Dado un número entero N , la tarea es imprimir todos los números semiprimos ≤ N .
Un número semiprimo es un número entero que se puede expresar como el producto de dos números primos distintos.
Por ejemplo, 15 = 3 * 5 es un número semiprimo pero 9 = 3 * 3 no lo es .
Ejemplos:
Entrada: N = 20
Salida: 6 10 14 15Entrada: N = 50
Salida: 6 10 14 15 21 22 26 33 34 35 38 39 46
requisitos previos:
Enfoque: Para cada número < N , cuente el número de factores primos que tiene. Si el número de factores primos es 2 , entonces el número es un número semiprimo ya que todos los números semiprimos tienen solo 2 factores primos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to create Sieve for Semi Prime Numbers
vector<int> createSemiPrimeSieve(int n)
{
int v[n + 1];
// This array will initially store the indexes
// After performing below operations if any
// element of array becomes 1 this means
// that the given index is a semi-prime number
// Storing indices in each element of vector
for (int i = 1; i <= n; i++)
v[i] = i;
int countDivision[n + 1];
for (int i = 0; i < n + 1; i++)
countDivision[i] = 2;
// This array will initially be initialized by 2 and
// will just count the divisions of a number
// As a semiprime number has only 2 prime factors
// which means after dividing by the 2 prime numbers
// if the index countDivision[x] = 0 and v[x] = 1
// this means that x is a semiprime number
// If number a is prime then its
// countDivision[a] = 2 and v[a] = a
for (int i = 2; i <= n; i++) {
// If v[i] != i this means that it is
// not a prime number as it contains
// a divisor which has already divided it
// same reason if countDivision[i] != 2
if (v[i] == i && countDivision[i] == 2) {
// j goes for each factor of i
for (int j = 2 * i; j <= n; j += i) {
if (countDivision[j] > 0) {
// Dividing the number by i
// and storing the dividend
v[j] = v[j] / i;
// Decreasing the countDivision
countDivision[j]--;
}
}
}
}
// A new vector to store all Semi Primes
vector<int> res;
for (int i = 2; i <= n; i++) {
// If a number becomes one and
// its countDivision becomes 0
// it means the number has
// two prime divisors
if (v[i] == 1 && countDivision[i] == 0)
res.push_back(i);
}
return res;
}
// Driver code
int main()
{
int n = 16;
vector<int> semiPrime = createSemiPrimeSieve(n);
// Print all semi-primes
for (int i = 0; i < semiPrime.size(); i++)
cout << semiPrime[i] << " ";
return 0;
}
Java
import java.util.*;
// Java implementation of the approach
class GFG
{
// Function to create Sieve for Semi Prime Numbers
static Vector<Integer> createSemiPrimeSieve(int n)
{
int v[] = new int[n + 1];
// This array will initially store the indexes
// After performing below operations if any
// element of array becomes 1 this means
// that the given index is a semi-prime number
// Storing indices in each element of vector
for (int i = 1; i <= n; i++)
{
v[i] = i;
}
int countDivision[] = new int[n + 1];
for (int i = 0; i < n + 1; i++)
{
countDivision[i] = 2;
}
// This array will initially be initialized by 2 and
// will just count the divisions of a number
// As a semiprime number has only 2 prime factors
// which means after dividing by the 2 prime numbers
// if the index countDivision[x] = 0 and v[x] = 1
// this means that x is a semiprime number
// If number a is prime then its
// countDivision[a] = 2 and v[a] = a
for (int i = 2; i <= n; i++)
{
// If v[i] != i this means that it is
// not a prime number as it contains
// a divisor which has already divided it
// same reason if countDivision[i] != 2
if (v[i] == i && countDivision[i] == 2)
{
// j goes for each factor of i
for (int j = 2 * i; j <= n; j += i)
{
if (countDivision[j] > 0)
{
// Dividing the number by i
// and storing the dividend
v[j] = v[j] / i;
// Decreasing the countDivision
countDivision[j]--;
}
}
}
}
// A new vector to store all Semi Primes
Vector<Integer> res = new Vector<>();
for (int i = 2; i <= n; i++)
{
// If a number becomes one and
// its countDivision becomes 0
// it means the number has
// two prime divisors
if (v[i] == 1 && countDivision[i] == 0) {
res.add(i);
}
}
return res;
}
// Driver code
public static void main(String[] args)
{
int n = 16;
Vector<Integer> semiPrime = createSemiPrimeSieve(n);
// Print all semi-primes
for (int i = 0; i < semiPrime.size(); i++)
{
System.out.print(semiPrime.get(i) + " ");
}
}
}
/* This code contributed by PrinciRaj1992 */
Python3
# Python 3 implementation of the approach # Function to create Sieve for Semi Prime Numbers def createSemiPrimeSieve(n): v = [0 for i in range(n + 1)] # This array will initially store the indexes # After performing below operations if any # element of array becomes 1 this means # that the given index is a semi-prime number # Storing indices in each element of vector for i in range(1, n + 1): v[i] = i countDivision = [0 for i in range(n + 1)] for i in range(n + 1): countDivision[i] = 2 # This array will initially be initialized by 2 and # will just count the divisions of a number # As a semiprime number has only 2 prime factors # which means after dividing by the 2 prime numbers # if the index countDivision[x] = 0 and v[x] = 1 # this means that x is a semiprime number # If number a is prime then its # countDivision[a] = 2 and v[a] = a for i in range(2, n + 1, 1): # If v[i] != i this means that it is # not a prime number as it contains # a divisor which has already divided it # same reason if countDivision[i] != 2 if (v[i] == i and countDivision[i] == 2): # j goes for each factor of i for j in range(2 * i, n + 1, i): if (countDivision[j] > 0): # Dividing the number by i # and storing the dividend v[j] = int(v[j] / i) # Decreasing the countDivision countDivision[j] -= 1 # A new vector to store all Semi Primes res = [] for i in range(2, n + 1, 1): # If a number becomes one and # its countDivision becomes 0 # it means the number has # two prime divisors if (v[i] == 1 and countDivision[i] == 0): res.append(i) return res # Driver code if __name__ == '__main__': n = 16 semiPrime = createSemiPrimeSieve(n) # Print all semi-primes for i in range(len(semiPrime)): print(semiPrime[i], end = " ") # This code is contributed by # Surendra_Gangwar
C#
// C# implementation of the approach
using System;
using System.Collections;
class GFG
{
// Function to create Sieve for Semi Prime Numbers
static ArrayList createSemiPrimeSieve(int n)
{
int[] v = new int[n + 1];
// This array will initially store the indexes
// After performing below operations if any
// element of array becomes 1 this means
// that the given index is a semi-prime number
// Storing indices in each element of vector
for (int i = 1; i <= n; i++)
v[i] = i;
int[] countDivision = new int[n + 1];
for (int i = 0; i < n + 1; i++)
countDivision[i] = 2;
// This array will initially be initialized by 2 and
// will just count the divisions of a number
// As a semiprime number has only 2 prime factors
// which means after dividing by the 2 prime numbers
// if the index countDivision[x] = 0 and v[x] = 1
// this means that x is a semiprime number
// If number a is prime then its
// countDivision[a] = 2 and v[a] = a
for (int i = 2; i <= n; i++)
{
// If v[i] != i this means that it is
// not a prime number as it contains
// a divisor which has already divided it
// same reason if countDivision[i] != 2
if (v[i] == i && countDivision[i] == 2)
{
// j goes for each factor of i
for (int j = 2 * i; j <= n; j += i)
{
if (countDivision[j] > 0)
{
// Dividing the number by i
// and storing the dividend
v[j] = v[j] / i;
// Decreasing the countDivision
countDivision[j]--;
}
}
}
}
// A new vector to store all Semi Primes
ArrayList res = new ArrayList();
for (int i = 2; i <= n; i++)
{
// If a number becomes one and
// its countDivision becomes 0
// it means the number has
// two prime divisors
if (v[i] == 1 && countDivision[i] == 0)
res.Add(i);
}
return res;
}
// Driver code
static void Main()
{
int n = 16;
ArrayList semiPrime = createSemiPrimeSieve(n);
// Print all semi-primes
for (int i = 0; i < semiPrime.Count; i++)
Console.Write((int)semiPrime[i]+" ");
}
}
// This code is contributed by mits
PHP
<?php
// PHP implementation of the approach
// Function to create Sieve for Semi Prime Numbers
function createSemiPrimeSieve($n)
{
$v = array_fill(0, $n + 1, 0);
// This array will initially store the indexes
// After performing below operations if any
// element of array becomes 1 this means
// that the given index is a semi-prime number
// Storing indices in each element of vector
for ($i = 1; $i <= $n; $i++)
$v[$i] = $i;
$countDivision = array_fill(0, $n + 1, 0);
for ($i = 0; $i < $n + 1; $i++)
$countDivision[$i] = 2;
// This array will initially be initialized by 2 and
// will just count the divisions of a number
// As a semiprime number has only 2 prime factors
// which means after dividing by the 2 prime numbers
// if the index countDivision[x] = 0 and $v[x] = 1
// this means that x is a semiprime number
// If number a is prime then its
// countDivision[a] = 2 and $v[a] = a
for ($i = 2; $i <= $n; $i++)
{
// If v[i] != i this means that it is
// not a prime number as it contains
// a divisor which has already divided it
// same reason if countDivision[i] != 2
if ($v[$i] == $i && $countDivision[$i] == 2)
{
// j goes for each factor of i
for ($j = 2 * $i; $j <= $n; $j += $i)
{
if ($countDivision[$j] > 0)
{
// Dividing the number by i
// and storing the dividend
$v[$j] = $v[$j] / $i;
// Decreasing the countDivision
$countDivision[$j]--;
}
}
}
}
// A new vector to store all Semi Primes
$res = array();
for ($i = 2; $i <= $n; $i++)
{
// If a number becomes one and
// its countDivision becomes 0
// it means the number has
// two prime divisors
if ($v[$i] == 1 && $countDivision[$i] == 0)
array_push($res, $i);
}
return $res;
}
// Driver code
$n = 16;
$semiPrime= array();
$semiPrime = createSemiPrimeSieve($n);
// Print all semi-primes
for ($i = 0; $i < count($semiPrime); $i++)
echo $semiPrime[$i], " ";
// This code is contributed by ihritik
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to create Sieve for Semi Prime Numbers
function createSemiPrimeSieve(n)
{
let v = new Array(n + 1).fill(0);
// This array will initially store the indexes
// After performing below operations if any
// element of array becomes 1 this means
// that the given index is a semi-prime number
// Storing indices in each element of vector
for (let i = 1; i <= n; i++)
v[i] = i;
let countDivision = new Array(n + 1).fill(0);
for (let i = 0; i < n + 1; i++)
countDivision[i] = 2;
// This array will initially be initialized by 2 and
// will just count the divisions of a number
// As a semiprime number has only 2 prime factors
// which means after dividing by the 2 prime numbers
// if the index countDivision[x] = 0 and v[x] = 1
// this means that x is a semiprime number
// If number a is prime then its
// countDivision[a] = 2 and v[a] = a
for (let i = 2; i <= n; i++)
{
// If v[i] != i this means that it is
// not a prime number as it contains
// a divisor which has already divided it
// same reason if countDivision[i] != 2
if (v[i] == i && countDivision[i] == 2)
{
// j goes for each factor of i
for (let j = 2 * i; j <= n; j += i)
{
if (countDivision[j] > 0)
{
// Dividing the number by i
// and storing the dividend
v[j] = v[j] / i;
// Decreasing the countDivision
countDivision[j]--;
}
}
}
}
// A new vector to store all Semi Primes
let res = new Array();
for (let i = 2; i <= n; i++)
{
// If a number becomes one and
// its countDivision becomes 0
// it means the number has
// two prime divisors
if (v[i] == 1 && countDivision[i] == 0)
res.push(i);
}
return res;
}
// Driver code
let n = 16;
let semiPrime= new Array();
semiPrime = createSemiPrimeSieve(n);
// Print all semi-primes
for (let i = 0; i < semiPrime.length; i++)
document.write(semiPrime[i] + " ");
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
6 10 14 15
Publicación traducida automáticamente
Artículo escrito por RishabhKejariwal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA