计算数组中的对，使得 LCM(arr[i], arr[j]) > min(arr[i],arr[j])

给定一个数组arr[] ，任务是找到数组中对的数量，使得LCM(arr[i], arr[j]) > min(arr[i], arr[j])
注意：对(arr[i], arr[j])和(arr[j], arr[i])被认为是相同的，并且只会被计算一次。

例子：

Input: arr[] = {1, 1, 4, 9} 
Output: 5 
All valid pairs are (1, 4), (1, 9), (1, 4), (1, 9) and (4, 9).

Input: arr[] = {2, 4, 5, 2, 5, 7, 2, 8} 
Output: 24 

方法：据观察，只有形式为(arr[i], arr[j])其中arr[i] = arr[j] 的对不满足给定条件。因此，问题现在简化为找到满足arr[i] != arr[j]的对(arr[i], arr[j]) 的数量。

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 count_pairs(int n, int a[])
{
    // Store frequencies of array elements
    unordered_map frequency;
    for (int i = 0; i < n; i++) {
        frequency[a[i]]++;
    }
 
    int count = 0;
 
    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    for (auto x : frequency) {
        int f = x.second;
        count += f * (f - 1) / 2;
    }
 
    // Count of pairs (arr[i], arr[j]) where
    // arr[i] != arr[j], i.e. Total pairs - pairs
    // where arr[i] = arr[j]
    return ((n * (n - 1)) / 2) - count;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 4, 5, 2, 5, 7, 2, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << count_pairs(n, arr);
    return 0;
}

Java

// Java implementation of the approach
import java.util.HashMap;
import java.util.Map;
 
class GfG
{
 
    // Function to return the count of valid pairs
    static int count_pairs(int n, int a[])
    {
        // Store frequencies of array elements
        HashMap frequency = new HashMap<>();
        for (int i = 0; i < n; i++)
        {
             
            if (!frequency.containsKey(a[i]))
                frequency.put(a[i], 0);
            frequency.put(a[i], frequency.get(a[i])+1);
        }
     
        int count = 0;
     
        // Count of pairs (arr[i], arr[j])
        // where arr[i] = arr[j]
        for (Map.Entry x: frequency.entrySet())
        {
            int f = x.getValue();
            count += f * (f - 1) / 2;
        }
     
        // Count of pairs (arr[i], arr[j]) where
        // arr[i] != arr[j], i.e. Total pairs - pairs
        // where arr[i] = arr[j]
        return ((n * (n - 1)) / 2) - count;
    }
 
    // Driver code
    public static void main(String []args)
    {
         
        int arr[] = { 2, 4, 5, 2, 5, 7, 2, 8 };
        int n = arr.length;
        System.out.println(count_pairs(n, arr));
    }
}
     
// This code is contributed by Rituraj Jain

Python3

# Python3 implementation of the approach
 
# Function to return the count
# of valid pairs
def count_pairs(n, a) :
 
    # Store frequencies of array elements
    frequency = dict.fromkeys(a, 0)
    for i in range(n) :
        frequency[a[i]] += 1
 
    count = 0
 
    # Count of pairs (arr[i], arr[j])
    # where arr[i] = arr[j]
    for f in frequency.values() :
        count += f * (f - 1) // 2
     
    # Count of pairs (arr[i], arr[j]) where
    # arr[i] != arr[j], i.e. Total pairs - pairs
    # where arr[i] = arr[j]
    return ((n * (n - 1)) // 2) - count
 
# Driver Code
if __name__ == "__main__" :
     
    arr = [ 2, 4, 5, 2,
            5, 7, 2, 8 ]
    n = len(arr)
    print(count_pairs(n, arr))
 
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GfG
{
 
    // Function to return the count of valid pairs
    static int count_pairs(int n, int []arr)
    {
        // Store frequencies of array elements
        Dictionary mp = new Dictionary();
        for (int i = 0 ; i < n; i++)
        {
            if(mp.ContainsKey(arr[i]))
            {
                var val = mp[arr[i]];
                mp.Remove(arr[i]);
                mp.Add(arr[i], val + 1);
            }
            else
            {
                mp.Add(arr[i], 1);
            }
        }
        int count = 0;
     
        // Count of pairs (arr[i], arr[j])
        // where arr[i] = arr[j]
        foreach(KeyValuePair x in mp)
        {
            int f = x.Value;
            count += f * (f - 1) / 2;
        }
     
        // Count of pairs (arr[i], arr[j]) where
        // arr[i] != arr[j], i.e. Total pairs - pairs
        // where arr[i] = arr[j]
        return ((n * (n - 1)) / 2) - count;
    }
 
    // Driver code
    public static void Main(String []args)
    {
         
        int []arr = { 2, 4, 5, 2, 5, 7, 2, 8 };
        int n = arr.Length;
        Console.WriteLine(count_pairs(n, arr));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

 
// Javascript implementation of the approach
 
// Function to return the count of valid pairs
function count_pairs(n, a)
{
    // Store frequencies of array elements
    var frequency = new Map();
    for (var i = 0; i < n; i++) {
        if(frequency.has(a[i]))
            frequency.set(a[i], frequency.get(a[i])+1)
        else
            frequency.set(a[i], 1)
    }
 
    var count = 0;
 
    // Count of pairs (arr[i], arr[j])
    // where arr[i] = arr[j]
    frequency.forEach((value, key) => {
         
        var f = value;
        count += f * (f - 1) / 2;
    });
 
    // Count of pairs (arr[i], arr[j]) where
    // arr[i] != arr[j], i.e. Total pairs - pairs
    // where arr[i] = arr[j]
    return ((n * (n - 1)) / 2) - count;
}
 
// Driver Code
var arr = [2, 4, 5, 2, 5, 7, 2, 8];
var n = arr.length;
document.write( count_pairs(n, arr));
 
 

输出：

24

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