Codeforces 1165D. Almost All Divisors Codeforces Round #560 (Div. 3)

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Suppose the given list of divisors is a list of almost all divisors of some x (in other words, suppose that the answer exists). Then the minimum divisor multiplied by maximum divisor should be x. This is true because if we have a divisor i we also have a divisor ni. Let's sort all divisors and let x=d1⋅dn.

Now we need to check if all divisors of x except 1 and x are the permutation of the array d (check that our answer is really correct). We can find all divisors of x in O(x−−√), sort them and compare with the array d. If arrays are equal then the answer is x otherwise the answer is -1.

A solution in c++