配对，使得一个是另一个的幂倍数

您将获得n个元素的数组A []和一个正整数k。现在，您已经找到了对Ai，Aj的对数，使得Ai = Aj *(k ^x ) ，其中x是整数。
注意：(Ai，Aj)和(Aj，Ai)必须计数一次。

例子 ：

Input : A[] = {3, 6, 4, 2},  k = 2
Output : 2
Explanation : We have only two pairs 
(4, 2) and (3, 6)

Input : A[] = {2, 2, 2},   k = 2
Output : 3
Explanation : (2, 2), (2, 2), (2, 2) 
that are (A1, A2), (A2, A3) and (A1, A3) are 
total three pairs where Ai = Aj * (k^0) 

为了解决这个问题，我们首先对给定的数组进行排序，然后对于每个元素Ai，对于不同的x值，找到等于值Ai * k ^ x的元素数量，直到Ai * k ^ x小于或等于最大i
算法：

// sort the given array
    sort(A, A+n);

    // for each A[i] traverse rest array
    for (int i=0; i

C++

// Program to find pairs count
#include 
using namespace std;
  
// function to count the required pairs
int countPairs(int A[], int n, int k) {
  int ans = 0;
  // sort the given array
  sort(A, A + n);
  
  // for each A[i] traverse rest array
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
  
      // count Aj such that Ai*k^x = Aj
      int x = 0;
  
      // increase x till Ai * k^x <= largest element
      while ((A[i] * pow(k, x)) <= A[j]) {
        if ((A[i] * pow(k, x)) == A[j]) {
          ans++;
          break;
        }
        x++;
      }
    }
  }
  return ans;
}
  
// driver program
int main() {
  int A[] = {3, 8, 9, 12, 18, 4, 24, 2, 6};
  int n = sizeof(A) / sizeof(A[0]);
  int k = 3;
  cout << countPairs(A, n, k);
  return 0;
}

Java

// Java program to find pairs count
import java.io.*;
import java .util.*;
  
class GFG {
      
    // function to count the required pairs
    static int countPairs(int A[], int n, int k) 
    {
        int ans = 0;
          
        // sort the given array
        Arrays.sort(A);
          
        // for each A[i] traverse rest array
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) 
            {
          
                // count Aj such that Ai*k^x = Aj
                int x = 0;
              
                // increase x till Ai * k^x <= largest element
                while ((A[i] * Math.pow(k, x)) <= A[j]) 
                {
                    if ((A[i] * Math.pow(k, x)) == A[j]) 
                    {
                        ans++;
                        break;
                    }
                    x++;
                }
            }
        }
        return ans;
    }
      
    // Driver program
    public static void main (String[] args) 
    {
        int A[] = {3, 8, 9, 12, 18, 4, 24, 2, 6};
        int n = A.length;
        int k = 3;
        System.out.println (countPairs(A, n, k));
          
    }
}
  
// This code is contributed by vt_m.

Python3

# Program to find pairs count
import math
  
# function to count the required pairs
def countPairs(A, n, k): 
    ans = 0
  
    # sort the given array
    A.sort()
      
    # for each A[i] traverse rest array
    for i in range(0,n): 
  
        for j in range(i + 1, n):
  
            # count Aj such that Ai*k^x = Aj
            x = 0
  
            # increase x till Ai * k^x <= largest element
            while ((A[i] * math.pow(k, x)) <= A[j]) :
                if ((A[i] * math.pow(k, x)) == A[j]) :
                    ans+=1
                    break
                x+=1
    return ans
  
  
# driver program
A = [3, 8, 9, 12, 18, 4, 24, 2, 6]
n = len(A)
k = 3
  
print(countPairs(A, n, k))
  
# This code is contributed by
# Smitha Dinesh Semwal

C#

// C# program to find pairs count
using System;
  
class GFG {
      
    // function to count the required pairs
    static int countPairs(int []A, int n, int k) 
    {
        int ans = 0;
          
        // sort the given array
        Array.Sort(A);
          
        // for each A[i] traverse rest array
        for (int i = 0; i < n; i++) 
        {
            for (int j = i + 1; j < n; j++) 
            {
          
                // count Aj such that Ai*k^x = Aj
                int x = 0;
              
                // increase x till Ai * k^x <= largest element
                while ((A[i] * Math.Pow(k, x)) <= A[j]) 
                {
                    if ((A[i] * Math.Pow(k, x)) == A[j]) 
                    {
                        ans++;
                        break;
                    }
                    x++;
                }
            }
        }
        return ans;
    }
      
    // Driver program
    public static void Main () 
    {
        int []A = {3, 8, 9, 12, 18, 4, 24, 2, 6};
        int n = A.Length;
        int k = 3;
        Console.WriteLine(countPairs(A, n, k));
          
    }
}
  
// This code is contributed by vt_m.

输出 ：

6