数组中的对数，以使2的最高乘方除以1的乘积

给定N个正整数的数组arr [] 。任务是找到对数(arr [i]，arr [j]) ，以使将arr [i] * arr [j]除以2的最大幂为1 。

例子：

Input: arr[] = {3, 5, 2, 8}
Output: 3
(3, 2), (5, 2) and (3, 5) are the only valid pairs.
Since the power of 2 that divides 3 * 2 = 6 is 1,
5 * 2 = 10 is 1 and 3 * 5 = 15 is 0.

Input: arr[] = {4, 2, 7, 11, 14, 15, 18}
Output: 12

方法：由于除以arr [i] * arr [j]的2的最大幂为max 1 ，这意味着如果P是乘积，则它必须是奇数或2是P的唯一偶数因子。
这意味着arr [i]和arr [j]都必须是奇数，或者恰好其中一个是偶数，而2是该数字的唯一偶数因子。
如果奇数是奇数，偶数是奇数，使得2是该数字的唯一偶数，则答案将是奇*偶+奇*(奇– 1)/ 2 。

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 pairs
int cntPairs(int a[], int n)
{
  
    // To store the count of odd numbers and
    // the count of even numbers such that 2
    // is the only even factor of that number
    int odd = 0, even = 0;
  
    for (int i = 0; i < n; i++) {
  
        // If current number is odd
        if (a[i] % 2 == 1)
            odd++;
  
        // If current number is even and 2
        // is the only even factor of it
        else if ((a[i] / 2) % 2 == 1)
            even++;
    }
  
    // Calculate total number of valid pairs
    int ans = odd * even + (odd * (odd - 1)) / 2;
  
    return ans;
}
  
// Driver code
int main()
{
  
    int a[] = { 4, 2, 7, 11, 14, 15, 18 };
    int n = sizeof(a) / sizeof(a[0]);
  
    cout << cntPairs(a, n);
  
    return 0;
}

Java

// Java implementation of the approach
class GFG 
{
  
// Function to return the count of valid pairs
static int cntPairs(int a[], int n)
{
  
    // To store the count of odd numbers and
    // the count of even numbers such that 2
    // is the only even factor of that number
    int odd = 0, even = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // If current number is odd
        if (a[i] % 2 == 1)
            odd++;
  
        // If current number is even and 2
        // is the only even factor of it
        else if ((a[i] / 2) % 2 == 1)
            even++;
    }
  
    // Calculate total number of valid pairs
    int ans = odd * even + (odd * (odd - 1)) / 2;
  
    return ans;
}
  
// Driver code
public static void main(String []args) 
{
    int a[] = { 4, 2, 7, 11, 14, 15, 18 };
    int n = a.length;
  
    System.out.println(cntPairs(a, n));
}
}
  
// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach 
  
# Function to return the count of valid pairs 
def cntPairs(a, n) : 
  
    # To store the count of odd numbers and 
    # the count of even numbers such that 2 
    # is the only even factor of that number 
    odd = 0; even = 0; 
  
    for i in range(n) :
  
        # If current number is odd 
        if (a[i] % 2 == 1) :
            odd += 1; 
  
        # If current number is even and 2 
        # is the only even factor of it 
        elif ((a[i] / 2) % 2 == 1) :
            even += 1; 
      
    # Calculate total number of valid pairs 
    ans = odd * even + (odd * (odd - 1)) // 2; 
  
    return ans; 
  
# Driver code 
if __name__ == "__main__" :
  
    a = [ 4, 2, 7, 11, 14, 15, 18 ]; 
    n = len(a); 
  
    print(cntPairs(a, n)); 
      
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
      
class GFG 
{
  
// Function to return the count of valid pairs
static int cntPairs(int []a, int n)
{
  
    // To store the count of odd numbers and
    // the count of even numbers such that 2
    // is the only even factor of that number
    int odd = 0, even = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // If current number is odd
        if (a[i] % 2 == 1)
            odd++;
  
        // If current number is even and 2
        // is the only even factor of it
        else if ((a[i] / 2) % 2 == 1)
            even++;
    }
  
    // Calculate total number of valid pairs
    int ans = odd * even + (odd * (odd - 1)) / 2;
  
    return ans;
}
  
// Driver code
public static void Main(String []args) 
{
    int []a = { 4, 2, 7, 11, 14, 15, 18 };
    int n = a.Length;
  
    Console.WriteLine(cntPairs(a, n));
}
}
  
// This code is contributed by Ajay KUmar

输出：

12

时间复杂度： O(N)