Task A. Div2.

It's easy to see that number *x* can appear in column *i* only once — in row *x* / *i*. For every column *i*, let's check that *x* divides *i* and *x* / *i* ≤ *n*. If all requirements are met, we'll update the answer.

The complexity is *O*(*n*)

Task B. Div2.

Let's consider two cases: *n* > *m* and *n* ≤ *m*.

If *n* > *m*, let's look at prefix sums. By pigeonhole principle, there are two equals sums modulo *m*. Assume *S*_{lmodm} = *S*_{rmodm}. Then the sum on segment [*l* + 1, *r*] equals zero modulo *m*, that means the answer is definitely "YES".

If *n* ≤ *m*, we'll solve this task using dynamic programming in *O*(*m*^{2}) time. Assume *can*[*i*][*r*] means if we can achieve the sum equal to *r* modulo *m* using only first *i* - 1 items. The updates in this dynamic programming are obvious: we either take number *a*_{i} and go to the state *can*[*i* + 1][(*r* + *a*_{i}) *mod* *m*] or not, then we'll get to the state *can*[*i* + 1][*r*].

The complexity is *O*(*m*^{2}).

Task A. Div1.

If Petya didn't ask *p*^{k}, where *p* is prime and *k* ≥ 1, he would not be able to distinguish *p*^{k - 1} and *p*^{k}.

That means, he should ask all the numbers *p*^{k}. It's easy to prove that this sequence actually guesses all the numbers from 1 to *n*

The complexity is *O*(*N*^{1.5}) or *O*(*NloglogN*) depending on primality test.

Task B. Div1.

Let's look at the answer. It's easy to notice, that centers of that tree must turn into centers after applying the permutation. That means, permutation must have cycle with length 1 or 2 since there're at most two centers.

If permutation has cycle with length 1, we can connect all the other vertices to it.

For example, let's look at the permutation (4, 2, 1, 3). 2 is a cycle with length 2, so let's connect all the other vertices to it. The resulting tree edges would be (1, 2), (2, 3), (2, 4).

If answer has two centers, let's remove the edge between them. The tree will split into two connected parts. It's easy to see that they will turn into each other after applying permutation. That means, all cycles should be even.

It's easy to come up with answer with these restrictions. Let's connect vertices from the cycles with length 2. Then, let's connect vertices with odd position in cycles to first of these and vetices with even cycles to second one.

For example, let's consider permutation (6, 5, 4, 3, 1, 2). There are two cycles: (3, 4) и (1, 6, 2, 5). We add edge (3, 4), all other vertices we connect to these two, obtaining edges (1, 3), (6, 4), (2, 3), (5, 4).

The complexity is *O*(*N*).

Task C. Div1.

Let's split rectangle 10^{6} × 10^{6} by vertical lines into 1000 rectangles 10^{3} × 10^{6}. Let's number them from left to right. We're going to pass through points rectangle by rectangle. Inside the rectangle we're going to pass the points in increasing order of *y*-coordinate if the number of rectangle is even and in decreasing if it's odd.

Let's calculate the maximum length of such a way. The coordinates are independent. By *y*-coordinate we're passing 1000 rectangles from 0 to 10^{6}, 10^{9} in total. By *x*-coordinate we're spending 1000 to get to the next point of current rectangle and 2000 to get to next rectangle. That means, 2 * 10^{9} + 2000000 in total, which perfectly fits.

The complexity is *O*(*n* * *log*(*n*))

Task D. Div1.

Let's optimize the first solution that comes to mind: *O*(*m* * *d*_{max}), let's calculate *can*[*t*][*v*] — can we get to the vertice *v*, while passing exactly *t* edges.

Now, it's easy to find out that the set of edges we are able to go through changed only *m* times. Let's sort these edges in increasing order of *d*_{i}, that means for each *i* *d*_{i} ≤ *d*_{i + 1}. Let's calculate *can*[*t*][*v*] only for *t* = *d*_{i}. We can calculate *can*[*d*_{i + 1}] using *can*[*d*_{i}] by raising the adjacency matrix to the *d*_{i + 1} - *d*_{i} power and applying it to *can*[*d*_{i}].

Next step is to fix an edge with maximal *d*_{i} on our shortest path, let it be *i*. We know all the vertices we can be at moment *d*_{i}, so we need to calculate the shortest path to *n* - 1 using edges we can go through. We can even use Floyd algorithm to calculate that.

The complexity of this solution is *O*(*m* * *n*^{3} * *log*(*d*_{max})) and it's not enough.

Next observation is that adjacency matrix contains only zeroes or ones, so we can multiply these matrixes using bitsets in *O*(*n*^{3} / 32).

This makes complexity *O*(*m* * *n*^{3} * *log*(*d*_{max}) / 32), which gets accepted.

Task E. Div1.

Let's solve an easier task first: independent of bipartivity, the color of edge changes.

Then we could write a solution, which is pretty similar to solution of Dynamic Connectivity Offline task in *O*(*nlog*^{2n}).

Let's consider only cases, where edges are not being deleted from the color graph. Then, we could use DSU with storing parity of the way to the parent along with parent. Now we can find parity of the path to the root by jumping through parents and xoring the parities. Also, we can connect two components by accurately calculating the parity of the path from one root to another.

Now edges are being deleted. For each edge and each color we know the segments of requests, when this edge will be in the graph of specified color. Let's build a segment tree on requests, in each vertex of the tree we store list of edges which exist on the subsegment. Every segment will be split into *log* parts, so, totally there would be *n* * *log* small parts.

Now we can dfs this segment tree with DSU. We get inside the vertex, apply all the requests inside it, go through the children and revert DSU to initial state. We also answer requests in leafs of the segment tree.

Let's return to initial task. We can't use this technique, because we don't know the exact segments of edge existence.

Instead, let's do following. Initially we add each edge right until the first appearance of this edge in requests. Now, when we're in some leaf, we found out which color this edge would be right until the next appearance of this edge. So, let's update this edge on that segment.

For each leaf we're going to make an update at most once, so the complexity is *O*(*nlog*^{2}*n*).

If anything is unclear or bad-written, ask any questions.

thanks for fast Editorial ;))

in

problem B Div2, Modulo sum,for n > m, can someone elaborate on how pigeonhole principle implies there exists 2 sums with same value in prefix sums.

Also can't we use the DP for n > m as well?

UPD: Understood and got AC. Here's my AC solution according to editorial above.12948929

We can't use DP if n > m since it will be performance for O(n^2) which really slow.

We have a1, a2, ... an. Make array Si = sum ak from 1 to i (S1 = a1; S2 = a1 + a2; ... ).

If n > m, there are Si and Sj which Si % m = Sj % m and j > i. So (Sj — Si) % m = 0. Since Sj — Si = a1+a2+...+Aj — (a1+a2+...+ai) = a(i+1) + ... + aj. So we've got answer.

it will be O(n*m) not O(n^2), but still slow

but if you teminate as soon as you find there is a solution, the complexity would be O(m^2) due to the conclusion mentioned above.

I solved this by Dynamic programming:

First sort all the values in non-decreasing order then apply dynamic programming on first 10^4 values only, and this is should work :D

And why only first 10^4 ?

Ok, there is no proof for this, but when you're inside a contest and you will not lose any thing if you send a wrong answer, then try different ways to solve.

10^4 * 10^3 for the state of dynamic programming, so I reduced the time from (10^6 * 10^3 to 10^4 * 10^3) but I sorted the values! and it worked.

Can we not use the prefix sum solution for "n <= m"?

No because the pigeonhole principle can't be applied there and hence there is no guarantee that there exists two sums Si, Sj such that Si % m = Sj % m

Could you show me a case when n == m the answer is 'no'? I understood for n > m, but why not n >= m?

Actually it can be applied here. SUM%m only has m states and 0 is a valid state so we only have m-1 invalid states. Thus since each added number can reach at least one new state its impossible to find m numbers which can't generate a multiple of m.

Also the only cases where m-1 numbers can't generate a multiple of m is when we have m-1 copies of a number num relatively prime to m. We can see that this is true since each new number will generate at least two new states if it is different from all of the past numbers and if num is not relatively prime to m then it will reach a multiple of m in less than m steps.

I get it now. Thanks for the explanation!

Thanks for the reply balalaika

I am still unable to prove using pigeonhole principle that there will definitely exist two sums Si, Sj such that Si % m = Sj % m.

It would be very kind of you if you could help me with that. :)

The author is talking about prefix sum modulo m. So this value must be less than m for all prefix sums. Now there are only m such distinct values possible from 0 to m-1. So if we find modulo for n prefix sums, at least two values must be same.

Thanks sahil23. I got it. :)

Thanks, that was helpful!

Thanks for explanation !

Thanks for the great explanation.

thanks a lot, great help.

This explanation is understandable for everyone.

thanks...

best explanation

sorry but i didn't understood that part So (Sj — Si) % m = 0. Since Sj — Si = a1+a2+...+Aj — (a1+a2+...+ai) = a(i+1) + ... + aj. So we've got answer.

psum[i] % m = c (c is constant) .......(1)

psum[j] % m = c .......................(2)

subtract eq(1) by eq(2)

(psum[j] — psum[i]) % m = c — c

(psum[j] — psum[i]) % m = 0

this is what we want.

But WHY there is two equal sums modulo M ?

UPD:nicely explainedgot it.Thanku balalaika

I did use DP for

`n > m`

actually.With knapsack optimization, you can achieve something like O(

M^{2}logN) with pretty small constants.Submission

I understand your ans is always yes when n > m, pls explain.

`if(n > m) //pigeonhole principle { //answer is always yes cout << "YES\n"; return 0; }`

for n > m, we can calculate prefix sum array, Si = (a0 + a1 + .. + ai ) % m

and always find Si and Sj such that Si % m = Sj % m and j > i.

Since this is always true, there is no need to do all this calculation, and we can directly ouptut YES for n > m.

Why this is always true? Simple: Pigeonhole Principle :)

Very well explained.

Of course, we can use dp in

n>mcase.All we have to do is, turn

ainto [a]_{m}，than count each [a_{i}]_{m}into slots, eg.n= 5m= 5a: [1, 2, 6, 7, 3]turn into

a]_{m}: [1, 2, 1, 2, 3]and put number into its slot

cnt: [0, 2, 2, 1, 0]idx: [0, 1, 2, 3, 4]We had reduced the prolem into classic 0/1 Knapsack problem with each item's amount greater than 1.

Here is the code, but it would be a little slower:

Actually, we could solve this use Monotone priority queue. in:

O(m^{2})see if n>m then when we will process the array a then at every prefix the value of pref[i]%m would be there and there can be at max m distinct values and since n>m there would exist at least (n-m) duplicate values hence the solution would work because if pref[i]%m==x and pref[j]%m==x and j>i; then (pref[j]-pref[i-1])%m=(pref[j]%m-pref[i]%m)==(x-x)==0 hence there would be a subsegment with sum%m==0

Could someone explain to me Div 1 Problem C? How does the estimate of the length work? I think the worst case sum for the Y direction is 10^6 * 1000 = 10^9. What does "we are spending 1000 to get to the next point of current rectangle and 2000 to get to next rectangle" mean?

Within one rectangle, to jump from one point to other you spend 1000 for X direction but when you jump from one rectangle to another you can spend 2000 as in worst case the points can be 2000 units apart in X direction. So a rough estimate for X direction will be 10^3 * 10^6 + 2000 * 10^3 = 10^9 + 2000000

In total = Ydir + Xdir = 2*10^9 + 2000000

Ahhhh...I used Manhattan minimum spanning tree for solving problem E in Div2, which complexity is O(nlogn) with a big constant. It ran for nearly 2s but I still got AC :)

cool! can you suggest a good article or clip about it ? tanx.

here it is, https://www.topcoder.com/community/data-science/data-science-tutorials/line-sweep-algorithms/

see the last topic

A 2-approximation algorithm for TSP problem when the cost function satisfies the triangle inequality has been mentioned in

Introduction to Algorithms. First of all, take a vertex from the graph to be a "root". Then find the MST of the graph and process a DFS from the "root" on the MST. The order you visit all the vertices on the graph will be the same as on the MST. So we just need to get the Manhattan MST using the algorithm mentioned above.Why did this 2 solutions get TLE in Task C. Div#1 ?!

Sorting vector

Priority Queue

use scanf

My solution for Div.1 C :

Let's call partition(l , r , isVertical , isDescending) recursively and alternate isVertical between 0 and 1. On each call we sort the points from l to r based on isVertical and isDescending, and after that if l < r , we'll call:

partition(l , (l + r)/2 , 1 — isVertical , 0)

partition((l + r)/2 , r , 1 — isVertical , 1)

the first part is sorted in non-decreasing order and the second one in non-increasing order.

Overal complexity : O(nlg^2n)

For problem D there is an Solution

Edit: Actually its

Awesome! Could you describe it in few words?

Sort edges in increasing order of

d_{i}and add them one by one. After adding each edge i want to know the set of all vertices that we can reach them with using exactlyd_{i}edges. If we know this set of vertices after adding each edge, we can compute our answer by using BFS every time. Now consider we know a set of verticesSwhich we can reach them usingd_{i - 1}edges and now we want to find all vertices which we can reach them using exactlyd_{i}edges. Letdp[lvl][i][j] show us is there a path from i to j with exactly 2^{lvl}edges?If we have this dp we can compute our next set of vertices from set

Swith an algorithm (I'm not going into details). We do thismtimes so we get an complexity. The only problem remains is that how we can updatedptable after adding one edge. If we add an edge fromutovwe should setdp[0][u][v] to true and change some other values on upper levels ofdpstates. For this we can simply go through higher level and updatedpbut we stop when we reach some state indpthat was true before (See mydfsfunction to see what i exactly do). By doing this, each state ofdpgetting true value at most once and for each time we change something from false to true we do anO(n) iteration. The number of different states is so we get an complexity by amortized analyse.Unless I am mistaken, there is a solution O(n^3*log(dmax)).

I always feel very very happy to have a very very high spped tutorial...

thanks for the editorial....thanks for a good contest....

Can somebody tell why does pow(5,2) returns 24 in problem C Div 2. Here is my solution http://codeforces.com/contest/577/submission/129368833

use ceil(pow(5,2)).

pow(5, 2) doesn't return 24, but it may return something like 24.99999999999

use powl and be safe . it works like pow but without the pains of pow.pow has led me down quite sometimes now :|

But in ideone and other online judges it does return 25, why not in codeforces?

Why you don't use this : (int)(pow(5,2) + 0.5) :D

Edit : the solution had some mistakes , however it passed as a complete search !! :O :'D AC CODE : 12947069

What does "cur" means in your solution ?

May be I got it.idx is the sum,cur is the coin in coin change.

Yes cur means what coin I am adding right now .

I could say your solution is definitely a complete search solution you don't use

`arr`

at all maybe it passed because of weak test data here is your code without`arr`

.for div2 problem B

i have solve it using BFS

first of all i take mod m for all in input

and for every value i count how many times it occurs in array

and iterate over all unique values in BFS

state : vis[mod%m][indx of unique value][take]

take mean that i have take a number or not until now

my code : 12947179

That's an elegant solution.

I also thought of similar solution using BFS during the contest but could not come to the conclusion that the total number of states to be visited will fit within time limit.

The total number of states in your solution is 2*1e6. Right? Please correct me if I am wrong.

Yes ,you are right

in the while loop of BFS, you have used the breaking condition.

Why have you included take there? Shouldn't it be just

nope :)

in the first i push to the queue value of mod = 0 ,and this doesn't mean that there is a solution

I see that but doesn't that mean that the number associated with the last state visited(when returning true) should be compulsorily included in the subset?

Nope because the loop inside BFS start from 0 and that mean i didn't take that value

It seems to me that every Div2 B we can solve with BFS. It is very often that exits solution with simple idea, but there are always people who solves it with BFS.

What if they swapped it with C ? As you see problem c is easier than it!

please can you explain what cal() function do

its the BFS function !

tell me what is the thing that you didn't understand and i will explain it :)

these two lines can you explain them

if(indx==int(a.size()) ) continue;

int nxt=(mod + 111*a[indx] *i) m;

a is a vector of unique mod values

when indx == a.size() that mean no any value after that

int(a.size()) : this is for casting unsigned int to signed int

1ll = long long(1) ,(its LL in small case not 11)

i = how many times i will take this value

a[indx] = the value

nxt = next mod after take value[indx] i times

there is o(n) solutoin for div1-c 12947556

I don't sure but maybe your solution is wrong. Let there will be 1000 points in square 1000x1000: 500 in left top corner, 500 in right bottom corner. If two neighbours in path are from different corners distance between them will be from 1500 to 2000. We can fill 1000 squares: a[0][0], a[999][1], a[0][2], a[999][3],.. Distance between two squares is about 1e6. 1e6 * 1000 squares = 1e9. Sum of distances between neighbours in same squares > 1500 * 1000 * 1000 or > 1.5e9. So sum of all distances between neighbours in path > 1e9 + 1.5e9 or > 2.5e9. It's bad. Sorry for bad English.

Actually it's good solution, here is my submission: http://codeforces.com/contest/576/submission/12947556, it is more clear imho. First you clasterize the plane in sqares 1000x1000 and put point respectively in squares that they belong to.

The idea is that you traverse the clasters in a way that there is no distance more than 2000 between any 2 points(1 from first claster,1 from another). Also you can print the points from the same claster in any order because the max distance between any 2 points in a claster is <2500. So if you keep the max distance from any 2 points from the same claster less than 2500 and the distance between any 2 points(different clasters) less than 2500 then the result can not be more than 2500*10^6 or 25*10^8

When you go from one cluster to another, the distance can be up to 3000, so your proof doesn't work.

Yes I found the same case right now. Does the idea work with 800x800 squares?

That realy nice problems. Yes, some hard Bdiv2, but its nice CF round, realy new problem from author.

Your idea for Div 2 B was great. Thanks for the awesome problem set.

My doubt is, while doing square-root decomposition or similar like in problem DIV2 — E, what is a good interval size to take in similar situation usually, 1.1*root(N) behaves better than root(N). Is 1.2 root(N) better than 1.1*root(N) usually?

DIV2E gave ac if I used interval size = 1100 and no internal sorting whereas WA if I used 1000 as the Interval size.

I have a feeling that any space-filling curve should work well enough for Div. 1 C. For example, Z-order curve suffices to pass the system tests.

Can anybody squeeze out Div2 E in Python? The best I could get (12948859) runs for about 12.6 seconds on my machine under CPython. Which is weird, since it is essentially parsing + a call to

`sort`

.Can anyone explain Div 1 Problem B in more detail? Why the permutation must have cycle with length 1 or 2?

@malcolm Can you please add editorials to other Div 2 problems?

There is another explanation for div2 B that I couldn't prove, every number we add before we reach m elements there exist some combination that will produce new mod which it didn't appear before thus if

n≥mthere exist a solution. Correct me if I'm wrong.In

Problem A Div.1, If n is not divisible by p^k - 1, do we still need to check with p^k?For instance, if n is 15, shouldn't the output be

7

2 3 9 5 7 11 13

and not

9

2 4 8 3 9 5 7 11 13 ?

If n is divisible by p^k-1, it doesnt mean that x is.

Could someone explain why checking with the powers of prime is sufficient in Div 2 Problem C ?

If we check with each power of prime <= n, we get the power of that prime in x. If we do this for all primes, we get x.

~~Problem B Div 1 : Can anyone explain what he means by centers remain unchanged and what he means by applying permutation on the tree?~~Div#2, Task C:

Input

15

Participant's output

8 2 3 4 5 7 9 11 13

Jury's answer

9 2 4 8 3 9 5 7 11 13

Checker comment

wrong answer cannot distinguish 4 and 8

------------------------------My comment------------------------------ .....Please clarify this ....why we have "8" in expected answer

1) If you ask for 2 and 4 then why 8 ? 2) IF 2 -> NO and 4 -> NO then 8 -> NO IF 2 -> YES and 4 -> NO then 8 -> NO IF 2 -> YES and 4 -> YES then 8 -> YES

so if you can deduct if number is divisible by 9 from asking question "number divisible by 2 & 4"...why "8" is in expected anser

correction in last sentence so if you can deduct if number is divisible by

8from asking question "number divisible by 2 & 4"...why "8" is in expected anser4 is divisible by 2,4 but not by 8

Task C. Div1.How do we decide that we have to divide into 1000 rectangles ?

sqrt(N)

Can you give mathematical/intuitive logic behind taking sqrt(N) ?

Let's split the rows into K buckets so each bucket contains N/K columns . For each bucket sort the numbers by their column number . With processing this order,the total cost coming from the difference between adjacent points' column numbers can be maximum N , so we got a K*N cost. And notice that the difference between adjacent point row numbers' can be maximum N/K, we got an extra N*N/K cost. Total cost is (N*N/K+K*N) to minimize this number we should choose the K as sqrt(N). Note that you should visit first bucket by normal order but the second by reverse order ... I hope it helps :D

I think this contest is inclined to math theory,i dislike :(

I am poor in math , That's exactly why i love this type contests so much for improve myself.

same here

In Div 1 B => It's easy to see that they will turn into each other after applying permutation. That means, all cycles should be even . Can somebody explain this statement ??

Please help with my solution here which timed out

http://codeforces.com/contest/577/submission/12937540

I am doing same thing as (i.e. O(n)) editorial. Moreover by changing few things I optimized it to avoid "/" division by "++" and still my solution go timed out

Please I am not doubting anybody. As per my (biased)knowledge by replacing division operator by ++ in loop, efficiency should have been improved...What am I missing ?...need to learn

Your solution is O(x) while the editorial is O(n)

In your solution it is enough to iterate till sqrt(x) and if i*i=x add 1, else add 2 (ofcourse you should check that both numbers are <=n)

Why this solution 12955987 get WA while this one 12955979 got AC?!

I think the first should get AC too!!!

in the first code idx can reach up to n which might be 10^6, while your dp array is only 1010*1010

how could idx reach up to 10^6 while i use this condition in the function

and i only call the function if the n < m other wise i cout yes without call the function at all.

Sorry I did not notice that, however your dp state should have a 3rd dimension for

cntAC Code 12959591

yes that's right. thank you :D

In Problem C Div2, A Div1 Why we just have to ask about prim numbers and its powers while <=n what is the idea here why its enough to guess ?!

because any number n can be written as:

n=a_{1}^{p1}*a_{2}^{p2}* ... *a_{m}^{pk}Where

a_{i}is primewhy do we need to ask for all powers of a prime like if n = 15 and we divide it with 2 and if the reply is no then we are sure that no powers of this number is going to divide n , but we are asking for 4 and 8 too . why ?

Because you only get a reply after asking all questions, so you can't know that it is not divisible by 2.

Thanks I get it now .

I am trying to Submit DIV2E==DIV1C on Java, but I am getting TL on 6th test, despite the fact that the same solution on C++ got Accepted. I know that there is a problem that Java is pretty slow, but still it seems like my solution does not differs from author's (nlogn) [as far as I know Array.sort works for NlogN, isn't it?]. So is there anything I can do to speed up my solution? Thank you!

The problem is not with the algorithm, but with how you are reading input. Scanner is really slow for reading input. I would suggest using BufferedReader or some fast form of reading input for this problem.

I now used the method, used in fastest solution

But still have the same problem...

Output took a lot of time too. Reading and sorting took together 888 ms. All other time took output. Thank you: AC

Div-1 B / Div-2 D: Can someone explain why there can be at most two centers ?

[and what actually is meant by center]

Is it because the tree will remain disconnected in the other case ?

Centered Tree

thanks ... I now got the underlined graph theory

Can you please explain this => It's easy to see that they will turn into each other after applying permutation. That means, all cycles should be even.

Not necessarily. There could be just one center, in which case there is a fixed point of the permutation.

I got AC in div1 D without any bitwise operation but using the algorithm whose time complexity is O(N^3 log max(d_i)).

here is my code: http://codeforces.com/contest/576/submission/12935207

In this code, I used the following fact: The number of edges don't decrease. We can do doubling efficiently by seeing only the places which are added by each turn.

can someone help me estimating the complexity of this code for Div2-B problem ?

In problem Div 1 A, I was wondering... how about we do binary search for every prime? For example, to find out the biggest power of 2 that divides

xforn= 1000, we need 3 questions, not 10. Where is my reasoning wrong?EDIT: Ignore this question. I hadn't read that he must first ask all the questions before receiving any answer.`It's easy to notice, that centers of that tree must turn into centers after applying the permutation.`

Why must this be true?

Because if we take the tree and remove all the labels then it must look the same before and after the operation (since all edges are preserved), so all properties of each node that doesn't rely on labels must stay the same. This means that the center nodes must stay the same because you can find them without relying on labels.

how would you find the center without relying on labels? I still don't get this affirmation.

ps: sorry for the necroposting

In task A , DIV 1 ... can anyone please explain to me that for a number X , if X is not divisible by a prime P , then why are we still checking for higher powers of P . For eg if X = 103 , clearly X is not divisible by 3 , but why are we then asking whether X is divisible by 9 or not

Well, you don't know a particular number, you just know that number that you have to guess is less than or equal to N. So, lets say it is 10. Hidden number can be 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. So lets say Vasya though of number 9. So, if you say 3, the answer is yes, Vasya's number is divisible by 3. But you can't be sure that it is 3 or 9, so you have to ask whether the number is also divisible by 9. That is why you have to check all powers of all the primes.

Can anyone explain why "permutation must have cycle with length 1 or 2 since there're at most two centers. " in Div1B?

Since every center can only connect with node among the centers. Example: when we choose 2, 3, 1 as the centers, connecting (2, 3) will lead to the connection of (3, 1), and then lead to (1, 2), which form a loop.

WTF is 3 centers?

We cannot choose permutation with length greater than 2 as centers, but there can be many centers, consist of those with length 1 or 2.

I want to notice that

Div1 D. Flights for Regular Customerstask has weak tests. I have just found my accepted solution to be incorrect. I did Dijkstra to test ifn-thvertex is reachable after I have added new edges to existing graph when I did enough flights for them.For the test case below the solution will give 3 instead of correct 2.

4 5

1 1 0

1 2 0

2 3 0

3 4 0

1 4 1

I found other accepted solutions to be wrong as well as mine. One, Two.

Can someone explain Div D (Invariance of Tree) in details?

can anybody explain this very strange submission 22281552?

I solved Div1.A with the lcm of all of pair of numbers, and got AC, but reading editorial I am not very sure about why my code got AC, can anybody explain me?

here's is my code: https://codeforces.com/contest/576/submission/62085651

Div2 B can also be solved with bitsets

Accepted Solution : 80020525

I think the editorial's explanation for Div-2 Problem D/ Div1 Problem B about the necessity of existence cycles of length $$$1$$$ or $$$2$$$ in the permutation is bullshit (some center of tree crap).

Here is the correct proof: First of all I will call permutation cycle as

p-cycle.If

p-cycle$$$C1$$$ is connected to cycle $$$C2$$$, then length of $$$C1$$$ must divide length of $$$C2$$$ (we can prove that otherwise, the vertices will form a cycle, hence not a tree).We can, then, prove that if we interconnect vertices of $$$C2$$$, then we will have a cycle (again easy to prove simple maths).

And if we interconnect vertices of $$$C1$$$, we won't have a cycle, but we will have disconnect component

UNLESSthe length of cycle is either $$$1$$$ or $$$2$$$.So there you go :)

Can anyone explain div1/A (div2/C) in more detail ?

The method for n>m tells that we have continuous segment in the array whose sum is divisible. How it still holds if there exist a subsequence which gives the answer.

Can someone also provide with proof that why pigeonhole always gives "YES"