数组中最长完美数子序列的长度

给定数组arr []包含长度为N的非负整数，任务是打印数组中Perfect数字s的最长子序列的长度。

A number is a perfect number if it is equal to the sum of its proper divisors, that is, sum of its positive divisors excluding the number itself.

例子：

Input: arr[] = { 3, 6, 11, 2, 28, 21, 8128 }
Output: 3
Explanation:
Longest Perfect number subsequence is {6, 28, 8128} and hence the answer is 3.

Input:arr[] = { 6, 4, 10, 13, 9, 25 }
Output: 1
Explanation:
Longest Perfect number subsequence is {6} and hence the answer is 1.

方法：

要解决上述问题，请执行以下步骤：

• 遍历给定的数组，并检查该数组中的每个元素是否为完美数。
• 如果元素是一个完美数，它将在最长完美数子序列中。因此，将最长完美数子序列的所需长度增加1

下面是上述方法的实现：

C++

// C++ program to find the length of
// Longest Perfect number Subsequence in an Array
  
#include 
using namespace std;
  
// Function to check if
// the number is a Perfect number
bool isPerfect(long long int n)
{
    // To store sum of divisors
    long long int sum = 1;
  
    // Find all divisors and add them
    for (long long int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
    // Check if sum of divisors is equal to
    // n, then n is a perfect number
    if (sum == n && n != 1)
        return true;
  
    return false;
}
  
// Function to find the longest subsequence
// which contain all Perfect numbers
int longestPerfectSubsequence(int arr[], int n)
{
    int answer = 0;
  
    // Find the length of longest
    // Perfect number subsequence
    for (int i = 0; i < n; i++) {
        if (isPerfect(arr[i]))
            answer++;
    }
  
    return answer;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 6, 11, 2, 28, 21, 8128 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << longestPerfectSubsequence(arr, n) << endl;
  
    return 0;
}

Java

// Java program to find the length of 
// longest perfect number subsequence
// in an array 
class GFG {
      
// Function to check if the
// number is a perfect number 
static boolean isPerfect(long n) 
{ 
      
    // To store sum of divisors 
    long sum = 1; 
      
    // Find all divisors and add them 
    for(long i = 2; i * i <= n; i++)
    { 
       if (n % i == 0)
       { 
           if (i * i != n) 
               sum = sum + i + n / i; 
           else
               sum = sum + i; 
       } 
    } 
      
    // Check if sum of divisors is equal  
    // to n, then n is a perfect number 
    if (sum == n && n != 1)
    {
        return true;
    } 
    return false; 
} 
      
// Function to find the longest subsequence 
// which contain all Perfect numbers 
static int longestPerfectSubsequence(int arr[], 
                                     int n) 
{ 
    int answer = 0; 
      
    // Find the length of longest 
    // perfect number subsequence 
    for(int i = 0; i < n; i++)
    { 
       if (isPerfect(arr[i]) == true) 
           answer++; 
    } 
    return answer; 
} 
      
// Driver code 
public static void main (String[] args)
{ 
    int arr[] = { 3, 6, 11, 2, 28, 21, 8128 }; 
    int n = arr.length; 
      
    System.out.println(longestPerfectSubsequence(arr, n)); 
} 
}
  
// This code is contributed by AnkitRai01

Python3

# Python3 program to find the length of
# Longest Perfect number Subsequence in an Array
  
  
# Function to check if 
# the number is Perfect number
def isPerfect( n ): 
      
    # To store sum of divisors 
    sum = 1
      
    # Find all divisors and add them 
    i = 2
    while i * i <= n: 
        if n % i == 0: 
            sum = sum + i + n / i 
        i += 1
      
    # Check if sum of divisors is equal to 
    # n, then n is a perfect number 
      
    return (True if sum == n and n != 1 else False) 
  
# Function to find the longest subsequence
# which contain all Perfect numbers
def longestPerfectSubsequence( arr, n): 
      
    answer = 0
      
    # Find the length of longest 
    # Perfect number subsequence 
    for i in range (n): 
        if (isPerfect(arr[i])): 
            answer += 1
      
    return answer
  
# Driver code 
if __name__ == "__main__": 
    arr = [ 3, 6, 11, 2, 28, 21, 8128 ] 
    n = len(arr) 
      
    print (longestPerfectSubsequence(arr, n))

C#

// C# program to find the length of 
// longest perfect number subsequence
// in an array
using System;
  
class GFG {
      
// Function to check if the
// number is a perfect number 
static bool isPerfect(long n) 
{ 
          
    // To store sum of divisors 
    long sum = 1; 
          
    // Find all divisors and add them 
    for(long i = 2; i * i <= n; i++)
    { 
       if (n % i == 0)
       { 
           if (i * i != n) 
               sum = sum + i + n / i; 
           else
               sum = sum + i; 
       } 
    } 
      
    // Check if sum of divisors is equal 
    // to n, then n is a perfect number 
    if (sum == n && n != 1)
    {
        return true;
    } 
    return false;
} 
          
// Function to find the longest subsequence 
// which contain all perfect numbers 
static int longestPerfectSubsequence(int []arr, 
                                     int n) 
{ 
    int answer = 0; 
          
    // Find the length of longest 
    // perfect number subsequence 
    for(int i = 0; i < n; i++)
    { 
       if (isPerfect(arr[i]) == true) 
           answer++; 
    } 
    return answer; 
} 
          
// Driver code 
public static void Main (string[] args)
{ 
    int []arr = { 3, 6, 11, 2, 28, 21, 8128 }; 
    int n = arr.Length; 
          
    Console.WriteLine(longestPerfectSubsequence(arr, n)); 
} 
}
  
// This code is contributed by AnkitRai01

输出：

3

时间复杂度： O(N×√N)

辅助空间复杂度： O(1)

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。