# Prime Factorization using Sieve O(log n) for multiple queries in C++

C++Server Side ProgrammingProgramming

In this problem, we need to create a program to calculate Prime Factorization using Sieve O(log n) for multiple queries.

As the general method takes O(sqrt(n) ) time which will increase the time required to a huge extent from multiple queries.

Let’s recap first,

Prime factorization of a number includes ONLY the prime factors, not any products of those prime factors.

Sieve of Eratosthenes is an algorithm to generate all prime numbers within the given range.

## Solution Approach

The solution to the problem is found by finding the smallest factor that divides the number, saving it as a factor and updating the number by dividing it by the factor. This process is done recursively till the number becomes 1 after division, which means no other factors are possible.

The calculation is done using sieve of eratosthenes which reduces the time complexity in finding the smallest prime factor.

## Example

Live Demo

#include <iostream>
using namespace std;
int primes[100001];

void sieveOfEratosthenes(int N) {

N+=2;
primes[1] = 1;
for (int i=2; i<N; i++)
primes[i] = i;
for (int i=4; i<N; i+=2)
primes[i] = 2;
for (int i=3; i*i<N; i++) {
if (primes[i] == i) {
for (int j=i*i; j<N; j+=i)
if (primes[j]==j)
primes[j] = i;
}
}
}
void findPrimeFactors(int num) {

sieveOfEratosthenes(num);
int factor;
while (num != 1) {
factor = primes[num];
cout<<factor<<" ";
num /= factor;
}
}

int main() {
int N = 45214;
cout<<"Prime factorization of the number "<<N<<" using sieve is ";
findPrimeFactors(N);
return 0;
}

## Output

Prime factorization of the number 45214 using sieve is 2 13 37 47
Published on 27-Jan-2021 05:05:09