Given an integer N, the task is to print all the semi-prime numbers ? N.
A semi-prime number is an integer that can be expressed as a product of two distinct prime numbers.
For example, 15 = 3 * 5 is a semi-prime number but 9 = 3 * 3 is not.
Examples:
Input: N = 20
Output: 6 10 14 15Input: N = 50
Output: 6 10 14 15 21 22 26 33 34 35 38 39 46
Prerequisites:
Approach: For every number < N, count the number of prime factors it has. If the number of prime factors is 2 then the number is a semi-prime number as all the semi-prime numbers have only 2 prime factors.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to create Sieve for Semi Prime Numbersvector<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 codeint 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 approachclass 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 Numbersdef 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 codeif __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 approachusing System;using System.Collections;class GFG{ // Function to create Sieve for Semi Prime Numbersstatic 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 codestatic 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 Numbersfunction 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($n);// Print all semi-primesfor ($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 Numbersfunction 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 codelet n = 16;let semiPrime= new Array();semiPrime = createSemiPrimeSieve(n);// Print all semi-primesfor (let i = 0; i < semiPrime.length; i++) document.write(semiPrime[i] + " ");// This code is contributed by _saurabh_jaiswal</script> |
Output:
6 10 14 15
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