Un número afortunado es el entero más pequeño m > 1 tal que, para un entero positivo dado n, p n + m es un número primo. Aquí p n es el producto de los n primeros números primos, es decir, factores primos (o primoriales ) de orden n.
Por ejemplo :
p3 = 2 × 3 × 5 = 30 p4 = 2 × 3 × 5 × 7 = 210 p5 = 2 × 3 × 5 × 7 × 11 = 2310
Ahora bien, la menor diferencia m entre el factorial primo p n y el primer número primo mayor que p n para el cual (m > 1), es un número primo.
Ejemplos:
Input : n = 3 Output : 7 Explanation : 7 must be added to the product of first n prime numbers to make the product prime. 2 x 3 x 5 = 30, need to add 7 to make it 37, which is a prime Input : n = 5 Output : 23
Planteamiento : Para encontrar el enésimo número de la Fortuna, calcule el producto de los primeros n números primos (primoriales). Que este producto sea p. Luego encontramos un número primo mayor que p y devolvemos la diferencia entre el número primo encontrado y p.
p4 + 13 = 223, where m = 13, a fortunate number p5 + 23 = 2333, where m = 23, a fortunate number p6 + 17 = 30047, where m = 17, a fortunate number
C++
// C++ program to find n-th Fortunate number
#include <bits/stdc++.h>
using namespace std;
bool isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n%2 == 0 || n%3 == 0) return false;
for (int i=5; i*i<=n; i=i+6)
if (n%i == 0 || n%(i+2) == 0)
return false;
return true;
}
// Function to Find primorial of order n
// (product of first n prime numbers).
long long int primorial(long long int n)
{
long long int p = 2;
n--;
for (int i = 3; n != 0; i++) {
if (isPrime(i)) {
p = p * i;
n--;
}
i++;
}
return p;
}
// Function to find next prime number greater
// than n
long long int findNextPrime(long long int n)
{
// Note that difference (or m) should be
// greater than 1.
long long int nextPrime = n + 2;
// loop continuously until isPrime
// returns true for a number above n
while (true) {
// Ignoring the prime number that
// is 1 greater than n
if (isPrime(nextPrime))
break;
nextPrime++;
}
return nextPrime;
}
// Returns n-th Fortunate number
long long int fortunateNumber(int n)
{
long long int p = primorial(n);
return findNextPrime(p) - p;
}
// Driver function
int main()
{
long long int n = 5;
cout << fortunateNumber(n) << "\n";
return 0;
}
Java
// Java program to find n-th Fortunate number
import java.lang.*;
import java.util.*;
class GFG
{
public static boolean isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0) return false;
for (int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to Find primorial of order n
// (product of first n prime numbers).
public static int primorial(int n)
{
int p = 2;
n--;
for (int i = 3; n != 0; i++) {
if (isPrime(i) == true) {
p = p * i;
n--;
}
i++;
}
return p;
}
// Function to find next prime number greater
// than n
public static int findNextPrime(int n)
{
// Note that difference (or m) should be
// greater than 1.
int nextPrime = n + 2;
// loop continuously until isPrime
// returns true for a number above n
while (true) {
// Ignoring the prime number that
// is 1 greater than n
if (isPrime(nextPrime) == true)
break;
nextPrime++;
}
return nextPrime;
}
// Returns n-th Fortunate number
public static int fortunateNumber(int n)
{
int p = primorial(n);
return findNextPrime(p)-p;
}
//Driver function
public static void main (String[] args) {
int n = 5;
System.out.println(fortunateNumber(n));
}
}
/*This code is contributed by Akash Singh*/
Python3
# Python3 program to find # n-th Fortunate number def isPrime(n): # Corner cases if (n <= 1): return False if (n <= 3): return True # This is checked so that we can skip # middle five numbers in below loop if (n % 2 == 0 or n % 3 == 0): return False i = 5 while(i * i <= n): if (n % i == 0 or n % (i + 2) == 0): return False i += 6 return True # Function to Find primorial of order n # (product of first n prime numbers). def primorial(n): p = 2; n -= 1; i = 3 while(n != 0): if (isPrime(i)): p = p * i n -= 1 i += 1 return p # Function to find next prime # number greater than n def findNextPrime(n): # Note that difference (or m) # should be greater than 1. nextPrime = n + 2 # loop continuously until isPrime # returns true for a number above n while (True): # Ignoring the prime number that # is 1 greater than n if (isPrime(nextPrime)): break nextPrime += 1 return nextPrime # Returns n-th Fortunate number def fortunateNumber(n): p = primorial(n) return findNextPrime(p) - p # Driver Code n = 5 print(fortunateNumber(n)) # This code is contributed by Anant Agarwal.
C#
// C# program to find
// n-th Fortunate number
using System;
class GFG
{
public static bool isPrime(int n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that
// we can skip middle five
// numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false;
for (int i = 5;
i * i <= n; i = i + 6)
if (n % i == 0 ||
n % (i + 2) == 0)
return false;
return true;
}
// Function to Find primorial
// of order n (product of first
// n prime numbers).
public static int primorial(int n)
{
int p = 2;
n--;
for (int i = 3; n != 0; i++)
{
if (isPrime(i) == true)
{
p = p * i;
n--;
}
i++;
}
return p;
}
// Function to find next
// prime number greater than n
public static int findNextPrime(int n)
{
// Note that difference (or m)
// should be greater than 1.
int nextPrime = n + 2;
// loop continuously until
// isPrime returns true
// for a number above n
while (true)
{
// Ignoring the prime number
// that is 1 greater than n
if (isPrime(nextPrime) == true)
break;
nextPrime++;
}
return nextPrime;
}
// Returns n-th
// Fortunate number
public static int fortunateNumber(int n)
{
int p = primorial(n);
return findNextPrime(p) - p;
}
// Driver Code
public static void Main ()
{
int n = 5;
Console.WriteLine(fortunateNumber(n));
}
}
// This code is contributed
// by anuj_67.
PHP
<?php
// PHP program to find n-th
// Fortunate number
function isPrime($n)
{
// Corner cases
if ($n <= 1)
return false;
if ($n <= 3)
return true;
// This is checked so that we can skip
// middle five numbers in below loop
if ($n % 2 == 0 || $n % 3 == 0)
return false;
for($i = 5; $i * $i <= $n;
$i = $i + 6)
if ($n % $i == 0 || $n % ($i + 2) == 0)
return false;
return true;
}
// Function to Find primorial of order n
// (product of first n prime numbers).
function primorial($n)
{
$p = 2;
$n--;
for ($i = 3; $n != 0; $i++)
{
if (isPrime($i))
{
$p = $p * $i;
$n--;
}
$i++;
}
return $p;
}
// Function to find next prime
// number greater than n
function findNextPrime($n)
{
// Note that difference (or m)
// should be greater than 1.
$nextPrime = $n + 2;
// loop continuously until isPrime
// returns true for a number above n
while (true)
{
// Ignoring the prime number that
// is 1 greater than n
if (isPrime($nextPrime))
break;
$nextPrime++;
}
return $nextPrime;
}
// Returns n-th Fortunate number
function fortunateNumber($n)
{
$p = primorial($n);
return findNextPrime($p) - $p;
}
// Driver Code
$n = 5;
echo fortunateNumber($n) , "\n";
// This code is contributed by ajit
?>
Javascript
<script>
// JavaScript program to find n-th Fortunate number
function isPrime(n)
{
// Corner cases
if (n <= 1) return false;
if (n <= 3) return true;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0) return false;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false;
return true;
}
// Function to Find primorial of order n
// (product of first n prime numbers).
function primorial(n)
{
let p = 2;
n--;
for (let i = 3; n != 0; i++) {
if (isPrime(i) == true) {
p = p * i;
n--;
}
i++;
}
return p;
}
// Function to find next prime number greater
// than n
function findNextPrime(n)
{
// Note that difference (or m) should be
// greater than 1.
let nextPrime = n + 2;
// loop continuously until isPrime
// returns true for a number above n
while (true) {
// Ignoring the prime number that
// is 1 greater than n
if (isPrime(nextPrime) == true)
break;
nextPrime++;
}
return nextPrime;
}
// Returns n-th Fortunate number
function fortunateNumber(n)
{
let p = primorial(n);
return findNextPrime(p)-p;
}
// Driver Code
let n = 5;
document.write(fortunateNumber(n));
</script>
Producción:
23
Optimización : La solución anterior se puede optimizar utilizando Tamiz de Eratóstenes .
Publicación traducida automáticamente
Artículo escrito por SaagnikAdhikary y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA