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Axial-Tilted's blog

Today problem E , SPOILER
By Axial-Tilted, history, 5 hours ago, In English

Since the editorial hasn't dropped yet for today's problem E plz share your best proofs on why starting from the minimum value is optimal , couldn't prove it for a long time , iam really interested In hearing Your proofs

edit : I can't sleep without hearing an intuitive proof and it doesn't look like the editorial is dropping soon

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5 hours ago, # |
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Auto comment: topic has been updated by Axial-Tilted (previous revision, new revision, compare).

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5 hours ago, # |
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Auto comment: topic has been updated by Axial-Tilted (previous revision, new revision, compare).

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4 hours ago, # |
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Why do you need it? It's AC, move forward

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    4 hours ago, # ^ |
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    Why are you making useless accounts and is afraid to use your main leave cf and move on

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4 hours ago, # |
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i'm not an expert, but i was thinking search two coprimes, in there will be the start of the new array, because gcd(a, b)=1 with a and b are coprimes.

In case there are not coprimes, probably using dp for search for a pair with the minimun gdc, and put there at the start, but i have to process more this part :/

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    4 hours ago, # ^ |
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    Greedily choosing the number that makes me the smallest is always good meaning that at the beginning the smallest number will always have smallest prefix sum than any other number , how to prove it

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      4 hours ago, # ^ |
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      idk man, i'm not how to be formal but i will try.

      knowing that we want to minimize the sum of the array gcd(a1) + gcd(a1, a2) + gcd(a1, a2, a3) ..., we can select a pair that minimize gcd(a1, a2), because we can ensure that a result can be

      min(a1, a2) + gcd(a1, a2)*(n-1) >= ans, that can you give us an approximation of the answer, so we need continue searching minimizin the gcd...

      Note1: gdc(gdc(a1, a2), x) <= gdc(a1, a2), so we can assume if we have the first gcd(a1, a2) minimizes, no matter what nummber add next, will satisfy the previous property

      Note2: if you find a coprime pair, the answer will be min(a1, a2) + (n-1), because no matter what will be the next number, the gcd(a1, 1) = 1.

      -------------------------- I means, this can be a route?

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3 hours ago, # |
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Ok here's my take:

For n = 3, let a[1] < a[2] < a[3]. We will prove that sorting them is optimal.

Define f(p[]) = the value of the expression in the statement when we order them using the permutation p[].

Let p[] be the optimal permutation. We will show that it has to have p[1] = 1. In other words, having the smallest number first is the best.

If p[1] > p[2], f(p[]) > f(p[] with p[1] and p[2] swapped) for obvious reasons by comparing the sums term by term, there only being a difference in the first term. As such, p[1] < p[2] from now on.

If p[1] != 1, we must have p[3] = 1, so p[] = {2, 3, 1}. Then, a[2] >= a[1] + (a[1], a[2]), so, using that inequality, we get f(p[]) < f({1, 2, 3}) by writing it all out.

As such, for n = 3, leaving the smallest first is optimal.

For n = 4, we assume the same things, and we already know that p[1] is in {1, 2} because the minimum among the first three has to be in the first position. So, if p[1] = 2, we must have p[4] = 1. We will prove f(p[]) > f({1, 2, p[2], p[3]}):

By writing it out and removing the last, equal terms, we get a[2] + (a[2], a[p[2]]) + (a[2], a[p[2]], a[p[3]]) > a[1] + (a[1], a[2]), (a[1], a[2], a[p[2]]), which is true for the same reasons as n = 3's equation!

For the rest of the n, the proof is similar inductively. It all boils down to the following lemma:

"If a < b, then b >= a + (a, b)"

As such, for any length of the sequence n, the minimum number has to be placed first.

PS: I left a lot of things unproven. Try to prove them yourself (or ask ChatGPT idc). (This was definitely not because I was too lazy!)

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    3 hours ago, # ^ |
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    I came up with something alitte bit similar let's take any two numbers X , Y , X < Y and prove that taking X will atleast improve the answer by 1

    Let g denote the gcd(X,Y) and let's write X as p*g and Y as q*g since p < q then (p+1)*g <= Y meaning if we take X then Y we have a prefix already <= Y and ending with g which allows for better upcoming options with already less prefix sum

    proof that having g will open up for better options than Y , lets take any number and call it C lets also denote g1 , g2 as gcd(g , C) and gcd(Y , C) respectively since g is a subset of Y interms of the prime divisors then g1 will never have an extra prime divisor over g2 as everything that exists in g already exists in Y

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3 hours ago, # |
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Here's a good proof

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