Dado un número N , la tarea es verificar si N es un número anti-perfecto o no. Si N es un número anti-perfecto, imprima «Sí», de lo contrario, imprima «No» .
Un número antiperfecto es un número que es igual a la suma del reverso de sus divisores propios.
Ejemplos:
Entrada: N = 244
Salida: Sí
Explicación:
los divisores propios de 24 son 1, 2, 4, 61, 122
la suma de su reverso es 1 + 2 + 4 + 16 + 221 = 244 = N.
Entrada: N = 28
Salida: No
Enfoque La idea es encontrar la suma del reverso de los divisores propios del número N y verificar si la suma es igual a N o no. Si la suma es igual a N , entonces N es un número anti-perfecto, luego imprime «Sí» , de lo contrario, imprime «No» .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Iterative function to reverse
// digits of num
int rev(int num)
{
int rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10
+ num % 10;
num = num / 10;
}
// Return the reversed num
return rev_num;
}
// Function to calculate sum
// of reverse all proper divisors
int divSum(int num)
{
// Final result of summation
// of divisors
int result = 0;
// Find all divisors of num
for (int i = 2; i <= sqrt(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 += rev(i);
else
result += (rev(i)
+ rev(num / i));
}
}
// Add 1 to the result as 1
// is also a divisor
return (result + 1);
}
// Function to check if N is
// anti-perfect or not
bool isAntiPerfect(int n)
{
return divSum(n) == n;
}
// Driver Code
int main()
{
// Given Number N
int N = 244;
// Function Call
if (isAntiPerfect(N))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program for the above approach
class GFG{
// Iterative function to reverse
// digits of num
static int rev(int num)
{
int rev_num = 0;
while (num > 0)
{
rev_num = rev_num * 10 +
num % 10;
num = num / 10;
}
// Return the reversed num
return rev_num;
}
// Function to calculate sum
// of reverse all proper divisors
static int divSum(int num)
{
// Final result of summation
// of divisors
int result = 0;
// Find all divisors of num
for(int i = 2; i <= Math.sqrt(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 += rev(i);
else
result += (rev(i) +
rev(num / i));
}
}
// Add 1 to the result as 1
// is also a divisor
return (result + 1);
}
// Function to check if N is
// anti-perfect or not
static boolean isAntiPerfect(int n)
{
return divSum(n) == n;
}
// Driver Code
public static void main (String[] args)
{
// Given Number N
int N = 244;
// Function Call
if (isAntiPerfect(N))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by rock_cool
Python3
# Python3 program for the above approach
# Iterative function to reverse
# digits of num
def rev(num):
rev_num = 0
while (num > 0) :
rev_num = rev_num * 10 + num % 10
num = num // 10
# Return the reversed num
return rev_num
# Function to calculate sum
# of reverse all proper divisors
def divSum(num) :
# Final result of summation
# of divisors
result = 0
# Find all divisors of num
for i in range(2, int(num**0.5)):
# 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 += rev(i)
else:
result += (rev(i) + rev(num / i))
# Add 1 to the result as 1
# is also a divisor
return (result + 1)
# Function to check if N is
# anti-perfect or not
def isAntiPerfect(n):
return divSum(n) == n
# Driver Code
# Given Number N
N = 244
# Function Call
if (isAntiPerfect(N)):
print("Yes")
else:
print("No")
# This code is contributed by Vishal Maurya.
C#
// C# program for the above approach
using System;
class GFG{
// Iterative function to reverse
// digits of num
static int rev(int num)
{
int rev_num = 0;
while (num > 0)
{
rev_num = rev_num * 10 +
num % 10;
num = num / 10;
}
// Return the reversed num
return rev_num;
}
// Function to calculate sum
// of reverse all proper divisors
static int divSum(int num)
{
// Final result of summation
// of divisors
int result = 0;
// Find all divisors of num
for(int i = 2; i <= Math.Sqrt(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 += rev(i);
else
result += (rev(i) +
rev(num / i));
}
}
// Add 1 to the result as 1
// is also a divisor
return (result + 1);
}
// Function to check if N is
// anti-perfect or not
static Boolean isAntiPerfect(int n)
{
return divSum(n) == n;
}
// Driver Code
public static void Main (String[] args)
{
// Given Number N
int N = 244;
// Function Call
if (isAntiPerfect(N))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by shivanisinghss2110
Javascript
<script>
// Javascript implementation
// Iterative function to reverse
// digits of num
function rev(num)
{
var rev_num = 0;
while (num > 0) {
rev_num = rev_num * 10 + num % 10;
num = Math.floor(num / 10);
}
// Return the reversed num
return rev_num;
}
// Function to calculate sum
// of reverse all proper divisors
function divSum(num)
{
// Final result of summation
// of divisors
var result = 0;
// Find all divisors of num
for (var i = 2; i <= Math.floor(Math.sqrt(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 += rev(i);
else
result += (rev(i)
+ rev(num / i));
}
}
// Add 1 to the result as 1
// is also a divisor
result += 1;
return result;
}
// Function to check if N is
// anti-perfect or not
function isAntiPerfect(n)
{
return divSum(n) == n;
}
// Driver Code
// Given Number N
var N = 244;
// Function Call
if (isAntiPerfect(N))
document.write("Yes");
else
document.write("No");
// This code is contributed by shubhamsingh10
</script>
Producción:
Yes
Complejidad del tiempo: O(sqrt(N))