Un número es un número perfecto si es igual a la suma de sus divisores propios, es decir, la suma de sus divisores positivos excluyendo el propio número. Escribe una función para verificar si un número dado es perfecto o no.
Ejemplos:
Input: n = 15 Output: false Divisors of 15 are 1, 3 and 5. Sum of divisors is 9 which is not equal to 15. Input: n = 6 Output: true Divisors of 6 are 1, 2 and 3. Sum of divisors is 6.
Una solución simple es pasar por todos los números del 1 al n-1 y verificar si es un divisor. Mantener la suma de todos los divisores. Si la suma se vuelve igual a n, entonces devuelve verdadero, de lo contrario, devuelve falso.
Una solución eficiente es recorrer números hasta la raíz cuadrada de n. Si un número ‘i’ divide a n, suma tanto ‘i’ como n/i a la suma.
A continuación se muestra la implementación de la solución eficiente.
C++
// C++ program to check if a given number is perfect
// or not
#include<iostream>
using namespace std;
// Returns true if n is perfect
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;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
return true;
return false;
}
// Driver program
int main()
{
cout << "Below are all perfect numbers till 10000\n";
for (int n =2; n<10000; n++)
if (isPerfect(n))
cout << n << " is a perfect number\n";
return 0;
}
Java
// Java program to check if a given
// number is perfect or not
class GFG
{
// Returns true if n is perfect
static boolean isPerfect(int n)
{
// To store sum of divisors
int sum = 1;
// Find all divisors and add them
for (int i = 2; i * i <= n; i++)
{
if (n % i==0)
{
if(i * i != n)
sum = sum + i + n / i;
else
sum = sum + i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
return true;
return false;
}
// Driver program
public static void main (String[] args)
{
System.out.println("Below are all perfect" +
"numbers till 10000");
for (int n = 2; n < 10000; n++)
if (isPerfect(n))
System.out.println( n +
" is a perfect number");
}
}
// This code is contributed by mits
Python3
# Python3 code to check if a given
# number is perfect or not
# 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
# If sum of divisors is equal to
# n, then n is a perfect number
return (True if sum == n and n!=1 else False)
# Driver program
print("Below are all perfect numbers till 10000")
n = 2
for n in range (10000):
if isPerfect (n):
print(n , " is a perfect number")
# This code is contributed by "Sharad_Bhardwaj".
C#
// C# program to check if a given
// number is perfect or not
class GFG
{
// Returns true if n is perfect
static bool isPerfect(int n)
{
// To store sum of divisors
int sum = 1;
// Find all divisors and add them
for (int i = 2; i * i <= n; i++)
{
if (n % i==0)
{
if(i * i != n)
sum = sum + i + n / i;
else
sum = sum + i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
return true;
return false;
}
// Driver program
static void Main()
{
System.Console.WriteLine("Below are all perfect" +
"numbers till 10000");
for (int n = 2; n < 10000; n++)
if (isPerfect(n))
System.Console.WriteLine( n +
" is a perfect number");
}
}
// This code is contributed by chandan_jnu
PHP
<?php
// PHP program to check if a given number
// is perfect or not
// Returns true if n is perfect
function isPerfect($n)
{
// To store sum of divisors
$sum = 1;
// Find all divisors and add them
for ($i = 2; $i * $i <= $n; $i++)
{
if ($n % $i == 0)
{
if($i * $i != $n)
$sum = $sum + $i + (int)($n / $i);
else
$sum = $sum + $i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if ($sum == $n && $n != 1)
return true;
return false;
}
// Driver Code
echo "Below are all perfect numbers till 10000\n";
for ($n = 2; $n < 10000; $n++)
if (isPerfect($n))
echo "$n is a perfect number\n";
// This code is contributed by mits
?>
Javascript
<script>
// Javascript program to check if a given number is perfect
// or not
// Returns true if n is perfect
function isPerfect(n)
{
// To store sum of divisors
sum = 1;
// Find all divisors and add them
for (let i=2; i*i<=n; i++)
{
if (n%i==0)
{
if(i*i!=n)
sum = sum + i + n/i;
else
sum=sum+i;
}
}
// If sum of divisors is equal to
// n, then n is a perfect number
if (sum == n && n != 1)
return true;
return false;
}
// Driver program
document.write("Below are all perfect numbers till 10000" + "<br>");
for (let n =2; n<10000; n++)
if (isPerfect(n))
document.write(n + " is a perfect number" + "<br>");
// This code is contributed by Mayank Tyagi
</script>
Producción:
Below are all perfect numbers till 10000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number
Complejidad del tiempo: O(√n)
Espacio auxiliar: O(1)
A continuación se presentan algunos datos interesantes sobre los números perfectos:
1) Todo número par perfecto tiene la forma 2 p ?1 (2 p ? 1) donde 2 p ? 1 es primo.
2) Se desconoce si existen números perfectos impares.
Referencias:
https://en.wikipedia.org/wiki/Perfect_number
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA