Dado un arreglo arr[] de elementos enteros, la tarea es encontrar la longitud del subarreglo más grande de arr[] tal que todos los elementos del subarreglo sean números perfectos .
Un número perfecto es un entero positivo que es igual a la suma de sus divisores propios .
Ejemplos:
Entrada: arr[] = {1, 7, 36, 4, 6, 28, 4}
Salida: 2
Explicación:
la subarray de longitud máxima con todos los elementos como número perfecto es {6, 28}.
Entrada: arr[] = {25, 100, 2, 3, 9, 1}
Salida: 0
Explicación:
Ninguno de los números es un número perfecto
Acercarse:
- Recorra la array de izquierda a derecha e inicialice una variable max_length y current_length con 0.
- Si el elemento actual es un número perfecto, incremente la variable longitud_actual y continúe con el proceso. De lo contrario, establezca current_length en 0.
- En cada paso, asigne max_length como max_length = max(current_length, max_length) .
- Imprime el valor de max_length al final, ya que almacenará el resultado requerido.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the length of the // largest sub-array of an array every // element of whose is a perfect number #include <bits/stdc++.h> 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
Javascript
<script> // Javascript 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 function isPerfect(n) { // Variable to store sum of divisors let sum = 1; let 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 function contiguousPerfectNumber(arr, n) { let current_length = 0; let max_length = 0; let 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 let arr = [ 1, 7, 36, 4, 6, 28, 4 ]; let n = arr.length; document.write(contiguousPerfectNumber(arr, n)); </script>
2
Complejidad de tiempo: O(N×√N)
Complejidad de espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por saurabhshadow y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA