Nice.

Wow, so fast editorial, thanks!

As a tester, I solved problem E with complexity $$$O(100(n+q)\log n)$$$: write the formula as $$$f_i=a_ib_i/\gcd^2(a_i,b_i)$$$. Let's maintain the answer by maintaining segments of same $$$a$$$-s together. It's easy to see that the segments only change $$$O(n+q)$$$ times. In queries, we specially manage the segments the query crosses (i.e. it contains $$$l$$$ or $$$r$$$) and the segments in the middle can be maintained with segment tree.

Now, we essentially have to calculate $$$f(l,r,x)=\max_{l\le i\le r} x\cdot b_i/\gcd^2 (x,b_i)$$$ for $$$O(n+q)$$$ times. Let's enumerate all divisors of $$$x$$$ and try it as a candidate for $$$\gcd(x,b_i)$$$. It's easy to see that we should select the maximum $$$b_i$$$ that is a multiple of $$$x$$$. To maintain largest $$$b_i$$$ in a segment that is a multiple of $$$x$$$, we can use a sparse table or a segment tree for each $$$x$$$. The sum of sizes of the tables is $$$100n$$$, since numbers in $$$[1,50000]$$$ have at most 100 divisors.

177649266

queryforces

Another unbalanced round

How to solve C1? I understand f(l,r) <= f(l,r+1), but I am not getting how binary search will help here

I think you should explain time complexities properly instead of ending the editorial with

Therefore, this solution will work quite quickly.

It might not be obvious to some of us.

Can someone explain D2 problem, I can't understand, the author explanation.

Alternative solution for E (it's just worse than what's in the editorial, but wotever, it got AC):

To do range assign and point get in A, a VEBTree over "important" indices can be used to get O(log log n) updates and queries.

Divide the arrays into blocks of size R (constant we will fix later). Then inside each block, compute for every value $$$\leq 5 \cdot 10^4$$$ what is the smallest multiple of that value which occurs in the range.

Then to compute the answer when an entire block gets set to a value $$$v$$$, it's $$$\min(\texttt{smallest_multiple}[d] \cdot v / d^2 \enspace \vert \enspace d \vert v)$$$

When a an update or query touches but doesn't cover an entire block, just brute force.

$$$O(Rq \log \log n + nqd/R)$$$ where $$$d$$$ is the largest number of factors a number can have, which is something like $$$128$$$ for numbers this small. I chose $$$R=400$$$ from trial and error instead of calculating what block size would make sense.

submission

Still don’t get A.. If a[n] mod n != 0, then why can we just check if gcd(g, n)? I mean if a[n] cannot be changed to n, then why checking gcd(g, n) is correct?

Sorry for my poor English in advance.

I still cant understand Question B, can somebody explain?

In problem E, formula answer_x = min(answer_(x/p) * p) is just plain wrong, even with d not equal to x.

Let n = 1, x = 4, b_1 = 8, p = 2. Then answer_x = lcm(x, b_1) / gcd(x, b_1) = lcm(4, 8) / gcd(4, 8) = 8 / 4 = 2. BUt answer_(x/p) = lcm(x/p, b_1) / gcd(x/p, b_1) = lcm(4/2, 8) / gcd(4/2, 8) = lcm(2, 8) / gcd(2, 8) = 8 / 2 = 4. answer_x is not equal to p * answer(x/p), actually, answer(x/p) > answer_x.

Please show me where I am wrong.

Managed to solve both C2 and D2, lost 16 rating.

The complexity analysis for D1 and D2 are not so obvious for me. Can someone maybe provide a better explanation?

In C1 and C2, I forgot that $$$a_i$$$ could be 0, TLE (QAQ)

unordered_set/map and gp_hash_table are uphacked.

Wow, so fast editorial, thanks!

In editorial of D1,

"Note that if k is not one of the largest divisors of x, then x/k becomes greater than q"

Why?

Does anyone have a correct proof to the solution of D1?

Can anybody explain problem C2? I don't understand what does $$$A$$$ means in the editorial.

C1 can be solved by brute force with optimization of skipping zeros with O = O(15*15) = 225, since any non-zero element is obliged to reduce the number of bits in the current xor. Thus C2 can be solved by adding a segment tree (to find the xor of segment) with O = (225+log(n))*n.

can someone explain c2 please?

What does Hack verdict 'Unexpected verdict' mean? Problem E's hacks have so many this type.

Я думаю стоит не писать словосочетание "тупое решение" в едиториалах.

This is related to the complexity analysis of $$$D1$$$ and $$$D2$$$:

If we have arrays $$$arr_1$$$ and $$$arr_2$$$, each of them has $$$N$$$ distinct values, where $$$1\le N \le 10^5$$$ and $$$1\le arr_1[i], arr_2[i]\le A_{max}$$$.

Then for each $$$arr_1[i]$$$, we iterate on its positive multiples in ascending order until the first multiple not in $$$arr_2[i]$$$. If $$$A_{max}=N$$$, then the total number of iterations can be calculated as $$$N\cdot \sum_{i=1}^{N}{\dfrac{1}{i}}=N\cdot H(N)=N\cdot log(N)$$$.

If $$$A_{max}$$$ is an arbitrary value larger than $$$N$$$, e.g., $$$10^{18}$$$, is there a way to prove the upper bound on the total number of iterations?

has anybody solved C1/C2 with segment tree ? i would like to see how the segments are Merged and stored

In C1 and C2, if I have a test like this : 0 8 2 0 and L = 1, R = 4

My output was "2 2" and Jury output was "1 1" and I got WA while segment [2,2] and [1,1] have the same length and f(l,r).

Could anybody explain to me why i'm WA, please :(

Problem D2 : I couldn't understand the last part. How we can recalculate these set in the case of an add/remove operation? Can anyone explain? TIA

74TrAkToR, In the editorial of D2 shouldn't it be:
*then we should already have added approximately $$$\sum\limits_{i = 1}^t\frac{x}{k_i}$$$.
At second to last line, since we are iterating over t divisors of x from $$${k_1}$$$ to $$${k_t}$$$.

Can someone please explain about problem E: how do we count answers for each x in blocks? Would apreciate it

Tried hard in D2 problem and at last accepted. I can't believe it was 2400 rated problem.

I have an alternate solution for D2 that uses sqrt optimization. Store a block of at max size $$$\sqrt n$$$ delete operations. For all queries run through all the erased numbers in the current block, if $$$x$$$ divides a number in the block ($$$b$$$) just min the answer you had kept previously with $$$b$$$.

If the block reaches size $$$\sqrt n$$$, clear the block and run D1's solution from 0 for all currently stored queries. Complexity over queries amortized is $$$O(n \sqrt n \log(n) + n \log(n)) = O(n \sqrt n \log(n))$$$. Because the clear and reset operation runs at max $$$\sqrt n$$$ times the total time complexity of that is also the same $$$O(n \sqrt n \log n)$$$. Needed some gp_hash_table optimization to remove the $$$\log n$$$ factor and make it pass. 178023023

Time limit on problem E is tooooooo tight. The overall complexity is still $$$\mathcal{O}(X\sqrt{X}\log X)$$$ where $$$X = 5 \times 10^4$$$. From the complexity perspective the fansy $$$\mathcal{O}(X\log \log X)$$$ optimization in the precalculation seems not necessary.

I've also tried a $$$\mathcal{O}(100n\log n)$$$ segment tree solution and it still can't pass the time limit.

Anyone got a solution to C1 using two pointers? Would help a lot! Thanks in advance!

In fact,I don't understand problem B and this tutorial o(╥﹏╥)o

There needs to be a proof for the upper bound time complexity on D1

why this givving me TL for D1 ?

https://codeforces.com/contest/1732/submission/214043354

Time complexity pf problem D1 should be O(qlogq). explanation : consider first q queries of insertion of 1 to q numbers and then for next q queries getting mex of 1 to q numbers . Insertion of 1 to q numbers takes O(qlogq) time . for finding mex of each element from 1 to q it will take (q/1+q/2+q/3 +q/4+....q/q)==qlogq Total time complexity : O(qlogq)+O(qlogq)==O(qlogq)

Honestly this is a very funny round if your rating does not depend on it.

The formal proof for D2 is really tedious, I could not normally prove it.

I passed E with a constant equal to 105, if it is 100 or 110 it TL's.

This is so fucking funny it is unbelievable.

Can someone please explain the proof for time complexity for problem D1. The editorial says If k is not one of the largest divisors of x, then x/k becomes greater than q. Therefore, this solution will work quite quickly

There could be case like x = 10^6 and y = 10. This could run for 10^5 times.