You would'n come into a situation to divide by ex. 9 or 14, cause you would've already divided the number by their prime decomposers 2, 3 and 7. Thus no need to check for primality.

board is unsigned int. what happened when unsigned int 0 minus by one?

hint: x|=1 << (board[row][col] -1); is not shifting left by -1. it's something else

The for loop starts from the first prime number (2) and if n would be divisible by non-prime i, it also means that it is divisible by i prime factors. So n would be reduced (n /= i) in earlier loop steps, n % i == 0 can be true only for prime i because of it.

Yes, ease on the eyes

Agree.
v[i+2] will be out of bounds when i >= v.size() - 2;
for loop condition is i<v.size(); I wonder how this code even passed all the tests!???

x |= 1 << (board[row][col] - 1);
when board[row][col] == 0 you are shifting left by -1. This is UB

Sorry sir, i don't think this is best practrice code. You are using loop and this wierd if() at the end...
Why not just format each component once and then just add them? Look at my solution and say what you think.

+1 Clever for recursion
-1 Best practices for the same reason )))

Why don't you check if i is prime or not before n % i == 0 check?
Kata says: "Given a positive number n > 1 find the prime factor decomposition of n."

So i must be prime... Am i right?

Can never overuse templates. 😉😂

Why this terribly slow solution with unordered_map got so many votes?

Template overuse detected!
Also this solution is not so good because of 55

op2 for jnz could be reg. It is not guaranteed to be a constant.
quote from description:"jnz x y - jumps to an instruction y steps away (positive means forward, negative means backward, y can be a register or a constant), but only if x (a constant or a register) is not zero"