Ought to be fixed now

Yes your solution is correct, there seems to be an issue with the code. I'll fix it sometime soon

One operation is combining any two adjacent numbers by adding them, so only addition is allowed. The answer for that particular one is adding all the 1s to get [3,3,3,4,4,4,4,4]

Removed using namespace std and passing n to the function.

As for the vector<ull>, it became a habit of mine to always take in inputs as ull or ll after overflows on CF and the like. But I changed it to vector<int> as you suggested.

I hadn't thought of that. Mb, switched to vector<pair> instead

@poltrona14 my solution is O(1) with no brute forcing involved uwu

Eh sometimes you gotta use it, not here ig. This can be done in O(1) without any loops

fixed

You might wanna try again, It wont work for n <= 1 and n == 4

yea, I changed the while to a for loop and the 6k one is faster almost always but the time difference is still in the same range and fluctuates wildly. I never realized that while loops are slower than for loops in python, ty for letting me know

also I have no idea why my reply got posted twice

hmm, the results of comparing the two often turns out to be wildly inconsistent in time difference but remains in the 1-10µs range, tho I found that the n = 19 and 1009 takes longer with 2k, n = 10000000019 takes less with the 2k one most of the time. Code used for testing:

import timeit
SETUP6K = '''
def prime_checker6k(n):
    if n < 4:
        return n > 1
    if not (n % 2 and n % 3):
        return False
    for i in range(5, int(n ** 0.5)+1, 6):
        if not ((n % i) and (n % (i + 2))):
            return False
    return True
'''
SETUP2K = '''
def prime_checker2k(n):
    if n == 2: return True
    if n%2 == 0 or n < 2: return False
    for i in range(3, int(n**.5)+1, 2):
        if n%i == 0: return False
    return True
'''
def bench(n):
    #print(f'Testing prime_checker2k and prime_checker6k with arg: {n}')
    a = timeit.timeit(stmt=f"prime_checker2k({n})", setup = SETUP2K, number = 1)
    b = timeit.timeit(stmt=f"prime_checker6k({n})", setup = SETUP6K, number = 1)
    #print(f'Time taken for prime_checker2k = {a}\t\tTime taken for prime_checker6k = {b}\t\t2k-6k = {a-b}\n')
    print(f'{"prime_checker2k" if a<b else "prime_checker6k"} is faster')
    print(f'6k-2k = {round(b-a,9)}')

for _ in range(10):
  #bench(19)
  #bench(1009)
  bench(10000000019)
  print()