Dado un entero n. Comprueba si el número n es un número superperfecto o no. Un número superperfecto es un entero positivo que satisface σ 2 (n) = σ(σ(n)) = 2n, donde σ es la función sumatoria del divisor .
Input: n = 16 Output: yes Explanation: 16 is a superperfect number as σ(16) = 1 + 2 + 4 + 8 + 16 = 31, and σ(31) = 1 + 31 = 32, thus σ(σ(16)) = 32 = 2 × 16. Input: n = 8 Output: no Explanation: σ(8) = 1 + 2 + 4 + 8 = 15 and σ(15) = 1 + 3 + 5 + 15 = 24 thus ( σ(σ(8)) = 24 ) ≠ (2 * 8 = 26)
Le recomendamos encarecidamente que haga clic aquí y lo practique antes de pasar a la solución.
La idea es sencillamente sencilla. Simplemente iteramos de 1 a sqrt (n) y encontramos la suma de todos los divisores de n, llamemos a esta suma como n1. Ahora nuevamente necesitamos iterar de 1 a sqrt(n1) y encontrar la suma de todos los divisores. Después de eso, solo tenemos que comprobar si la suma resultante es igual a 2*n o no.
C++
// C++ program to check whether number is
// superperfect or not
#include<bits/stdc++.h>
using namespace std;
// Function to calculate sum of all divisors
int divSum(int num)
{
// Final result of summation of divisors
int result = 0;
// find all divisors which divides 'num'
for (int i=1; i*i <= num; ++i)
{
// if 'i' is divisor of 'num'
if (num%i == 0)
{
// if both divisors are same then add
// it only once else add both
if (i == (num/i))
result += i;
else
result += (i + num/i);
}
}
return result;
}
// Returns true if n is Super Perfect else false.
bool isSuperPerfect(int n)
{
// Find the sum of all divisors of number n
int n1 = divSum(n);
// Again find the sum of all divisors of n1
// and check if sum is equal to n1
return (2*n == divSum(n1));
}
//Driver code
int main()
{
int n = 16;
cout << (isSuperPerfect(n) ? "Yes\n" : "No\n");
n = 6;
cout << (isSuperPerfect(n) ? "Yes\n" : "No\n");
return 0;
}
Java
// Java program to check whether number is
// superperfect or not
public class Divisors
{
// Function to calculate sum of all divisors
static int divSum(int num)
{
// Final result of summation of divisors
int result = 0;
// find all divisors which divides 'num'
for (int i=1; i*i <= num; ++i)
{
// if 'i' is divisor of 'num'
if (num%i == 0)
{
// if both divisors are same then add
// it only once else add both
if (i == (num/i))
result += i;
else
result += (i + num/i);
}
}
return result;
}
// Returns true if n is Super Perfect else false.
static boolean isSuperPerfect(int n)
{
// Find the sum of all divisors of number n
int n1 = divSum(n);
// Again find the sum of all divisors of n1
// and check if sum is equal to n1
return (2*n == divSum(n1));
}
public static void main (String[] args)
{
int n = 16;
System.out.printf((isSuperPerfect(n) ? "Yes\n" : "No\n"));
n = 6;
System.out.printf((isSuperPerfect(n) ? "Yes\n" : "No\n"));
}
}
// This code is contributed by Saket Kumar
Python3
# Python program to check whether number
# is superperfect or not
import math
# Function to calculate sum of all divisors
def divSum(num):
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
sq = int(math.sqrt(num))
for i in range(1, sq+1):
# if 'i' is divisor of 'num'
if num %i == 0:
# if both divisors are same then add
# it only once else add both
if i == (num//i):
result += i
else:
result += (i + num//i)
return result
# Returns true if n is superperfect else false
def isSuperPerfect(n):
# Find the sum of all divisors of number n
n1 = divSum(n)
# Again find the sum of all divisors of n1
return divSum(n1) == 2*n
#Driver code
n = 16
print ('Yes' if isSuperPerfect(n) else 'No')
n = 6
print ('Yes' if isSuperPerfect(n) else 'No')
C#
// C# program to check whether number is
// superperfect or not
using System;
class Divisors
{
// Function to calculate sum of all divisors
static int divSum(int num)
{
// Final result of summation of divisors
int result = 0;
// find all divisors which divides 'num'
for (int i = 1; i * i <= num; ++i)
{
// if 'i' is divisor of 'num'
if (num % i == 0)
{
// if both divisors are same then add
// it only once else add both
if (i == (num / i))
result += i;
else
result += (i + num / i);
}
}
return result;
}
// Returns true if n is Super Perfect else false.
static bool isSuperPerfect(int n)
{
// Find the sum of all divisors of number n
int n1 = divSum(n);
// Again find the sum of all divisors of n1
// and check if sum is equal to n1
return (2 * n == divSum(n1));
}
public static void Main ()
{
int n = 16;
Console.WriteLine((isSuperPerfect(n) ? "Yes" : "No"));
n = 6;
Console.WriteLine((isSuperPerfect(n) ? "Yes" : "No"));
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to check whether
// number is superperfect or not
// Function to calculate
// sum of all divisors
function divSum($num)
{
// Final result of
// summation of divisors
$result = 0;
// find all divisors
// which divides 'num'
for ($i = 1; $i * $i <= $num; ++$i)
{
// if 'i' is divisor
// of 'num'
if ($num % $i == 0)
{
// if both divisors
// are same then add
// it only once else
// add both
if ($i == ($num / $i))
$result += $i;
else
$result += ($i + $num/$i);
}
}
return $result;
}
// Returns true if n is
// Super Perfect else false.
function isSuperPerfect($n)
{
// Find the sum of all
// divisors of number n
$n1 = divSum($n);
// Again find the sum
// of all divisors of n1
// and check if sum is
// equal to n1
return (2 * $n == divSum($n1));
}
// Driver code
$n = 16;
$hh = (isSuperPerfect($n) ? "Yes\n" : "No\n");
echo($hh);
$n = 6;
$hh=(isSuperPerfect($n) ? "Yes\n" : "No\n");
echo($hh);
// This code is contributed by AJit
?>
Javascript
<script>
// JavaScript program for the above approach
// Function to calculate sum of all divisors
function divSum(num)
{
// Final result of summation of divisors
let result = 0;
// find all divisors which divides 'num'
for (let i = 1; i * i <= num; ++i)
{
// if 'i' is divisor of 'num'
if (num % i == 0)
{
// if both divisors are same then add
// it only once else add both
if (i == (num/i))
result += i;
else
result += (i + num/i);
}
}
return result;
}
// Returns true if n is Super Perfect else false.
function isSuperPerfect(n)
{
// Find the sum of all divisors of number n
let n1 = divSum(n);
// Again find the sum of all divisors of n1
// and check if sum is equal to n1
return (2*n == divSum(n1));
}
// Driver Code
let n = 16;
document.write((isSuperPerfect(n) ? "Yes\n" : "No\n") + "<br />");
n = 6;
document.write((isSuperPerfect(n) ? "Yes\n" : "No\n") + "<br />");
// This code is contributed by splevel62.
</script>
Output: Yes No
Complejidad temporal: O(sqrt(n + n1)) donde n1 es la suma de los divisores de n.
Espacio auxiliar: O(1)
Datos sobre los supernúmeros:
- Si n es un número superperfecto par, entonces n debe ser una potencia de 2, es decir, 2 k tal que 2 k+1 – 1 es un primo de Mersenne .
- No se sabe si existen números superperfectos impares. Un número superperfecto impar n tendría que ser un número cuadrado tal que n o σ(n) sea divisible por al menos tres números primos distintos. No hay números superperfectos impares por debajo de 7×10 24
Referencia:
https://en.wikipedia.org/wiki/Superperfect_number
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA