Find Largest Prime Factor - Complexity of Code - c++

Question

I tried a code on a coding website to find the largest prime factor of a number and it's exceeding the time limit for the last test case where probably they are using a large prime number. Can you please help me to reduce the complexity of the following code?
int main()
{
long n;
long int lar, fact;
long int sqroot;
int flag;
cin >> n;
lar=2, fact=2;
sqroot = sqrt(n);
flag = 0;
while(n>1)
{
if((fact > sqroot) && (flag == 0)) //Checking only upto Square Root
{
cout << n << endl;
break;
}
if(n%fact == 0)
{
flag = 1;
lar = fact;
while(n%fact == 0)
n = n/fact;
}
fact++;
}
if(flag == 1) //Don't display if loop fact reached squareroot value
cout << lar << endl;
}
Here I've also taken care of the loop checking till Square Root value. Still, how can I reduce its complexity further?

Answer 1

You can speed things up (if not reduce the complexity) by supplying a hard-coded list of the first N primes to use for the initial values of fact, since using composite values of fact are a waste of time. After that, avoid the obviously composite values of fact (like even numbers).

Answer 2

You can reduce the number of tests by skipping even numbers larger than 2, and stopping sooner if you have found smaller factors. Here is a simpler and faster version:
int main() {
unsigned long long n, lar, fact, sqroot;
cin >> n;
lar = 0;
while (n && n % 2 == 0) {
lar = 2;
n /= 2;
}
fact = 3;
sqroot = sqrt(n);
while (fact <= sqroot) {
if (n % fact == 0) {
lar = fact;
do { n /= fact; } while (n % fact == 0);
sqroot = sqrt(n);
}
fact += 2;
}
if (lar < n)
lar = n;
cout << lar << endl;
return 0;
}
I am not sure how large the input numbers may become, using the larger type unsigned long long for these computations will get you farther than long. Using a precomputed array of primes would help further, but not by a large factor.

Answer 3

The better result I've obtained is using the function below (lpf5()). It's based on the primality() function (below) that uses the formulas 6k+1, 6k-1 to individuate prime numbers. All prime numbers >= 5 may be expressed in one of the forms p=k*6+1 or p=k*6-1 with k>0 (but not all the numbers having such a forms are primes). Developing these formulas we can see a sequence like the following:
k=1 5,7
k=2 11,13
k=3 17,19
k=4 23,25*
k=5 29,31
.
.
.
k=10 59,61
k=11 65*,67
k=12 71,73
...
5,7,11,13,17,19,23,25,29,31,...,59,61,65,67,71,73,...
We observe that the difference between the terms is alternatively 2 and 4. Such a results may be obtained also using simple math. Is obvious that the difference between k*6+1 and k*6-1 is 2. It's simple to note that the difference between k*6+1 and (k+1)*6-1 is 4.
The function primality(x) returns x when x is prime (or 0 - take care) and the first divisor occurs when x is not prime.
I think you may obtain a better result inlining the primality() function inside the lpf5() function.
I've also tried to insert a table with some primes (from 1 to 383 - the primes in the first 128 results of the indicated formulas) inside the primality function, but the speed difference is unappreciable.
Here the code:
#include <stdio.h>
#include <math.h>
typedef long long unsigned int uint64;
uint64 lpf5(uint64 x);
uint64 primality(uint64 x);
uint64 lpf5(uint64 x)
{
uint64 x_=x;
while ( (x_=primality(x))!=x)
x=x/x_;
return x;
}
uint64 primality(uint64 x)
{
uint64 div=7,f=2,q;
if (x<4 || x==5)
return x;
if (!(x&1))
return 2;
if (!(x%3))
return 3;
if (!(x%5))
return 5;
q=sqrt(x);
while(div<=q) {
if (!(x%div)) {
return div;
}
f=6-f;
div+=f;
}
return x;
}
int main(void) {
uint64 x,k;
do {
printf("Input long int: ");
if (scanf("%llu",&x)<1)
break;
printf("Largest Prime Factor: %llu\n",lpf5(x));
} while(x!=0);
return 0;
}