按位 XOR 值大于其按位 AND 值的对的计数

给定一个包含N个正整数的数组arr 。查找按位XOR值大于按位AND值的所有可能对的计数

例子：

Input : arr[]={ 12, 4, 15}
Output:  2
Explanation: 12 ^ 4 = 8, 12 & 4 = 4. so 12 ^ 4 > 12 & 4
                       4 ^ 15 = 11,   4 & 15 = 4. so 4 ^ 15  > 4 & 15 . 
So, above two are valid pairs { 12, 4 }, {4, 15}  .

Input:  arr[] = { 1, 1, 3, 3 }
Output: 4

方法：参考下面的思路来解决这个问题：

We know that, for an array of length N, there will be a total (N*(N-1))/2 pairs. 

Therefore check for every pair, and increase count by one if a pair satisfy the given condition in the question. 

请按照以下步骤解决此问题：

• 检查每一对按位异或值是否大于按位与。
• 如果是，则增加计数
• 返回计数。

以下是上述方法的实现：

C++

// C++ program to Count number of pairs
// whose bit wise XOR value is greater
// than bit wise AND value
 
#include 
using namespace std;
 
// Function that return count of pairs
// whose bitwise xor >= bitwise
long long valid_pairs(int arr[], int N)
{
    long long cnt = 0;
    // check for all possible pairs
    for (int i = 0; i < N; i++) {
        for (int j = i + 1; j < N; j++) {
            if ((arr[i] ^ arr[j]) >= (arr[i] & arr[j]))
                cnt++;
        }
    }
 
    return cnt;
}
 
// Driver Code
int main()
{
    int N = 3;
    int arr[] = { 12, 4, 15 };
    cout << valid_pairs(arr, N);
}

Java

// Java program to Count number of pairs
// whose bit wise XOR value is greater
// than bit wise AND value
import java.io.*;
 
class GFG
{
 
  // Function that return count of pairs
  // whose bitwise xor >= bitwise
  public static long valid_pairs(int arr[], int N)
  {
    long cnt = 0;
 
    // check for all possible pairs
    for (int i = 0; i < N; i++) {
      for (int j = i + 1; j < N; j++) {
        if ((arr[i] ^ arr[j]) >= (arr[i] & arr[j]))
          cnt++;
      }
    }
 
    return cnt;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 3;
    int arr[] = { 12, 4, 15 };
    System.out.print(valid_pairs(arr, N));
  }
}
 
// This code is contributed by Rohit Pradhan.

Python3

# python3 program to Count number of pairs
# whose bit wise XOR value is greater
# than bit wise AND value
 
# Function that return count of pairs
# whose bitwise xor >= bitwise
def valid_pairs(arr, N):
 
    cnt = 0
     
    # check for all possible pairs
    for i in range(0, N):
        for j in range(i + 1, N):
            if ((arr[i] ^ arr[j]) >= (arr[i] & arr[j])):
                cnt += 1
 
    return cnt
 
# Driver Code
if __name__ == "__main__":
 
    N = 3
    arr = [12, 4, 15]
    print(valid_pairs(arr, N))
 
    # This code is contributed by rakeshsahni

Javascript

       // JavaScript code for the above approach
 
 
       // Function that return count of pairs
       // whose bitwise xor >= bitwise
       function valid_pairs(arr, N) {
           let cnt = 0;
           // check for all possible pairs
           for (let i = 0; i < N; i++) {
               for (let j = i + 1; j < N; j++) {
                   if ((arr[i] ^ arr[j]) >= (arr[i] & arr[j]))
                       cnt++;
               }
           }
 
           return cnt;
       }
 
       // Driver Code
 
       let N = 3;
       let arr = [12, 4, 15];
       document.write(valid_pairs(arr, N));
 
   // This code is contributed by Potta Lokesh
   

输出

2

时间复杂度：O(N*N)
辅助空间：O(1)