 + 0 = 7
8. (7 | 0) + (7 & 0) = 7 + 0 = 7

天真的方法：最简单的方法是在[0，N]范围内进行迭代并打印满足条件(A | B)+(A＆B)= N的那些对(A，B) 。

时间复杂度： O(N ² )
辅助空间： O(1)

高效方法：为了优化上述方法，该思想基于以下观察结果：所有总和等于N的非负对都满足给定条件。因此，使用变量i在[0，N]范围内进行迭代，并打印对i和(N – i) 。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to print all pairs whose
// sum of Bitwise OR and AND is N
void findPairs(int N)
{
    // Iterate from i = 0 to N
    for (int i = 0; i <= N; i++) {
 
        // Print pairs (i, N - i)
        cout << "(" << i << ", "
             << N - i << "), ";
    }
}
 
// Driver Code
int main()
{
    int N = 5;
    findPairs(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
import java.lang.*;
 
class GFG{
     
// Function to print all pairs whose
// sum of Bitwise OR and AND is N
static void findPairs(int N)
{
     
    // Iterate from i = 0 to N
    for(int i = 0; i <= N; i++)
    {
         
        // Print pairs (i, N - i)  
        System.out.print( "(" + i + ", " +
                          (N - i) + "), ");
   }
}
 
// Driver code
public static void main(String[] args)
{
    int N = 5;
     
    findPairs(N);
}
}
 
// This code is contributed by ajaykr00kj

Python3

# Python3 program for the above approach
 
# Function to print all pairs whose
# sum of Bitwise OR and AND is N
def findPairs(N):
     
     # Iterate from i = 0 to N
     for i in range(0, N + 1):
         
        # Print pairs (i, N - i)
        print("(", i, ",",
              N - i, "), ", end = "")
 
# Driver code
if __name__ == "__main__":
     
    N = 5
     
    findPairs(N)
 
# This code is contributed by ajaykr00kj

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to print all pairs whose
// sum of Bitwise OR and AND is N
static void findPairs(int N)
{
     
    // Iterate from i = 0 to N
    for(int i = 0; i <= N; i++)
    {
         
        // Print pairs (i, N - i)  
        Console.Write( "(" + i + ", " +
                        (N - i) + "), ");
   }
}
 
// Driver code
public static void Main()
{
    int N = 5;
      
    findPairs(N);
}
}
 
// This code is contributed by sanjoy_62

Javascript

// JavaScript program for the above approach
 
// Function to print all pairs whose
// sum of Bitwise OR and AND is N
function findPairs(N)
{
      
    // Iterate from i = 0 to N
    for(let i = 0; i <= N; i++)
    {
          
        // Print pairs (i, N - i) 
        document.write( "(" + i + ", " +
                          (N - i) + "), ");
   }
}
 
// Driver Code
  let N = 5;
  findPairs(N);
  
 // This code is contributed by avijitmondal1998.

输出：

(0, 5), (1, 4), (2, 3), (3, 2), (4, 1), (5, 0),

时间复杂度： O(N)
辅助空间： O(1)