The prime factors (Editorial)

Written by @Yusif_M and @Tuncay_Pashayev.

In this problem there is a number $n$ and we have to show number $n$ with the product of prime numbers.

First we have to find prime factors of this number, then we'll divide $n$ to prime factor while it is divisible to prime factor. When it is not divisible we'll move to next number and becuase we are dividing $n$ to prime factors while it divisible we won't face with the numbers which are not prime. At last we'll check if $n$ is greater than $1$ or not if it is greater then it means that it is divided to all possible prime numbers and remaining $n$ is prime and it can't be divided to any other prime number. This is the steps that we have to do to find prime factors of number $n$.

Implementation

vector <int> v;

Creating a vector called $v$.

for (int i = 2; i <= sqrt(n); i++)
{
	while (n % i == 0)
	{
		v.push_back(i);
		n /= i;
	}
}

Checking if number is divisible by any prime factor. If it is then add this number to vector. Their number can be more than $1$, so adding all of them.

if (n > 1)
{
	v.push_back(n);
}

Checking if remaining number after divided by all prime numbers is bigger than $1$, that means the remaining number is also one of prime factor of $n$.

for (int i=0; i < v.size(); i++)
{
	cout << v[i];
	if (i < v.size()-1)
	{
		cout << "*";
	}
}

Printing all prime factors of $n$ by '*'. If $v[i]$ is equals to v[v.size()-1], that means we mustn't output '*'.