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151 lines (102 loc) · 2.35 KB
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#include <bits/stdc++.h>
using namespace std;
// Prime Check & Prime Factorisation
int main() {
/*
n is a prime if its divisors are only 1 and n
n --> 1, n
*/
// bool is_prime = true;
// int n;
// cin>>n;
// for (int i = 2; i < n; i++)
// {
// if(n % i == 0){
// is_prime = false;
// break;
// }
// }
// // O(n)
// cout<<is_prime<<endl;
// this is brute force method
/*
SQRT Method
1 * 24 1 * 36
2 * 12 2 * 18
3 * 8 3 * 12
4 * 6 4 * 9
6 * 4 6 * 6
8 * 3 9 * 4
12 * 2 12 * 3
24 * 1 18 * 2
36 * 1
Thus we can convert the loop from O(n) to O(sqrt(n))
if a number is divided by 2 then it is fact then it is also divided by 18
also if a number is not divided by 2 then it is also not divided by 18
thus we can check upto sqrt(n)
*/
// int n;
// cin>>n;
// if(n ==1){
// cout<<0<<endl;
// return 0;
// }
// bool is_prime = true;
// for (int i = 2; i * i <= n; i++)
// {
// if(n % i == 0){
// is_prime = false;
// break;
// }
// }
// // O(n)
// cout<<is_prime<<endl;
/*
Finding Prime Factorisation
## in any number the smallest divisor other than 1 is a prime number
ex. for 36 = 2 is prime
25 = 5 is a prime
91 = 7 is a prime
## 36 --2--> 18 --2--> 9 --3--> 3 --3--> 1
*/
// int n;
// cin>>n;
// vector<int> prime_factors;
// for (int i = 2; i <= n; i++)
// {
// while (n%i ==0)
// {
// prime_factors.push_back(i);
// n /= i;
// }
// }
// // O(n)
// for (int &prime : prime_factors)
// {
// cout<<prime<<" ";
// }
/*
for a composite number
atleast one prime factor will lie from [2, sqrt(n)]
*/
int n;
cin>>n;
vector<int> prime_factors;
for (int i = 2; i*i <= n; i++)
{
while (n%i ==0)
{
prime_factors.push_back(i);
n /= i;
}
}
if(n > 1){
prime_factors.push_back(n);
}
// TC:- O(sqrt(n))
for (int &prime : prime_factors)
{
cout<<prime<<" ";
}
return 0;
}