从一个数组中打印所有完全数，其数字总和也是一个完全数

给定一个大小为N的数组arr[] ，任务是从一个数组中打印所有完全数，该数组的数字总和也是一个完全数。

例子：

Input: arr[] = { 3, 8, 12, 28, 6 }
Output: 6
Explanation: The array element arr[4] (= 6) is a perfect number. The array element arr[3] (= 28) is a perfect number but its sum of digits (= 10) is not a perfect number.

Input: arr[] = { 1, 2, 3 }
Output: 1

处理方法：按照以下步骤解决问题：

1. 声明一个函数isPerfect()来检查数字是否是完美数字。
2. 声明另一个函数sumOfDigits()来计算一个数字的所有数字的总和。
3. 遍历数组arr[] ：
   • 如果arr[i]是一个完美数：
     • 初始化一个变量，比如digitSum，以存储当前数组元素的数字总和。
     • 如果digitSum也是一个完全数，则打印该数字。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if a number
// is perfect number or not
int isPerfect(int N)
{
    // Stores sum of proper divisors
    int sumOfDivisors = 1;
    for (int i = 2; i <= N / 2; ++i) {
       
        if (N % i == 0) {
            sumOfDivisors += i;
        }
    }
 
    // If sum of digits is equal to N,
    // then it's a perfect number
    if (sumOfDivisors == N) {
        return 1;
    }
 
    // Otherwise, not a perfect number
    else
        return 0;
}
 
// Function to find the
// sum of digits of a number
int sumOfDigits(int N)
{
    // Stores sum of digits
    int sum = 0;
 
    while (N != 0) {
        sum += (N % 10);
        N = N / 10;
    }
 
    // Return sum of digits
    return sum;
}
 
// Function to count perfect numners from
// an array whose sum of digits is also perfect
void countPerfectNumbers(int arr[], int N)
{
    // Traverse the array
    for (int i = 0; i < N; ++i) {
       
        // If number is perfect
        if (isPerfect(arr[i])) {
           
            // Stores sum of digits
            // of the number
            int sum = sumOfDigits(arr[i]);
 
            // If that is also perfect number
            if (isPerfect(sum)) {
               
                // Print that number
                cout << arr[i] << " ";
            }
        }
    }
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 3, 8, 12, 28, 6 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call to count perfect numbers
    // having sum of digits also perfect
    countPerfectNumbers(arr, N);
 
    return 0;
}

Java

// Java Program to implement
// the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
  // Function to check if a number
  // is perfect number or not
  static boolean isPerfect(int N)
  {
    // Stores sum of proper divisors
    int sumOfDivisors = 1;
    for (int i = 2; i <= N / 2; ++i) {
 
      if (N % i == 0) {
        sumOfDivisors += i;
      }
    }
 
    // If sum of digits is equal to N,
    // then it's a perfect number
    if (sumOfDivisors == N) {
      return true;
    }
 
    // Otherwise, not a perfect number
    else
      return false;
  }
 
  // Function to find the
  // sum of digits of a number
  static int sumOfDigits(int N)
  {
    // Stores sum of digits
    int sum = 0;
 
    while (N != 0) {
      sum += (N % 10);
      N = N / 10;
    }
 
    // Return sum of digits
    return sum;
  }
 
  // Function to count perfect numners from
  // an array whose sum of digits is also perfect
  static void countPerfectNumbers(int arr[], int N)
  {
    // Traverse the array
    for (int i = 0; i < N; ++i) {
 
      // If number is perfect
      if (isPerfect(arr[i])) {
 
        // Stores sum of digits
        // of the number
        int sum = sumOfDigits(arr[i]);
 
        // If that is also perfect number
        if (isPerfect(sum)) {
 
          // Print that number
          System.out.print(arr[i] + " ");
        }
      }
    }
  }
 
  // Driver Code
  public static void main(String[] args)
  {
     
    // Given array
    int arr[] = { 3, 8, 12, 28, 6 };
 
    // Size of the array
    int N = arr.length;
 
    // Function call to count perfect numbers
    // having sum of digits also perfect
    countPerfectNumbers(arr, N);
  }
}
 
// This code is contributed by Kingash.

Python3

# Python Program to implement
# the above approach
 
# Function to check if a number
# is perfect number or not
def isPerfect(N):
   
    # Stores sum of proper divisors
    sumOfDivisors = 1;
    for i in range(2, int(N / 2) + 1):
 
        if (N % i == 0):
            sumOfDivisors += i;
 
    # If sum of digits is equal to N,
    # then it's a perfect number
    if (sumOfDivisors == N):
        return True;
 
    # Otherwise, not a perfect number
    else:
        return False;
 
# Function to find the
# sum of digits of a number
def sumOfDigits(N):
   
    # Stores sum of digits
    sum = 0;
 
    while (N != 0):
        sum += (N % 10);
        N = N // 10;
 
    # Return sum of digits
    return sum;
 
# Function to count perfect numners from
# an array whose sum of digits is also perfect
def countPerfectNumbers(arr, N):
   
    # Traverse the array
    for i in range(N):
 
        # If number is perfect
        if (isPerfect(arr[i])):
 
            # Stores sum of digits
            # of the number
            sum = sumOfDigits(arr[i]);
 
            # If that is also perfect number
            if (isPerfect(sum)):
               
                # Prthat number
                print(arr[i], end=" ");
 
# Driver Code
if __name__ == '__main__':
   
    # Given array
    arr = [3, 8, 12, 28, 6];
 
    # Size of the array
    N = len(arr);
 
    # Function call to count perfect numbers
    # having sum of digits also perfect
    countPerfectNumbers(arr, N);
 
    # This code is contributed by 29AjayKumar

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to check if a number
  // is perfect number or not
  static bool isPerfect(int N)
  {
     
    // Stores sum of proper divisors
    int sumOfDivisors = 1;
    for (int i = 2; i <= N / 2; ++i) {
 
      if (N % i == 0) {
        sumOfDivisors += i;
      }
    }
 
    // If sum of digits is equal to N,
    // then it's a perfect number
    if (sumOfDivisors == N) {
      return true;
    }
 
    // Otherwise, not a perfect number
    else
      return false;
  }
 
  // Function to find the
  // sum of digits of a number
  static int sumOfDigits(int N)
  {
     
    // Stores sum of digits
    int sum = 0;
 
    while (N != 0) {
      sum += (N % 10);
      N = N / 10;
    }
 
    // Return sum of digits
    return sum;
  }
 
  // Function to count perfect numners from
  // an array whose sum of digits is also perfect
  static void countPerfectNumbers(int []arr, int N)
  {
     
    // Traverse the array
    for (int i = 0; i < N; ++i) {
 
      // If number is perfect
      if (isPerfect(arr[i])) {
 
        // Stores sum of digits
        // of the number
        int sum = sumOfDigits(arr[i]);
 
        // If that is also perfect number
        if (isPerfect(sum)) {
 
          // Print that number
          Console.Write(arr[i] + " ");
        }
      }
    }
  }
 
// Driver Code
static public void Main()
{
   
    // Given array
    int []arr = { 3, 8, 12, 28, 6 };
 
    // Size of the array
    int N = arr.Length;
 
    // Function call to count perfect numbers
    // having sum of digits also perfect
    countPerfectNumbers(arr, N);
}
}
 
// This code is contributed by jana_sayantan.

Javascript

// JavaScript program for the above approach
 
// Function to check if a number
// is perfect number or not
function isPerfect(N)
{
 
    // Stores sum of proper divisors
    let sumOfDivisors = 1;
    for (let i = 2; i <= Math.floor(N / 2); ++i)
    {
     
        if (N % i === 0)
        {
            sumOfDivisors += i;
        }
    }
 
    // If sum of digits is equal to N,
    // then it's a perfect number
    if (sumOfDivisors === N) {
        return 1;
    }
 
    // Otherwise, not a perfect number
    else
        return 0;
}
 
// Function to find the
// sum of digits of a number
function sumOfDigits(N)
{
    // Stores sum of digits
    let sum = 0;
 
    while (N != 0) {
        sum += (N % 10);
        N = Math.floor(N / 10);
    }
 
    // Return sum of digits
    return sum;
}
 
// Function to count perfect numners from
// an array whose sum of digits is also perfect
function countPerfectNumbers(arr, N)
{
    // Traverse the array
    for (let i = 0; i < N; ++i) {
     
        // If number is perfect
        if (isPerfect(arr[i])) {
         
            // Stores sum of digits
            // of the number
            let sum = sumOfDigits(arr[i]);
 
            // If that is also perfect number
            if (isPerfect(sum)) {
             
                // Prlet that number
                document.write(arr[i] + " ");
            }
        }
    }
}
 
// Driver Code
 
    // Given array
    let arr = [ 3, 8, 12, 28, 6 ];
 
    // Size of the array
    let N = arr.length;
 
    // Function call to count perfect numbers
    // having sum of digits also perfect
    countPerfectNumbers(arr, N);
 
// This code is contributed by Surbhi Tyagi.

输出：

6

时间复杂度： O(N ³ * log N)
辅助空间： O(1)

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