Finding GCD of a set of numbers?

Question

So, I was asked this question in an interview. Given a group of numbers (not necessarily distinct), I have to find the multiplication of GCD\'s of all possible subsets of th

Answer 1

Pre - Requisite :

Fermat's little theorem (there is a generalised theorem too) , simple maths , Modular exponentiation

Explanation : Notations : A[] stands for our input array

Clearly the constraints 1<=N<=10^5 , tell me that either you need a O(N * LOG N ) solution , dont try to think DP as its complexity according to me will be N * max(A[i]) i.e. approx. 10^5 * 10 ^ 6 . Why? because you need the GCD of the subsets to make a transition.

Ok , moving on

We can think of clubbing the subsets with the same GCD so as to make the complexity.

So , lets decrement an iterator i from 10^6 to 1 trying to make the set with GCD i !

Now to make the subset with GCD(i) I can club it with any i*j where j is a non negative Integer. Why ?

GCD(i , i*j ) = i

Now ,

We can build a frequency table for any element as the number is quite reachable!

Now , during the contest what I did was , keep the number of subsets with gcd(i) at f2[i]

hence what we do is sum frequency of all elements from j*i where j varies from 1 to floor(i/j) now the subsets with a common divisor(and not GCD) as i is (2^sum - 1) .

Now we have to subtract from this sum the subsets with GCD greater than i and having i as a common divisor of gcd as i.

This can also be done within the same loop by taking summation of f2[i*j] where j varies from 1 to floor(i/j)

Now the subsets with GCD i equal to 2^sum -1 - summation of f2[ij] Just multiply i ( No . of subsets with GCD i times ) i.e. power ( i , 2^sum -1 - summation of f2[ij] ) . But now to calculate this the exponent part can overflow but you can take its % with given MOD-1 as MOD was prime! (Fermat little theorem) using modular exponentiation

Here is a snippet of my code as I am unsure that can we post the code now!

for(i=max_ele; i >= 1;--i)
            {
                to_add=F[i];
                to_subtract = 0 ;
                for(j=2 ;j*i <= max_ele;++j)
                    {
                        to_add+=F[j*i];
                        to_subtract+=F2[j*i];
                        to_subtract>=(MOD-1)?(to_subtract%=(MOD-1)):0;
                    }

                subsets = (((power(2 , to_add , MOD-1) ) - 1) - to_subtract)%(MOD-1) ;

            if(subsets<0)
                subsets = (subsets%(MOD-1) +MOD-1)%(MOD-1);

            ans  = ans * power(i , subsets , MOD);
            F2[i]= subsets;
            ans %=MOD;
        }

I feel like I had complicated the things by using F2, I feel like we can do it without F2 by not taking j = 1. but it's okay I haven't thought about it and this is how I managed to get AC .