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Very interesting problems)

How to solve

E (Usmjeri)andF (Garaža)?Root the tree. Let path

a-bbep_{1}=a,p_{2},p_{3}, ...,p_{k}, ...,p_{t}=bwherep_{k}is their lca. Ifk= 1 ork=tthen all edges (p_{i},p_{i + 1}), (1 ≤i<t) must be either all directed down the root or all up the root. Suppose now this is not the case, then the edges (p_{i},p_{i + 1}), (1 ≤i<k) must have the same direction (down or up the root) and also the edges (p_{i},p_{i + 1}), (k≤i<n) must have the same direction while the edges (p_{k - 1},p_{k}) and (p_{k},p_{k + 1}) must have different directions so now you obtained a simple 2-sat problem for which you can find if there is a solution inO(n) and the number of solutions is 2^{number of conex components in the 2 - sat graph}.Oh 2-sat. Very beautiful solution.

Nice. Can you post your code, please?

I'll post it after it gets AC.

Mine is more or less the same big idea as reality's, but the details might be different. I only managed to finish coding after the contest was over :(

Here's my solution accepted in analysis mode (runs in 0.72s): Click

Can you please elaborate on how you check if the answer is 0? I am getting 4 wrong answers because my code thinks it's non-zero.

If (v, 0) and (v, 1) are in the same component.

I understood everything you said but I couldn't figure out during the contest just how I maintain the subsets of edges that should have the same direction. Mind explaining a little further?

You will maintain edges in a compressed form. Let the

k^{th}ancestor of vertexxbef[k][x]. You will maintain an arrayE[x][y] that is marked if all the edges on the path from vertexyto vertexf[2^{x}][y] are equal. Now if you have that all edges from vertexxto vertexf[k][x] are equal then pick up the largesttwith 2^{t}≤kand markE[t][x] andE[t][f[k- 2^{t}][x]]. Now ifE[t][x] is marked you can say thatE[t- 1][x] andE[t- 1][f[2^{t - 1}][x]] are marked and the edges (f[2^{t - 1}- 1][x],f[2^{t - 1}][x]) and (f[2^{t - 1}][x],f[2^{t - 1}+ 1][x]) have the same direction. This way you can reduce the edges in .How can you check if answer is 0? And also what are variables in 2-SAT problem (what do variables represent in our task)? I have never solved any 2-SAT problem, so I would really appreciate more detailed description of solution?

hint for F (it can be solved with different data structures after that):

Imagine we change the value of position

i. Let's find the leftmost positionjsuch thatgcd(j,i) =a_{i}(heregcd(l,r) is the GCD of the rangea_{l},a_{l + 1}, ...,a_{r}). Now we know that thegcd(j- 1,r) will decrease. Well let's find the leftmost positionj_{2}, such thatgcd(j_{2},i) =gcd(j- 1,i).If we do this until

gcd(j_{k}- 1,i) = 1 how many such ranges can we have in worst case?Well .

I already knew that fact. But I have no idea what to do next.

Well if we had no updates we could solve the problem like that:

For every position

ifind rightmost positionr_{i}such thatgcd(i,r_{i}) > 1.Now the answer for a query [

L;R] is( (sum of.rvalues in [L;R]) — (sum ofrvalues in [L;R] greater thanR) + (count ofrvalues in [L;R] greater thanR) *R—R* (R+ 1) / 2 + (L* (L- 1) / 2))Well this can be done with a merge sort tree in .

Now how to handle updates?

Well we go through the

O(log(10^{9})) ranges from the previous comment. Now with binary search in segment tree we find for each of the ranges the correspondingvalue. You can see that thisrrvalue will be the same for all of the positions in a given range. Well now we simply set the values in the range tor.The data structure we will use will be merge sort tree with treap in every node, because using it the set operation of an interval can be done easily.

The complexity will be .

How to understand this?

UPDUnderstood. Thanks. But I guess treap is too heavy and, probably, a simpler solution exists.Well time limit is 4 sec so this should fit in TL without a problem.

Another possibility is to use SQRT decomposition (if you don't like treaps :D) and get a solution which also should pass if implemented well.

I think your solution is too complex. With one more observation, you can simplify your solution a lot.

Suppose, f(i) = maximum index j, such that gcd[i, j] > 1. Note that, f(1) <= f(2) <= ...

So for a query, [L, R], you can binary search and find, largest i (L <= i <= R) such that, f(i) <= R. Then rest is just application of simple segment tree.

Oh nice. I feel stupid now :D

How can you calculate

j_{1},j_{2}, ...? I cannot find a suitable time complexity to solve this. I only know aO(log(10^{9}) *log(N)^{2}) solution per update by using a segment tree of GCD and binary search. However, it seems that such a solution is not fast enough to pass the testcases.OPS I explained how to find the closest position to the right, not to the left, but the idea to find the one to the left is basically the same.You do the binary search inside the segment tree:

Let's get the nodes in the segment tree which the query [

i;N] will cover. They will obviously be at most . Now we begin from the leftmost such interval and go through these nodes.If the interval of the current node is [

L;R] we can easily findgcd(i,R) inO(1) using thegcdvalues of the previous intervals.We find the first interval, such that

gcd(i,R) = 1. This will be done in time.Well now we know that the position which we are searching for (current

j) is in a given interval [L;R] (we also know the number of this node in the segment tree).Well we just start going down from our current node:

If the GCD of the current gcd and the gcd of the left child is not 1, we go to the right child with the new gcd. Else we go to the left child with the same gcd.

Using this in the end we will be at the first position

xwithgcd(i,x) = 1. The complexity will be .So overall complexity for finding the closest position to the right of a given position will be .

In Problem E I found solution with HLD. But i haven't coded HLD before. So i can't.

My SolutionFor each (ai,bi) we check is this path's contain any visited edge. if it isn't ans*=2. if it's contain any visited edge answer remains constant. Because we direct this edge before. Then We update all edges which within this path. Then we check all edge's in tree if it isn't visited yet ans*=2.

Am i wrong ? Sorry for my bad English :D

I've had something like that. But without hld. I even have code, but it doesn't work. It seems to be correct.

i found my mistake. But it will work

Yeah, nice problems

How to solve C? I solved D with meet-in-the-middle but not C :P

You only need to consider subarrays that if you reverse it one of its endpoints will become fixed point. So only o(n) subarrays to check.

Yes, subarrays to check. But, every subarray is checked in , no? Am I blind? :D

UPDNo question anymore.What about

N,N- 1, ..., 1 test?UPDNo question anymore.For each i, calculate i+a[i], and put it into buckets. The number of fixed points in the flipped subarray [L, R] is the number of indices between L and R that are in bucket L+R. (you can use binary search). You can see it in my code that I posted down.

r — (i — l + 1) + 1 = p[i]

r — i + l — 1 + 1 = p[i]

r + l = p[i] + i

also don't forget about fixed points that are not inside [L,R]

Can you describe this point in more details, please? :)

UPDUndestood. Good :)Calculate prefix sum of elements that are at begining in right position. For every element

xthere is only oney(element or position between 2 elements) so one of the options is to reverse array around y where x is last element we reverse.Now for everyywe try solutions and possiblexare sorted (you processed from 1 to n) so just iterrate that list and substract number of elements that are at right position in that interval.That is what I implemented but I didn't think it would pass with N up to 500,000. How did you go about checking each of the O(n) subarrays?

This is what I did, I think it's correct: Basically, it only makes sense to swap [L, R] if a[L] = R or a[R] = L, therefore you only have N subarrays to check. codeEdit: too late :P

EDIT : Wrong, only 80/100 on systest :(

How to solve D? I guess it is solved using meet-in-the middle but couldn't figure out the solution.

You keep an array of pairs for the first half {sum, height of last one}, and for the second you do {sum, height of first one}, then you place the second ones into buckets based on height, then using binary search you find out how many will make a sum over K.

Like this: https://pastebin.com/64uCcnfC

Approximately when the results will be posted?

Near 1 hour. It is tеsting now.

Result are out in evaluator!

Out in evaluator for outsiders but for some reason (KHMjerking off) it is not out for Croatian high school students who wrote the contest in the morning.

How to solve problem

D(San)?