Dado un número entero N , la tarea es comprobar si N es un Número Admirable.
Un número admirable es un número, si existe un divisor propio D’ de N tal que sigma(N)-2D’ = 2N, donde sigma(N) es la suma de todos los divisores de N
Ejemplos:
Entrada: N = 12
Salida: Sí
Explicación:
Los divisores propios de 12 son 1, 2, 3, 4, 6 y 12
sigma(N) = 1 + 2 + 3 + 4 + 6 + 12 = 28
sigma(N) – 2D ‘ = 2N
28 – 2*2 = 2*12
24 == 24Entrada: N = 28
Salida: No
Enfoque: La idea es encontrar la suma de todos los factores de un número que es sigma(N). Y luego encontraremos todo divisor propio de un número D’ y comprobaremos si existe un divisor propio D’ de N tal que
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to check if
// N is an admirable number
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the sum of
// all divisors of a given number
int divSum(int n)
{
// Sum of divisors
int result = 0;
// Find all divisors
// which divides 'num'
for (int i = 2; i <= sqrt(n); i++) {
// if 'i' is divisor of 'n'
if (n % i == 0) {
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
}
// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
bool check(int num)
{
int sigmaN = divSum(num);
// Find all divisors which divides '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)) {
if (sigmaN - 2 * i == 2 * num)
return true;
}
else {
if (sigmaN - 2 * i == 2 * num)
return true;
if (sigmaN - 2 * (num / i) == 2 * num)
return true;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true;
return false;
}
// Function to check if N
// is an admirable number
bool isAdmirableNum(int N)
{
return check(N);
}
// Driver code
int main()
{
int n = 12;
if (isAdmirableNum(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation to check if N
// is a admirable number
class GFG{
// Function to calculate the sum of
// all divisors of a given number
static int divSum(int n)
{
// Sum of divisors
int result = 0;
// Find all divisors
// which divides 'num'
for (int i = 2; i <= Math.sqrt(n); i++)
{
// if 'i' is divisor of 'n'
if (n % i == 0)
{
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
}
// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
static boolean check(int num)
{
int sigmaN = divSum(num);
// Find all divisors which divides '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))
{
if (sigmaN - 2 * i == 2 * num)
return true;
}
else
{
if (sigmaN - 2 * i == 2 * num)
return true;
if (sigmaN - 2 * (num / i) == 2 * num)
return true;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true;
return false;
}
// Function to check if N
// is an admirable number
static boolean isAdmirableNum(int N)
{
return check(N);
}
// Driver code
public static void main(String[] args)
{
int n = 12;
if (isAdmirableNum(n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by shubham
Python3
# Python3 implementation to check if
# N is an admirable number
import math
# Function to calculate the sum of
# all divisors of a given number
def divSum(n):
# Sum of divisors
result = 0
# Find all divisors
# which divides 'num'
for i in range(2, int(math.sqrt(n)) + 1):
# If 'i' is divisor of 'n'
if (n % i == 0):
# If both divisors are same
# then add it once else add
if (i == (n // i)):
result += i
else:
result += (i + n // i)
# Add 1 and n to result as above loop
# considers proper divisors greater
return (result + n + 1)
# Function to check if there
# exists a proper divisor
# D' of N such that sigma(n)-2D' = 2N
def check(num):
sigmaN = divSum(num)
# Find all divisors which divides 'num'
for i in range(2, int(math.sqrt(num)) + 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)):
if (sigmaN - 2 * i == 2 * num):
return True
else:
if (sigmaN - 2 * i == 2 * num):
return True
if (sigmaN - 2 * (num // i) == 2 * num):
return True
# Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num):
return True
return False
# Function to check if N
# is an admirable number
def isAdmirableNum(N):
return check(N)
# Driver code
n = 12
if (isAdmirableNum(n)):
print("Yes")
else:
print("No")
# This code is contributed by divyeshrabadiya07
C#
// C# implementation to check if N
// is a admirable number
using System;
class GFG{
// Function to calculate the sum of
// all divisors of a given number
static int divSum(int n)
{
// Sum of divisors
int result = 0;
// Find all divisors
// which divides 'num'
for(int i = 2; i <= Math.Sqrt(n); i++)
{
// If 'i' is divisor of 'n'
if (n % i == 0)
{
// If both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
}
// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
static bool check(int num)
{
int sigmaN = divSum(num);
// Find all divisors which divides '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))
{
if (sigmaN - 2 * i == 2 * num)
return true;
}
else
{
if (sigmaN - 2 * i == 2 * num)
return true;
if (sigmaN - 2 * (num / i) == 2 * num)
return true;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true;
return false;
}
// Function to check if N
// is an admirable number
static bool isAdmirableNum(int N)
{
return check(N);
}
// Driver code
public static void Main()
{
int n = 12;
if (isAdmirableNum(n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Code_Mech
Javascript
<script>
// Javascript implementation to check if N
// is a admirable number
// Function to calculate the sum of
// all divisors of a given number
function divSum( n)
{
// Sum of divisors
let result = 0;
// Find all divisors
// which divides 'num'
for ( let i = 2; i <= Math.sqrt(n); i++)
{
// if 'i' is divisor of 'n'
if (n % i == 0)
{
// if both divisors are same
// then add it once else add
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above loop
// considers proper divisors greater
return (result + n + 1);
}
// Function to check if there
// exists a proper divisor
// D' of N such that sigma(n)-2D' = 2N
function check( num) {
let sigmaN = divSum(num);
// Find all divisors which divides 'num'
for (let 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)) {
if (sigmaN - 2 * i == 2 * num)
return true;
} else {
if (sigmaN - 2 * i == 2 * num)
return true;
if (sigmaN - 2 * (num / i) == 2 * num)
return true;
}
}
}
// Check 1 since 1 is also a divisor
if (sigmaN - 2 * 1 == 2 * num)
return true;
return false;
}
// Function to check if N
// is an admirable number
function isAdmirableNum( N) {
return check(N);
}
// Driver code
let n = 12;
if (isAdmirableNum(n))
document.write("Yes");
else
document.write("No");
// This code is contributed by todaysgaurav
</script>
Producción:
Yes
Complejidad de tiempo: O(N 1/2 )
Referencias: OEIS