所有元素均为完美数的最大子数组的长度

给定整数元素的数组arr [] ，任务是找到arr []的最大子数组的长度，以使子数组的所有元素均为Perfect number 。

A perfect number is a positive integer that is equal to the sum of its proper divisors.

例子：

Input: arr[] = {1, 7, 36, 4, 6, 28, 4}
Output: 2
Explanation:
Maximum length sub-array with all elements as perfect number is {6, 28}.

Input: arr[] = {25, 100, 2, 3, 9, 1}
Output: 0
Explanation:
None of the number is a perfect number

方法：

• 从左到右遍历数组，并将max_length和current_length变量初始化为0。
• 如果当前元素是一个完美数字，则增加current_length变量并继续该过程。否则，将current_length设置为0。
• 在每个步骤中，将max_length分配为max_length = max(current_length，max_length) 。
• 最后打印max_length的值，因为它将存储所需的结果。

下面是上述方法的实现：

C++

// C++ program to find the length of the
// largest sub-array of an array every
// element of whose is a perfect number
  
#include 
using namespace std;
  
// Function that returns true if n is perfect
bool isPerfect(long long int n)
{
    // Variable 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 return the length of the
// largest sub-array of an array every
// element of whose is a perfect number
int contiguousPerfectNumber(int arr[], int n)
{
  
    int current_length = 0;
    int max_length = 0;
  
    for (int i = 0; i < n; i++) {
  
        // Check if arr[i] is a perfect number
        if (isPerfect(arr[i]))
            current_length++;
        else
            current_length = 0;
  
        max_length = max(max_length,
                         current_length);
    }
  
    return max_length;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 7, 36, 4, 6, 28, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << contiguousPerfectNumber(arr, n);
  
    return 0;
}

Java

// Java program to find the length of the
// largest sub-array of an array every
// element of whose is a perfect number
  
import java.util.*;     
  
class GFG
{
    // Function that returns true if n is perfect
    static boolean isPerfect(int n)
    {
        // Variable to store sum of divisors
        int sum = 1;
        int i;
          
        // Find all divisors and add them
        for ( 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 return the length of the
    // largest sub-array of an array every
    // element of whose is a perfect number
    static int contiguousPerfectNumber(int arr[], int n)
    {
      
        int current_length = 0;
        int max_length = 0;
        int i;
        for (i = 0; i < n; i++) {
      
            // Check if arr[i] is a perfect number
            if (isPerfect(arr[i]))
                current_length++;
            else
                current_length = 0;
      
            max_length = Math.max(max_length,
                            current_length);
        }
      
        return max_length;
    }
      
    // Driver code
    public static void main(String []args)
    {
        int arr[] = { 1, 7, 36, 4, 6, 28, 4 };
        int n = arr.length;
      
        System.out.print(contiguousPerfectNumber(arr, n));
      
    }
} 
  
//This code is contributed by chitranayal

Python3

# Python 3 program to find the length of 
# the largest sub-array of an array every 
# element of whose is a perfect number 
  
  
# Function that returns true if n is perfect 
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 the 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 return the length of the 
# largest sub-array of an array every 
# element of whose is a perfect number
def contiguousPerfectNumber(arr, n): 
    current_length = 0
    max_length = 0
  
    for i in range(0, n, 1): 
          
        # check if arr[i] is a perfect number
        if (isPerfect(arr[i])): 
            current_length += 1
        else: 
            current_length = 0
  
        max_length = max(max_length, 
                        current_length) 
      
    return max_length 
  
# Driver code 
if __name__ == '__main__': 
    arr = [1, 7, 36, 4, 6, 28, 4]
    n = len(arr) 
  
    print(contiguousPerfectNumber(arr, n))

C#

// C# program to find the length of the
// largest sub-array of an array every
// element of whose is a perfect number
using System;
  
class GFG{
      
// Function that returns true if n is perfect
static bool isPerfect(int n)
{
      
    // Variable to store sum of divisors
    int sum = 1;
    int i;
          
    // Find all divisors and add them
    for(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 return the length of the
// largest sub-array of an array every
// element of whose is a perfect number
static int contiguousPerfectNumber(int []arr,
                                   int n)
{
    int current_length = 0;
    int max_length = 0;
    int i;
    for(i = 0; i < n; i++)
    {
         
       // Check if arr[i] is a perfect number
       if (isPerfect(arr[i]))
       {
           current_length++;
       }
       else
       {
           current_length = 0;
       }
       max_length = Math.Max(max_length,
                             current_length);
    }
    return max_length;
}
      
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 7, 36, 4, 6, 28, 4 };
    int n = arr.Length;
      
    Console.Write(contiguousPerfectNumber(arr, n));
}
}
  
// This code is contributed by sapnasingh4991

输出：

2

时间复杂度： O(N×√N)

辅助空间复杂度： O(1)