计算无序对(i，j)，以使a [i]和a [j]的乘积为2的幂

给定N个元素的数组。任务是对数组中的无序对(i，j)进行计数，以便a [i]和a [j]的乘积可以表示为2的幂。

例子：

Input : arr[] = {2, 3, 4, 8, 10}
Output : 3
Explanation: The pair of array element will be 
(2, 4), (2, 8), (4, 8) whose product are 
8, 16, 32 respectively which can be expressed 
as power of 2, like 2^3, 2^4, 2^5.

Input : arr[] = { 2, 5, 8, 16, 128 }
Output : 6

如果乘$x$和$y$他们的产品成为$z$ ，则z = x * y，如果可以表达的话$z$作为两个的幂，那么可以证明$x$和$y$可以表示为2的幂。基本上z = 2 ^a = 2 ^{(b + c)} = 2 ^b * 2 ^c = x * y $b$和$c$两个都
可以保持最小值0。

因此，现在我们必须计算数组中可以表示为2的幂的元素的数量。如果计数为k，则答案将为kC2 = k *(k-1)/ 2，因为我们需要无序对的计数。

下面是上述方法的实现：

C++

// C++ program to Count unordered pairs (i, j)
// in array such that product of a[i] and a[j]
// can be expressed as power of two
#include 
using namespace std;
  
/* Function to check if x is power of 2*/
bool isPowerOfTwo(int x) 
{ 
  /* First x in the below expression is
     for the case when x is 0 */
  return x && (!(x&(x-1))); 
} 
  
// Function to Count unordered pairs
void Count_pairs(int a[], int n)
{
    int count = 0;
  
    for (int i = 0; i < n; i++) {
  
        // is a number can be expressed
        // as power of two
        if (isPowerOfTwo(a[i]))
            count++;
    }
  
    // count total number
    // of unordered pairs
    int ans = (count * (count - 1)) / 2;
  
    cout << ans << "\n";
}
  
// Driver code
int main()
{
    int a[] = { 2, 5, 8, 16, 128 };
  
    int n = sizeof(a) / sizeof(a[0]);
  
    Count_pairs(a, n);
  
    return 0;
}

Java

// Java program to Count unordered pairs (i, j)
// in array such that product of a[i] and a[j]
// can be expressed as power of two
  
import java.io.*;
  
class GFG {
  
  
/* Function to check if x is power of 2*/
static boolean isPowerOfTwo(int x) 
{ 
/* First x in the below expression is
    for the case when x is 0 */
return (x >0&& (!((x&(x-1))>0))); 
} 
  
// Function to Count unordered pairs
static void Count_pairs(int a[], int n)
{
    int count = 0;
  
    for (int i = 0; i < n; i++) {
  
        // is a number can be expressed
        // as power of two
        if (isPowerOfTwo(a[i]))
            count++;
    }
  
    // count total number
    // of unordered pairs
    int ans = (count * (count - 1)) / 2;
  
    System.out.println( ans);
}
  
// Driver code
  
    public static void main (String[] args) {
            int a[] = { 2, 5, 8, 16, 128 };
  
    int n = a.length;
    Count_pairs(a, n);
  
    }
}
  
// This code is contributed
// by shs

Python 3

# Python3 program to Count unordered pairs 
# (i, j) in array such that product of a[i] 
# and a[j] can be expressed as power of two 
  
# Function to check if x is power of 2
def isPowerOfTwo(x) :
  
    # First x in the below expression 
    # is for the case when x is 0 
    return (x and(not(x & (x - 1))))
  
# Function to Count unordered pairs 
def Count_pairs(a, n) :
  
    count = 0
  
    for i in range(n) :
  
        # is a number can be expressed 
        # as power of two 
        if isPowerOfTwo(a[i]) :
            count += 1
  
    # count total number 
    # of unordered pairs
    ans = (count * (count - 1)) / 2
  
    print(ans)
  
# Driver code     
if __name__ == "__main__" :
  
    a = [ 2, 5, 8, 16, 128]
  
    n = len(a)
  
    Count_pairs(a, n)
                  
# This code is contributed by ANKITRAI1

C#

// C# program to Count unordered pairs (i, j) 
// in array such that product of a[i] and a[j] 
// can be expressed as power of two 
  
using System;
  
public class GFG{
      
      
/* Function to check if x is power of 2*/
static bool isPowerOfTwo(int x) 
{ 
/* First x in the below expression is 
    for the case when x is 0 */
return (x >0&& (!((x&(x-1))>0))); 
} 
  
// Function to Count unordered pairs 
static void Count_pairs(int []a, int n) 
{ 
    int count = 0; 
  
    for (int i = 0; i < n; i++) { 
  
        // is a number can be expressed 
        // as power of two 
        if (isPowerOfTwo(a[i])) 
            count++; 
    } 
  
    // count total number 
    // of unordered pairs 
    int ans = (count * (count - 1)) / 2; 
  
    Console.WriteLine( ans); 
} 
  
// Driver code 
  
    static public void Main (){
            int []a = { 2, 5, 8, 16, 128 }; 
  
    int n = a.Length; 
    Count_pairs(a, n); 
  
    } 
} 
  
// This code is contributed 
// by Sach_Code

输出：

6

时间复杂度： O(N)，其中N是数组中元素的数量。