数组中具有偶数和奇数 LCM 的对数

给定一个大小为N的数组arr[] ，任务是计算具有偶数 LCM 和奇数 LCM 的对的数量。

例子：

Input: arr[] = {3, 6, 5, 4}
Output: Even = 5, Odd = 1
Explanation: LCM of (3, 6) is 6, LCM of (3, 5) is 15, LCM of (3, 4) is 12, LCM of (6, 5) is 30, LCM of (6, 4) is 12, LCM of (5, 4) is 20.
Pairs having even LCM are (3, 6), (3, 4), (6, 5), (6, 4) and (5, 4).
Only pair having odd LCM is (3, 5).

Input: arr[] = {4, 7, 2, 12}
Output: Even = 6, Odd = 0
Explanation: Pairs having even LCM = (4, 7), (4, 2), (4, 12), (7, 2), (7, 12), (2, 12).
No pair has odd LCM.

朴素的方法：最简单的方法是生成所有可能的对以获得所有不同的对，并为每对计算它们的 LCM。如果他们的 LCM 是偶数，则增加偶数的计数。否则，增加奇数。最后，分别打印它们的计数。
时间复杂度： O((N ² )*log(M))，其中 M 是数组中的最小元素 
辅助空间： O(1)

有效的方法：为了优化上述方法，这个想法是基于这样一个事实，即2 个数字的LCM 是奇数当且仅当两个数字都是奇数。因此，找到数组中的总奇数对并获得偶数 LCM 对的计数，从可能对的总数中减去奇数对的计数。

请按照以下步骤解决问题：

• 将对的总数存储在一个变量中，比如totalPairs 。将totalPairs初始化为(N*(N – 1))/2 。
• 将数组中奇数元素的数量存储在一个变量中，比如cnt 。
• 将仅由奇数组成的对的计数存储在一个变量中，比如odd 。因此，奇数 = (cnt*(cnt – 1))/2 。
• 完成以上步骤后，用odd LCM打印odd作为对的计数值。打印(totalPairs –odd)作为具有偶数 LCM 的对的计数。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find count of distinct
// pairs having even LCM and odd LCM
void LCMPairs(int arr[], int N)
{
    // Store the total number of pairs
    int total_pairs = (N * (N - 1)) / 2;
 
    // Stores the count of odd
    // numbers in the array
    int odd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
        if (arr[i] & 1)
            odd++;
    }
 
    // Update the count of pairs with odd LCM
    odd = (odd * (odd - 1)) / 2;
 
    // Print the count of required pairs
    cout << "Even = " << total_pairs - odd
         << ", Odd = " << odd;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 6, 5, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
    LCMPairs(arr, N);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG
{
  
// Function to find count of distinct
// pairs having even LCM and odd LCM
static void LCMPairs(int arr[], int N)
{
   
    // Store the total number of pairs
    int total_pairs = (N * (N - 1)) / 2;
 
    // Stores the count of odd
    // numbers in the array
    int odd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
        if ((arr[i] & 1) != 0)
            odd++;
    }
 
    // Update the count of pairs with odd LCM
    odd = (odd * (odd - 1)) / 2;
 
    // Print the count of required pairs
    System.out.println("Even = " +
                       (total_pairs - odd)  +
                       ", Odd = " + odd);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 6, 5, 4 };
    int N = arr.length;
    LCMPairs(arr, N);
}
}
  
// This code is contributed by splevel62.

Python3

# Python 3 program for the above approach
 
# Function to find count of distinct
# pairs having even LCM and odd LCM
def LCMPairs(arr, N):
    # Store the total number of pairs
    total_pairs = (N * (N - 1)) / 2
 
    # Stores the count of odd
    # numbers in the array
    odd = 0
 
    # Traverse the array arr[]
    for i in range(N):
        if (arr[i] & 1):
            odd += 1
 
    # Update the count of pairs with odd LCM
    odd = (odd * (odd - 1)) // 2
 
    # Print the count of required pairs
    print("Even =",int(total_pairs - odd),","," Odd =",odd)
 
# Driver Code
if __name__ == '__main__':
    arr = [3, 6, 5, 4]
    N = len(arr)
    LCMPairs(arr, N)
 
     # This code is contributed by ipg2016107.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
 
  // Function to find count of distinct
  // pairs having even LCM and odd LCM
  static void LCMPairs(int[] arr, int N)
  {
    // Store the total number of pairs
    int total_pairs = (N * (N - 1)) / 2;
 
    // Stores the count of odd
    // numbers in the array
    int odd = 0;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
      if ((arr[i] & 1) != 0)
        odd++;
    }
 
    // Update the count of pairs with odd LCM
    odd = (odd * (odd - 1)) / 2;
 
    // Print the count of required pairs
    Console.Write("Even = " + (total_pairs - odd) + ", Odd = " + odd);
  }
 
  // Driver code
  static void Main()
  {
    int[] arr = { 3, 6, 5, 4 };
    int N = arr.Length;
    LCMPairs(arr, N);
  }
}
 
// This code is contributed by divyeshrabadiya07.

Javascript

// javascript program to implement
// the above approach
 
    // Function to find count of distinct
    // pairs having even LCM and odd LCM
    function LCMPairs(arr , N) {
 
        // Store the total number of pairs
        var total_pairs = (N * (N - 1)) / 2;
 
        // Stores the count of odd
        // numbers in the array
        var odd = 0;
 
        // Traverse the array arr
        for (i = 0; i < N; i++) {
            if ((arr[i] & 1) != 0)
                odd++;
        }
 
        // Update the count of pairs with odd LCM
        odd = (odd * (odd - 1)) / 2;
 
        // Prvar the count of required pairs
        document.write("Even = " + (total_pairs - odd) + ", Odd = " + odd);
    }
 
    // Driver Code
     
        var arr = [ 3, 6, 5, 4 ];
        var N = arr.length;
        LCMPairs(arr, N);
 
// This code contributed by aashish1995

输出：

Even = 5, Odd = 1

时间复杂度： O(N)
辅助空间： O(1)

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