Thanks for participating in my contest!

It was my first round, I hope you enjoyed.

**Editorial**:

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I enjoy this contest very much. I hope I will see you as a problemsetter soon!

bu cu

D can also be solved with suffix arrays + BWT (see this reference: http://blog.avadis-ngs.com/2012/04/elegant-exact-string-match-using-bwt-2/ ).

First you should know how to do it with suffix arrays, in that you can binary search for first and last entry so you can get all occurrences of a word in log(n) per word.

You can speed this up with BWT to take O(|w|) time for a word w (so total is O(sum of lengths of words) rather than O((number of words) * log(n))). The basic idea is you can maintain an interval on the suffix array which contains a suffix of the pattern, and you reduce this interval's size in constant time every time you add another letter. You can see my implementation here: http://codeforces.com/contest/963/submission/37445232

We can do it using only suffix arrays too. No need for optimization using bwt.my submission: http://codeforces.com/contest/963/submission/37425722

Didn't you have to iterate through all occurrences? Wouldn't that still be ? Or am I missing something?

I have another solution to problem D with suffix automation + dfs. This is my AC code: submission:37454639(http://codeforces.com/contest/963/submission/37454639)

For string

s, we build a suffix automaton(sam), and for each node in sam, we can also know its matching position in strings. For each query, we find the nodexin sam which represent stringm. By dfs the fail tree of sam, we can get a sequence of the positions of each node in the fail tree of nodex. We sort this sequence and traversal it to get the answer.In the worst case, the complexity of this sulotion is O(N·sqrt(N)·log(N)).

But when I solving this problem, I got a trouble. If I swap line 114 and line 115 in my AC code, I will receive a TLE, I don't know why, maybe the compiling optimize? Can anybody explain, thanks.

maybe UB. try drmemory.

Maybe O(N·sqrt(N)·log(N)) is too large. In test 47, s[]="aaa..a", so the fail tree is a chain. If Pre-order traversal, order[l...r) is increasing before sort. So maybe the compiling optimize make it faser.

You could use an O(n) linear sorting algorithm for an array whose elements are not very large.It could get rid of the "log" time complexity to construct a more fast solution.

My solution involved precalculating all of a & b. Then finding the sum takes O(n) time. This didn't pass, however I have solved n<10^9 problems with linear time before, is there a general rule as to what time is required? (I'm somewhat new to this)

Zcontains onlykelements, so you can calculate in complexityO(k)Sorry, I meant the full sum not Z(Its changed now)

can't understand "alternating sum " problem's solution . can anybody explain how it works. thanks.

and solve the last line using summation of geometric progression

in second line,second SUMMATION i think it will be=> S (in range from i=0 to i=k-1) = Si*a^(n-i-k)*b^(i+k)

you wrote just Si*a^(n-i-k)*b^(i). if i mistake please help.

my typo, sorry for that, it should be

We are basically taking advantage of the periodicity of s. So we first calculate the sum for the first k values.

()

can be seen as

()

where (

p) is ()Then can we multiply this sum by (n+1)/k i.e. number of this periodic strings to get the final result. Not quite. Why?

Because for calculating sum each subsequent periodic substring The power that you raise p will be more. But by how much?- Yes it is k.

How to get these (b/a) values. You need to learn inverse modulo if you haven't learned already.

trueglory How can I prove that the power is k ?

Why am i getting a runtime error in testcase 9?

http://codeforces.com/contest/964/submission/38454719

Because you are dividing by zero

Can you explain how you got the exponents in the last term in the second line? How you got a^(k-i-1)*b^(i+n-k+1). And also how you got the last term in the third line from that.

the general thought is to use the peroidicity

Can someone please explain the reason behind why the solution in Destruction Tree Question work ? I am not able to understand the tutorial solution.

My solution is slightly different from the tutorial, but the thoughts are the same.

It is obvious that the answer is `yes' if and only if the tree have odd numbers of nodes. Otherwise it has even nodes, and

n- 1 edges, which is odd. Each step we remove even edges, soodd-even-even- ... doesn't equal to 0, hence the edges cannot be removed completely.And then we use DFS to delete the nodes:

For a node

u, DFS(u) does:v) on any subtreevthat has a even sizeuv) on any subtreevthat has a odd sizeWhy this method is right? Proof using mathematical induction:

Base case:It is right when the tree has 1 or 3 nodes.Inductive step:Case 1: the (sub)tree has even nodes. For convenience we call it subtree u. Because it has even nodes, it must have a parent node (otherwise no solution). After removing u's `even subtrees', there remain even nodes in the tree u(

even-even=even). Leaving node u alone, there're odd nodes in the tree u. Since all the even subtrees of u was removed, the remaining subtrees have odd size.odd·odd=odd-> node u has odd numbers of odd-sized subtrees. (odd numbers of odd-sized subtrees) + parent -> the degree of u is even. So we can instantly remove u, and then recursively remove its odd-sized subtrees.Case 2: the (sub)tree has odd nodes. The proof is very similar with case 1, so omitted.

As shown above, this strategy can remove all the nodes.

Q.E.D.

Thank you!!

How can you remove u's

evensubtree?Thank you. It is helpful for me.

Thanks a lot. :)

I know a tree with even nodes can not be destructed, but how can you prove that there must exist a solution for an tree with odd nodes?

Cannot understand the solution of `Cutting rectangle'... can anyone explain the tutorial for it in detail?

Thanks in advance!

We'd like to combine the rectangles to a big one, and let's call a way to build the big rectangle A GOOD WAY. In a good way, we can move some small rectangles. For example:

It has four types of rectangles and shows a good way. We can also get the same big rectangle in this way:

It can be built by first constructing one part of them, then copy them several times. The Editorial states that every big rectangle contains the same part, like the picture above. More formally, we need to build one BASE PART, such that it contains all types of small rectangles, and the

gcdof the numbers of all types in a part is 1. So we need to divide everyc_{i}withgcd_{i = 1 to n}c_{i}(call itG), and check if base part can be built. If built, we will copy the base partGtimes to make a bigger rectangle, and the number of ways is the number of divisors of G (copyxtimes in one direction andytimes another,x*y=G). Else, the answer is 0.The last is how to check the base part. We divide the rectangles into different groups based on their

a, and discover that if a way is good, each line's ratio of number ofbis the same. For example, whena= 2, [b= 1]: [b= 2]: [b= 4] = 3: 5: 7, and whena= 3, [b= 1]: [b= 2]: [b= 4] must be 3: 5: 7. Based on this we just need to compare some ratios and check if they are the same.Thanks! a picture says more than thousands of words!

in problem c div2 i want to help how to calculate q when i saw one's solution i found him caculate it by this line long long q=poow(a,k*(MOD-2)%(MOD-1))*poow(b,k)%MOD;

but i can't understand why it calculate in this way as from eq i know that q = (b/a)^k but i can't match this line of code with eq of q can any one help me?

the code is a little weird(may contain typo) but i guess it's modulo inverse?

i got it in this way:

`% (MOD-1)`

sincei didn't know modulo inverse before so your code confusing me.as i can't understand why that's true and why i should do that could you help me to understand it by any source or detailed disscusion to that.

q = fastow(b * fastpow(a, MOD-2) % MOD, k)

that's more aceptable to me as i confused from why %Mod-1 not %MOD

all great thanks To you voidmax and panda_2134

i have another question in the same problem as i understand all above but i can't understand why i should multiply po(q-1,MD-2)%MD this term but in my answer as i understand from your equation above that the answer if q!=1 : cout<<ans*(po(q,(n+1)/k)-1)%MD; but i found answer: cout<<ans*po(q-1,MD-2)%MD*(po(q,(n+1)/k)-1)%MD; can you help me?!

https://stackoverflow.com/questions/43605542/how-to-find-modular-multiplicative-inverse-in-c

So is that problem 963E a Markov chain?

Can anyone please provide useful resources for aho-corasick as i am new to this topic ???

"Then we can replace all maximum numbers with twos and the rest we split into ones and weight will be the same."

Sorry I'm very new. Can you please explain the solution to Div2 problem A again? I understand every number n > 1 has at least two ways to be partitioned in a strict ordering ([n] and [1,1,1....1]). But what about the numbers in between? Merge a pair of 1's into 2's? Then what do we do about the pairs of 2's? Thank you for your patience.

"Then we can replace all maximum numbers with twos and the rest we split into ones and weight will be the same."

So [4, 4, 2, 1] equivalent to [2, 2, 1, 1, 1, 1, 1, 1, 1]

So we can use only ones and twos.

You replace 4's with 2's then where did the extra 1's come from? And what about odd numbers? What is base case?

Can anyone please explain why the Number of different lengths of mi is O(sqrt(M))? And How do we calculate the O(M*sqrt(M))? Many Thanks

Let M = m1 + m2 + ... + mk, then the number of distinct values of mi is O(sqrt(M)), because sum from integers from 1 to 2*sqrt(M) is equal to (2*sqrt(M) + 1)*sqrt(M) > M . Then, for every length Mi, we have O(M) occurrences in the text (note that this is only true because all query strings are different) and since we have O(sqrt(M)) different lengths, total complexity is O(M*sqrt(M)). Just leaving this comment here in case anyone has the same doubt as myself.

My approach in D was a bit different.I used a modified topological sort.I started with a queue and pushed a vertex with even degree in the queue and then just popped a vertex reduced the degree of it's adjacent vertices by one,checked if any of these vertices now have a even degree if so then push this vertex into the queue.I am not able to figure out what is wrong with my approach .Any help is much appreciated. Here is the link to my solution http://codeforces.com/contest/964/submission/37423833 . Thanking you in advance.

5 0 1 1 2 3 Have a try.

My code is giving "NO" and I think no is the correct answer

You can break node 2 first and then break node 3 and finally you got the answer 2 3 1 4 5

In Div 2D/Div 1B, problem tag contains dp. i do not understand how dp is used here. can someone pls explain.

it's tree dp

To late but ==>

dp[i][0] = Is it possible to destroy i-th subtree if we havent deleted the (par,i) edge dp[i][1] = same as above but we have deleted the (par,i) edge

How we update it ==>

so we do a dfs from 1 assume that you're now in vertex start, lets call the neighbours of start that dp[u][0] == false && dp[u][1] == true group 1. and similar call the neighbours of start that dp[u][0] == true && dp[u][1] == false group 2, also call the neighbours of start that dp[u][0] == true && dp[u][1] == true group 3 ( if for some vertex both of them are false the answer is NO).

So now it's obvious that you have to delete all the group 2 before deleting this vertex and similarly you have to delete all of the group 1 after deleting this vertex so now depends on your vertex degree and your DP state (0,1) you have to put some of the group 3 to make it correct ( you can put any vertex of group 3 after or before deletion of this vertex and it doesn't matter)

D can be solved simply by hashing lol

How will you do it, wouldn't it lead to a complexity of O(N^2), because you will have to check every substring of length p where p is the length of M_i for a given query.

Testcase 9 in DIV2C/DIV1A is (a^k/b^k)%mod = 1 but a != b.

Testcase 7 in DIV2D/DIV1B is "root of the tree is not 1" which means p1 is not zero

My solution for DIV2 D/DIV1 B

Deleting a node will toggle the degree of it's neighbours (even becomes odd, odd becomes even)

We find the deepest node that has even degree, delete the subtree of that node in a bfs manner( as all node should have odd degree, when we delete in bfs manner, they become even level by level).

The reason why this is optimal is, we have to delete the deepest node and it's subtree in the above mentioned manner, because if we delete it's parent first then this becomes odd degree and the entire subtree becomes odd. So it is always optimal to delete the deepest even degree node and it's subtree, that will not change the answer

Submission : 91716266