Dado un número entero N , la tarea es verificar si N es un número de Rhonda en base 10.
Los números de Rhonda en base 10 son números si el producto de sus dígitos es igual a 10*Suma de factores primos de N (incluida la multiplicidad).
Ejemplos:
Entrada: N = 1568
Salida: Sí
Explicación:
factorización prima de 1568 = 2 5 * 7 2 .
Suma de factores primos = 2*5+7*2=24.
Producto de dígitos de 1568 = 1*5*6*8=240 = 10*24.
Por lo tanto, 1568 es un número de Rhonda en base 10.Entrada: N = 28
Salida: No
Enfoque: La idea es encontrar la suma de todos los factores primos de N y verificar si diez veces la suma de los factores primos de N es igual al producto de los dígitos de N o no.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation to check if N
// is a Rhonda number
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// product of digits
int getProduct(int n)
{
int product = 1;
while (n != 0) {
product = product * (n % 10);
n = n / 10;
}
return product;
}
// Function to find sum of all prime
// factors of a given number N
int sumOfprimeFactors(int n)
{
int sum = 0;
// add the number of
// 2s that divide n
while (n % 2 == 0) {
sum += 2;
n = n / 2;
}
// N must be odd at this
// point. So we can skip
// one element
for (int i = 3; i <= sqrt(n);
i = i + 2) {
// While i divides n,
// add i and divide n
while (n % i == 0) {
sum += i;
n = n / i;
}
}
// Condition to handle the case when N
// is a prime number greater than 2
if (n > 2)
sum += n;
return sum;
}
// Function to check if n
// is Rhonda number
bool isRhondaNum(int N)
{
return 10 * sumOfprimeFactors(N) == getProduct(N);
}
// Driver code
int main()
{
int n = 1568;
if (isRhondaNum(n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java program to check if N
// is a Rhonda number
import java.util.*;
import java.lang.*;
// import Math;
class GFG{
// Function to find the
// product of digits
static int getProduct(int n)
{
int product = 1;
while (n != 0)
{
product = product * (n % 10);
n = n / 10;
}
return product;
}
// Function to find sum of all prime
// factors of a given number N
static int sumOfprimeFactors(int n)
{
int sum = 0;
// Add the number of
// 2s that divide n
while (n % 2 == 0)
{
sum += 2;
n = n / 2;
}
// N must be odd at this
// point. So we can skip
// one element
for(int i = 3; i <= Math.sqrt(n);
i = i + 2)
{
// While i divides n,
// add i and divide n
while (n % i == 0)
{
sum += i;
n = n / i;
}
}
// Condition to handle the case when N
// is a prime number greater than 2
if (n > 2)
sum += n;
return sum;
}
// Function to check if n
// is Rhonda number
static boolean isRhondaNum(int N)
{
return (10 * sumOfprimeFactors(N) ==
getProduct(N));
}
// Driver Code
public static void main(String[] args)
{
// Given Number n
int n = 1568;
if (isRhondaNum(n))
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
// This code is contributed by vikas_g
Python3
# Python3 implementation to check
# if N is a Rhonda number
import math
# Function to find the
# product of digits
def getProduct(n):
product = 1
while (n != 0):
product = product * (n % 10)
n = n // 10
return product
# Function to find sum of all prime
# factors of a given number N
def sumOfprimeFactors(n):
Sum = 0
# Add the number of
# 2s that divide n
while (n % 2 == 0):
Sum += 2
n = n // 2
# N must be odd at this
# point. So we can skip
# one element
for i in range(3, int(math.sqrt(n)) + 1, 2):
# While i divides n,
# add i and divide n
while (n % i == 0):
Sum += i
n = n // i
# Condition to handle the case when N
# is a prime number greater than 2
if (n > 2):
Sum += n
return Sum
# Function to check if n
# is Rhonda number
def isRhondaNum(N):
return bool(10 * sumOfprimeFactors(N) ==
getProduct(N))
# Driver code
n = 1568
if (isRhondaNum(n)):
print("Yes")
else:
print("No")
# This code is contributed by divyeshrabadiya07
C#
// C# implementation to check if N
// is a Rhonda number
using System;
class GFG{
// Function to find the
// product of digits
static int getProduct(int n)
{
int product = 1;
while (n != 0)
{
product = product * (n % 10);
n = n / 10;
}
return product;
}
// Function to find sum of all prime
// factors of a given number N
static int sumOfprimeFactors(int n)
{
int sum = 0;
// Add the number of
// 2s that divide n
while (n % 2 == 0)
{
sum += 2;
n = n / 2;
}
// N must be odd at this
// point. So we can skip
// one element
for(int i = 3; i <= Math.Sqrt(n);
i = i + 2)
{
// While i divides n,
// add i and divide n
while (n % i == 0)
{
sum += i;
n = n / i;
}
}
// Condition to handle the case when N
// is a prime number greater than 2
if (n > 2)
sum += n;
return sum;
}
// Function to check if n
// is Rhonda number
static bool isRhondaNum(int N)
{
return (10 * sumOfprimeFactors(N) ==
getProduct(N));
}
// Driver code
public static void Main(String []args)
{
int n = 1568;
if (isRhondaNum(n))
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
}
// This code is contributed by vikas_g
Javascript
<script>
// Javascript program to check if N
// is a Rhonda number
// Function to find the
// product of digits
function getProduct( n)
{
let product = 1;
while (n != 0)
{
product = product * (n % 10);
n = parseInt(n / 10);
}
return product;
}
// Function to find sum of all prime
// factors of a given number N
function sumOfprimeFactors( n) {
let sum = 0;
// Add the number of
// 2s that divide n
while (n % 2 == 0) {
sum += 2;
n = parseInt(n / 2);
}
// N must be odd at this
// point. So we can skip
// one element
for ( let i = 3; i <= Math.sqrt(n); i = i + 2)
{
// While i divides n,
// add i and divide n
while (n % i == 0)
{
sum += i;
n = parseInt(n / i);
}
}
// Condition to handle the case when N
// is a prime number greater than 2
if (n > 2)
sum += n;
return sum;
}
// Function to check if n
// is Rhonda number
function isRhondaNum( N) {
return (10 * sumOfprimeFactors(N) == getProduct(N));
}
// Driver Code
// Given Number n
let n = 1568;
if (isRhondaNum(n)) {
document.write("Yes");
} else {
document.write("No");
}
// This code is contributed by todaysgaurav
</script>
Producción:
Yes
Complejidad del tiempo: O(sqrt(N))
Referencias: OEIS