my solution is the following:
start with least common multiple = 1
make char array of 1000 000 to track forbidden divisors

let div be the divisor and ans the answer

if ans == 1 make new lcm of the previous lcm and current divisor. if it is over 10^12 - fail. find all the divisors of div, check that they are not forbidden in the forbidden array
if ans == 0 - check if lcm is divisible by the divisor and fail if it is.

complexitiy O(n * sqrt(1000000)) (sqrt from the "find all divisors)

my solution gets 0.5 sec (using cin,cout with tie(null))
what is the solution that gets less than 0.1 sec? Is there any hint?

no need to find all divisors. u can just use lcm and mod operations. i can send you easy solution if u r still interested