The 2018 Egyptian Collegiate Programming Contest was held yesterday this was a problem from it where he asks for

Where n<=10^5 and a[i]<=10^5 answer should be printed %10^9+7

I would appreciate any help in solving this problem and thanks in advance

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The 2018 Egyptian Collegiate Programming Contest was held yesterday this was a problem from it where he asks for

Where n<=10^5 and a[i]<=10^5 answer should be printed %10^9+7

I would appreciate any help in solving this problem and thanks in advance

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Auto comment: topic has been updated by Dijkstra. (previous revision, new revision, compare).hintinclusion exclusion

solution ideaFor a fixed

ghow many pairs that will have thisgas a gcd, it'll be all multiples ofg, it's easy to get the sum of multiplying those pairs.but there is a problem with this, there will be some pairs that will be calculated multiple times, so we have to exclude the answer of (2 *

g, 3 *g, ...), for example if we are solving wheng= 2 we know that all the numbers divisible by 4 is also divisible by 2, so we have to exclude the answer of 4, and so on.so the answer is iterate from the max number in descending order to 1 and solve for each fixed

g, the total complexity of this solution iscodeWhy do you need to, if for example g=2, to exclude twice the answer of 4?

EDIT: I got it. It turns out that in the calculation of g=2, I overcount the pairs that have 4 as a gcd and I count them as having 2 as a gcd... In the ans[4], they are divided by 4 (their gcd), so to remove the overcounting I multiply the ans[4] by 2.

Mhm... nice problem. Do you agree, mnbvmar?

<3

Story: We proposed this exact task (under a tiny bit different constraints, though) to our high school students back in 2015. I was quite surprised that apparently it has never appeared anywhere even though the statement feels really natural and obvious. So it seems it eventually did!

I think e-pluszak came up with another solution, involving...

Spoilerthe Möbius inversion formula

I can't remember any details, though. Mind sharing (if you still remember it)?

SolutionFirst I calculated an array

B[n] such, that (how: first set , then for eachifrom 1 toniterate over all multiples ofiand subtractB[i] fromB[di]).Now we can just iterate over all possible

dand add to accumulator sum of entries divisible bydsquared timesB[d]. This way for each paira[i],a[j] we have added , which is equal toThe only moment where Möbius inversion formula came into play was calculating

B[n] array. It can also be done using that formula, but the way I posted here is much easier. I think i used this approach on that contest back in 2015, but I'm not sure