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Разбор задач Codeforces Round #453 (Div. 1)

Разбор задач Codeforces Round #453 (Div. 2)

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Codeforces (c) Copyright 2010-2018 Михаил Мирзаянов

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Can anyone explain in detail problem c div2?

Wow you got downvoted hard.

Can Some Explain Div2 B with more detail and Example.

Div2 B: You must first build the tree using the normal process and then do a bfs from the root. We're doing a bfs for 2 reasons: 1. because we will colour a subtree only if its parent has a different colour and a bfs is the optimal way to decide that. 2.We need to minimize the number of colourings so its best if we go from top to bottom one layer at a time than any other order.

Is is possible to use strategy like this ? Do tree traversal root-left-right. If child's color different from parent's then increase answer.

Yes that's exactly the same as doing a bfs(breadth first search) :). But we cant be sure that a parent node will only have only two nodes, left and right. it can have any number of children.

Thank you !

SIMPLE APPROACH:without tree building it can be done just maintaining the two arrays at a time one for checking the parent and other for which colour it has earlier..and if the colour assigned to it due to its parent is not the required colour then increment the steps and set the colour value to the above assigned colour for the remaining subtrees...

so the crux is just check for every possible parent and iterate through its all possible childrens and do the above criteria :)

Nice Logic!! But i figured this approach out only after the contest. I guess contest pressure pushed me into implementing what came to my mind first and hence the tree implementation. :P

Thanks For Help :)

Much simpler code... without using bfs or dfs :) 33419770

Nicely Done!!! Easier the problem the more the ways of solving it i guess!

Maybe I am wrong, but I think you are doing a dfs in your code, you have built the edges and go throw them. You just did not create a separate function and marked vertex already visited.

No I am not using dfs , I am just checking for any given node if there is any first-level descendant which has a different color than its parent,if it is then I increment the answer. In dfs you traverse from one node to its descendant until no further descendants are left to traverse. Hope this helps :)

in the div2 if you want color the tree,you can color the tree from Leaf to root， if anyone node's color don't same as its father,you should color the father first, then color this node,so if c[i] != c[fa[i]] the ans need ++ans

finally, the root need color once,so you need ans + 1;

that's all

here is my code http://codeforces.com/contest/902/submission/33418162

Could anyone explain this sentence "if remainders sequence has

ksteps while you consider numbers by some modulo it will have at leastksteps in rational numbers" in problem 901B - НОД многочленов?You can perform Euclid's algorithm using field of remainders modulo some prime number instead of rationals. If at some point you will get

B= 0 in rationals, you will also haveB= 0 at this step using remainders field. It is possible though that you obtain 0 even earlier but we generate sequence in such way that you don't getB= 0 fornsteps modulo prime number 2 thus you won't getB= 0 in rationals during thesensteps also.Hey, what's wrong with this comment? I'm just explaining things :(

I don't know, it's entirely correct... this is how I solved the problem.

Let's say you do one step of the Euclidean algorithm, for polynomials p and q: p = aq+r, where a and r are polynomials with deg r<deg q.

Then, r=p-aq, so (r mod 2) = (p mod 2) — (a mod 2)(q mod 2).

Because the Euclidean algorithm gives unique remainders (up to constant multiples), this means that we can run the two in parallel, and at any given point, the rational polynomial will equal the mod2 polynomial, when reduced.

The algorithm ends when we get r=0. Obviously, if we reach 0 in the actual (rational) sequence, we will also get 0 in the reduced sequence. So the original algorithm will take at least as many steps as the reduced one.

There is a small subtlety — can we reduce any rational number mod 2, like we can with integers? In fact we can, if we don't have an even denominator. It's not hard to see that this never happens, so we are OK.

Do you mean that we will not receive even denominators for any sequence of polynomials or for the sequence of polynomials from the solution of the problem?

Apparently, the first variant is incorrect: . And when applying the Euclidean algorithm to the pairs like (

X^{3}+X+ 1, - 2X- 1) in and it doesn't seem like steps of the algorithm correspond to each other.And I still can't find the obvious reason why it is true for the sequence from the solution.

You're right. I'll have to think of it some more. Hopefully it's not a fatal error...

OK I think it still works.

The reason is that in the defined sequence, the degrees decrease by 1. So, you can get extra terms with even coefficients, but the leading term will always have an odd coefficient.

Oh, that makes sense. Thanks!

In Div 2 C, is there any solid prove to this: if there is such a height k that a[k] > 1 and a[k + 1] > 1, then we can create 2 non-isomorphic trees from the current sequence? I mean how can we know if 2 created trees from that sequence are non-isomorphic or not?

You have one longest path from root to farthest leaf. In first tree for each vertex either it is on the path or its parent is at such path. In second tree however it is possible that parent of some vertex still not on longest path.

Take all the nodes in height k+1 and set their parents to only one node of depth k (call it v). This is a rooted tree that has a[k+1]-1 leaves with height k+1.

Now, do the same thing, but take one node in height k+1 and set its parent to another node different from v. This is a rooted tree that has a[k+1]-2 leaves with height k+1. These two trees cannot be isomorphic because the number of leaves with height k+1 are different.

That's why there is the condition of the two consecutives a[i] such that a[i] > 1. If this doesn't happen in the sequence, we can guarantee that there isn't any non-isomorphic pair of trees because:

For every height k such that a[k] > 1 and a[k+1] = 1 (because there aren't consecutive terms greater than 1), picking a parent for this node with height k+1 will not change the isomorphism.

For every height k such that a[k] = 1 and a[k+1] > 1, there is only one possible parent for these nodes with height k+1.

`This is a rooted tree that has a[k+1]-2 leaves with height k+1. These two trees cannot be isomorphic because the number of leaves with height k+1 are different.`

.Sorry if i'm wrong, but i think they want to find 2 non-isomorphic trees with same sequence of vertices amount at heights, and i still don't understand why two trees cannot be isomorphic there is the condition of the two consecutives a[i] such that a[i] > 1.

Yes, they want you to give 2 trees (non-isomorphic) with the same sequence of vertices.

The thing is: If there are two consecutives a[i] such that a[i] > 1 then there

existstwo non-isomorphic trees with the given sequence. In fact, there are some trees with this sequence. Some of them are isomorphic among each other, some aren't. So you just need to pick a pair of non-isomorphic trees.If there aren't these consecutives a[i], all the possible trees are isomorphic (by the argument above).

To check if two trees are isomorphic, every node in the first tree must have one corresponding node in the second tree. In the example above, there is a node in the first tree with height k and a[k+1] children, and there isn't any corresponding node in the second tree satisfying this (looking at the nodes with height k, there are some with 0 children, one with 1 child and one with a[k+1]-1 children).

But what if in the second tree, there is a node with a[k + 1] children tbat isn't at height k?

The two trees still won't be isomorphic since these two nodes (the node from the first tree and the node you mentioned from the second tree) can't be matched (because they don't have the same height).

The height of a node depends on the node which you chose to be the root doesn't it?

Can we solve 1B through a random algorithm? As we all know, there are hundreds of thousands of possible solutions, and I have no idea on construction of such a sequence, so I used a random algorithm and passed system test.

Here is my code. Hack is welcomed.

Kinda hard to hack if there are 150 possible inputs and all of them can be tested in no time xD

Or even better, you only really need to solve this problem for

n= 150. Then you can just work your way down from there.Not really, intermediate polynomials don't have to have coefficients in {-1, 0, 1} which is kinda crucial here (I think).

I think not if you do it modulo 2

Edit: I agree, but when you figure out that you can do the everything modulo 2 you've practically solved the whole thing

How does the checker work for problem B div 1? It would seem that it might require large (how large?) bignums. I guess it can compute the GCD sequence with a random prime modulus. If it works then great. If not then try another modulus. Once the product of the primes you try is larger than the largest number you'd ever need, I guess you can declare it "good".

It computes pseudoremainders using big integers. Generally size of coefficients grows exponentially but if you divide each term by gcd of coefficients then size of coefficients will grow only linearly of

n. I wanted to use prime numbers at first but KAN decided it's not the best idea. However I suppose one needs much smaller number of primes than maximum number to have very high probability of correct verdict...can someone please explain the idea behind DIV2-D. How fibonacci polynomial leads to the desired answer?

Need it too QWQ

I was upsolving problem C in Div2, using PyPy2 and got a TLE on test case 14. The complexity is O(sum(a)) which is fine, but still got a TLE. However it was accepted when the same code was submitted in Python2 and Python3. I had always known PyPy to be faster than native Python. This has been true in Codechef platform. Can anyone reason as to why it TLE'd in PyPy2?

Python2 Accepted, Python3 Accepted, PyPy2 TLE (14)

This link might help you. Your solution seems to have a lot of string manipulation, and pypy might not be optimizing them as much as cpython, leading to TLE.

For Div1B:

Why does xp_n +/- p_{n-1} always have coefficients in {-1,0,1} for some choice of plus/minus? Is there a reason, or does it just happen to work for the given bounds?

That's tough one to prove strictly. You can check this mathoverflow question.

Spoiler: The answer is by Terry Tao.

Talk about bringing in the big guns...

Is this something you noticed from playing around with polynomials, or did it come up in research or something?

I was thinking on what's the worst case for Euclid's algorithm and apparently found out that this sequence seems to be infinite and has some peculiar properties (like you can choose + or - only at powers of 2, others are determined by that). Surely such approach shouldn't be model solution since it's hard to prove theoretically but there's modular solution which is just fine for this purpose.

I'm not doing any serious research now :)

I was just looking at this year's Putnam problems. http://kskedlaya.org/putnam-archive/2017.pdf

Look at A2... curious isn't it? This sequence appears everywhere it seems.

How is it same?

I guess you have to solve the problem to see that. :D

So was it supposed to just believe in this fact for the model solution? It was hard even to understand, don't think much people could prove this during the contest.

It was supposed to either guess the fact or to use sum modulo 2 which is easy to prove. I doubt that's hard to understand. You want exactly

nsteps so on each step you want to have . Obviously, you would like to construct such sequences and that they're answers fornand first step lead you from to . ThusBut this is the same as . And it's not about believe because it can be checked in no time.

Sorry, I somehow haven't noticed second solution in the editorial. Yes, I agree that it's easy to guess and to prove (ability to use - 1 misleads a bit though)

Could you tell me which knowledges are required to understand div2 D's solution?

To come up with the solution like in Div2D / Div1B, I would rather thinking like this:

We already know that it takes two steps to find (

x^{2}+ 1,x).Let's say we need to find the pair that takes three steps. If we call it (

A,B), such that the polynomial divisionA/Bwill have the divisor equal tox^{2}+ 1 and the remainder equal tox. By that way, we will find (A,B) by continuing finding (x^{2}+ 1,x), which takes two steps. Therefore, it takes three step to find (A,B). Now, as I stated, the divisor equal tox^{2}+ 1, thenB=x^{2}+ 1. The remaining task is to findAwhich has the form ofA= (x^{2}+ 1)Q+x.We can freely choose

Qto obtainA. However, to satisfy the restriction that coefficents must equal to - 1, 0, or 1,Qshould bexor -x. This solution is pretty close to what is stated in the tutorial. (Mine isp_{n + 1}= ±xp_{n}+p_{n - 1}. You can easily figure out that if the tutorial's solution satisfies the restriction then so does mine)Proof by Terry Tao

After we have

AandB, we continue applying this algorithm with four steps, five steps, ..., and finallynsteps.Using the characteristic of two consecutive Fibonacci numbers to propose a solution of two consecutive Fibonacci polynomials is, though correct, a little vague (!)

How can you decide when to choose -x and when to choose x?

Just try to assign

Q=xandQ= -x. Select the case that brings up a polynomialAthat satisfies the coefficents contraint. If both cases satisfy, choose either one of them.Is it easy to figure out why always either to assign Q=x,Q=-x will lead to a polynomial that satisfies the coefficients constant?

Not really. It’s pretty difficult. Did you check the proof by Terence Tao?

yep, I agree is too difficult. I finally decided to stick with the version of the problem with mod 2. I have heard that it is easier to prove yet I couldn't. Do you know how why the algorithm using mod 2 works?

Thanks a lot, this comment made it seem very simple! (Though, I would be scared to use it because I can't really understand the proof of why it always works.)

but does your method (pn + 1 = ± xpn + pn - 1) guarantee that the leading coefficient will always be 1 (not -1) ??

Unly nubz nid tutorial becoz they hev alwayz been spoonfed by their fat mommas

In div2B why dfs gives the minimum answer.when to apply dfs or bfs i am weak at it plzz help

Okay, I will try to explain. You need to understand that if there is a vertex A and a vertex B, and A is parent of B, you

ALWAYSneed to paint A first. Why? Because if we paint B, then A, the color of B becomes thesameas the color of A (don't forget that A is parent B). So you will need to paint it again, which is not optimal. That's why we need to start painting from the root of tree.Simple solution for 2B:

AC solution

HASHING TREES:why the output of 2 sample case is

Div1B can be solved with backtracking — http://codeforces.com/contest/901/submission/33452328

nicely done but in two loops everyhthing can be done easily bro can u plzz explain me how to solve C i am not getting the editorial there can be easy way to solve it ?

Can someone elaborate on the idea behind div2/E? I still don't understand after reading the editorial.

I too had some trouble understanding it. Convince yourself that there could be no intersecting cycles. You can prove this by contradiction and by considering the parity of length of intersection of the cycle. Now, for every cycle, find the maximum node index in it and the minimum node index in it. Then a segment is good only if it does not contain both the minimum and maximum indices of any cycle in it. So, basically, you have some intervals and now in every query, you are asked to find the number of segments which do not completely cover any of these given intervals.

For reference, I used this code: http://codeforces.com/contest/901/submission/33440062

My solution:

First of all, let's try to reformulate the problem statement using this observation: A bipartite graph is a graph with no odd cycles (Proof). Since the given graph has no even cycles by definition, the only cycles it (and its subgraphs) will have must be odd. Therefore, a subsegment of this graph is bipartite iff it is acyclic. A subsegment of the graph will only be acyclic if it is either a tree, or a forest. Therefore, the problem can be reduced to this: "Find the number of subsegments [

x,y] (l≤x≤y≤r) that are forests or trees.".The second observation is this: If we only take vertex

xfrom the graph and incrementally add verticesx+ 1,x+ 2,x+ 3 and so on, at some point the graph will become cyclic, and from that point on it can never be acyclic again. Let's call that pointm_{x}(that is, after the addition of vertexm_{x}, the subsegment [x,m_{x}] becomes cyclic). Note thatm_{1}≤m_{2}≤m_{3}≤ ... ≤m_{n}. Therefore, we can use a two pointer approach to calculatem(I used the solution to this problem to remove vertices, see my code: 33436708). The rest of the solution is identical to the one described in the editorial.Thanks a lot for your answer.

Can you please elaborate a bit more on how you are computing the m

_{x}? I referred to your code but could not quite get it. A link to some resource your code follows would also be fine.All editorials should be written with this level of precision.

Can you explain your implementation?

yep! wait some hours I will make you an explanation

That would be really kind. Thank you so much.

1] We read the tree, CONN vector array stores the connections

2] We run a

dfsto search for cycles. The trick is to maintain updated in arrayvisthe deep level of each node that is part of the current track the dfs algorithm analyzes. When we collide with a node member of the track we interpret there is a cycle and withvis[]array we re-build the cycle and detect the min-max values on it (that is what we will need later). We give each cycle a code We add tostart[]andfinish[]indexed by the min-max value the code of the cycle. (I will explain in 3 why we need them)3] We start a two-pointers processing to compute

m_{i}(that is defined in the editorial as the lower index such that [i,m_{i}] is cyclic) Note that in this case, likef2lk6wf90dsays, it is true that a sub-graph iscyclic<=>not bipartiteAlso, as f2lk6wf90d says, it can be proved that

m_{i}< =m_{i + 1}1 < =i< =N- 1 (for example by contradiction)This is the reason now we can use or two-pointers technique or iterate N binary searches. I decided to use two pointers method in the following way:

Use pointers

i,jmaintaining this invariant [i,j] is cyclic. Iterate overiand updatejpointer in each iteration by incrementing it until the sub-graph with nodes of codes [i,j] becomes cyclic thus by definitionm_{i}=j. To be able to check if the subgraph is cyclic on each moment we need to make use ofstart[] vector array,finish[] vector array and min-max values of cycles stored, that gives us the information needed to update if the subgraph [i,j] is or is not a cycle (the interval [i,j] is member of cyclekif and only ifi< =min_{k}< =max_{k}< =j).Each time when we increment

jwe check if a cyclekends in thatjvalue and have already started (min_{k}> =i) (using start,finish vector arrays and min-max cycles data). When this happens we know the graph is cycle.Also when we increment

iwe make sure we can check which open cycles stop from existing, usingstart[] vector array. So I explained almost all details, I assure you that by this process you computem_{i}4] Now we have

m_{i}computed 1 < =i< =Nand this is the hardest part. We precomputepre_{i}to make prefix sums.pre_{0}= 0pre_{i}=pre_{i - 1}+m_{i}1 < =i< =Nnow 1 < =

i< =j< =NWith this information, there is a way to answer the queries in

O(log(n)) each one. with tricks obviously.The query is an interval [

l,r] we need to check for everyx,y,l< =x< =y< =rif [x,y] is not cyclic (thus bipartite). We iterate overxand get the amounts ofyfor eachxsuch that [x,y] is not cyclicTo develop the formula we need to analyze two cases:

if for an

xm_{x}< =rthen there are (m_{x}- 1) -x+ 1 =m_{x}-xvalues ofysuch that [x,y] is not cyclic.if for an

xm_{x}>rthen we've gotr-x+ 1 values ofy.So for each

xwe need to compute indexusuch thatuis the lower value such thatm_{u}>rWe knowmis increasing the binary search makes this possible. (We could also use two pointers) And now the answer to the query should be+ This should be computed on

O(R-L).But by working algebraically we can show it is equivalent to

— (

r(r+ 1) -l(l- 1)) / 2 + (r+ 1)(r-u+ 1)Using pre-array we can compute the sum so this is

O(log(n)) per query that is more than enough to solve the problem.I have written lot of details. Tell me if I made some mistake. It was long and this could be a possibility.

Thank you very much! Your explanation and your submission were really helpful

In problem Div 1C / Div 2E, can someone please explain, how are we finding the value of mx, in detail. Details in the editorial are not clear to me.

Awesome dude! That's what we need!

In Editorial of Div 1.C, this sentence

To answer the query, we need to take the sum over mxi - i + 1 for those who have

mxi ≥ rand the sum over r - i + 1 for those who have mxi ≥ rI think it should be

To answer the query, we need to take the sum over mxi - i + 1 for those who have

mxi < rand the sum over r - i + 1 for those who have mxi ≥ r`then the vertex segment is good - if there is no loop`

What is 'loop'?

then the vertex range (l<=r) is good — if there is no cycle

Can you explain your implementation?

In DIV 1D's description，what does it mean "It is guaranteed that the given graph is connected and does not contain loops and multiple edges." especially 'loop' here, isn't it crashed with the sample?

'loop' which you mentioned is an edge that connect a vertex to itself. But 'loop' in the editorial is like circle.

When they say loop here, they actually mean cycle.

Div 1 B. Can anyone prove that mod 2 of the coefficients of series

`p(n+1) = x*p(n) + p(n-1)`

will give the correct answer? The given explanation isn't very clear. :/I am interested in seeing answer to this question.

problem A(div 2) says It is guaranteed that ai ≥ ai - 1 for every i (2 ≤ i ≤ n). but my submission failed at test 17 which was: 1 10 0 10 so is 0 greater than 1 ? link :http://codeforces.com/submissions/tibialAcorn98

1 10 0 10.

1 — the number of teleports;

10 — the point we want to reach;

0 — a[1]; 10 — b[1].

thnx i got my mistake .

I can not understand the isomorphic statement . i got WA on test 2 .

How these two trees are isomorphic ?This was a confusion I had during contest.If u read the question carefully,it is written that if u

REENUMERATEthe vertices in some way, hereREENUMERATEmeans that assigning the vertices in some order to make the trees equal.Now if u see ur output, if in the 2nd tree, if I interchange vertices

2and3, then it becomes exactly equal to the 1st tree, which shows that they are both isomorphic.In question U have to output two

non-isomorphictrees instead.Hope this helps :)

if I interchange vertices 2 and 3, then how it becomes exactly equal ? In my output 1st tree -> node 4, 5 connected with node 3and In 2nd tree node 4,5 connected with node 2 then how it becomes equal ?By interchanging or reenumerating the vertices I meant that

renamevertex 2 as vertex 3 AND vertex 3 as vertex 2.See imageTHANKS :)

But if we have 1child for root2 and root3 for first tree,i build second my tree as picture shows .Here,how two trees are equal?

How to Solve This Problems help me plz.

Can someone please some implementation details about Div 2E? It would be really helpful.

My thinking of why the mod approach works for div2 D:

consider always making A = B*x + C where degree of C is less than degree of B, the problem we will face is that sometimes the result will have coefficients greater than 1, but the leading term coefficient will never be greater than 1, because degree of B*x is higher than degree of C, so the leading term will always be existing once in the expression B*x+C (and so will not be affected by the mod).

now think of A%2, this is = (B*x + C)%2, and similarly B%2 = (C*x + D)%2, and C%2 = (D*x + E)%2, and so on till the rightmost term (E in the last expression) is 0. before we reach this point, all expressions A, B, C, ...etc will always have the leading element coefficient = 1 after and before the mod. so with the mod, we took the same number of steps we would take without the mod.