Dado un número N , la tarea es verificar si N es un Número Duffiniano o no. Si N es un número de Duffinian, escriba «Sí» , de lo contrario, escriba «No» .
Número de Duffinian es un número compuesto N tal que es relativamente primo a la suma de los divisores de N.
Ejemplos:
Entrada: N = 35
Salida: Sí
Explicación:
Suma de divisores de 35 = 1 + 5 + 7 + 35 = 48,
y 48 es primo relativo a 35.
Entrada: N = 28
Salida: No
Explicación:
Suma de divisores de 28 = 1 + 2 + 4 + 7 + 14 + 28 = 56,
y 56 no es primo relativo a 28.
Planteamiento: La idea es encontrar la suma de factores del número N . Si el gcd de N y la suma de los factores de N son primos relativos, entonces el número dado N es un número de Duffin , de lo contrario, N no es un número de Duffin .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#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 of 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
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
}
// Function to check if n is an
// Duffinian number
bool isDuffinian(int n)
{
// Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false;
// Find the gcd of n and sum of
// divisors of n
int hcf = __gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
}
// Driver Code
int main()
{
// Given Number
int n = 36;
// Function Call
if (isDuffinian(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program for the above approach
class GFG{
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// 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 of 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
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
}
// Function to check if n is an
// Duffinian number
static boolean isDuffinian(int n)
{
// Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false;
// Find the gcd of n and sum of
// divisors of n
int hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
}
// Driver code
public static void main(String[] args)
{
// Given Number
int n = 36;
// Function Call
if (isDuffinian(n))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by shubham
Python3
# Python3 program for the above approach
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 of 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
if (i == (n // i)):
result += i
else:
result += (i + n / i)
# Add 1 and n to result as above
# loop considers proper divisors
# greater than 1.
return (result + n + 1)
# Function to check if n is an
# Duffinian number
def isDuffinian(n):
# Calculate the sum of divisors
sumDivisors = int(divSum(n))
# If number is prime return false
if (sumDivisors == n + 1):
return False
# Find the gcd of n and sum of
# divisors of n
hcf = math.gcd(n, sumDivisors)
# Returns true if N and sumDivisors
# are relatively prime
return hcf == 1
# Driver Code
# Given number
n = 36
# Function call
if (isDuffinian(n)):
print("Yes")
else:
print("No")
# This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
class GFG{
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
// 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 of 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
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
}
// Function to check if n is an
// Duffinian number
static bool isDuffinian(int n)
{
// Calculate the sum of divisors
int sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false;
// Find the gcd of n and sum of
// divisors of n
int hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
}
// Driver code
public static void Main(string[] args)
{
// Given number
int n = 36;
// Function call
if (isDuffinian(n))
{
Console.Write("Yes");
}
else
{
Console.Write("No");
}
}
}
// This code is contributed by rock_cool
Javascript
<script>
// Javascript program for the above approach
// Recursive function to return
// gcd of a and b
function gcd( a , b) {
if (b == 0)
return a;
return gcd(b, a % b);
}
// Function to calculate the sum of
// all divisors of a given number
function divSum( n) {
// Sum of divisors
let result = 0;
// Find all divisors of 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
if (i == (n / i))
result += i;
else
result += (i + n / i);
}
}
// Add 1 and n to result as above
// loop considers proper divisors
// greater than 1.
return (result + n + 1);
}
// Function to check if n is an
// Duffinian number
function isDuffinian( n) {
// Calculate the sum of divisors
let sumDivisors = divSum(n);
// If number is prime return false
if (sumDivisors == n + 1)
return false;
// Find the gcd of n and sum of
// divisors of n
let hcf = gcd(n, sumDivisors);
// Returns true if N and sumDivisors
// are relatively prime
return hcf == 1;
}
// Driver code
// Given Number
let n = 36;
// Function Call
if (isDuffinian(n)) {
document.write("Yes");
} else {
document.write("No");
}
// This code contributed by Rajput-Ji
</script>
Producción:
Yes
Complejidad de tiempo: O(1)