## 550A - Two Substrings

There are many ways to solve this problem. Author solution does the following: check if substring "AB" goes before "BA", and then vice versa, if "BA" goes before "AB".

You can do it in the following way: find the first occurence of "AB" then check all substrings of length two to the right of it to check if substring "BA" also exists. Then do it vice versa.

Complexity of the solution is *O*(*n*), where *n* is the length of the given string.

## 550B - Preparing Olympiad

Because of the low constraints, this problem can be solved by complete search over all problem sets (there are 2^{n} of them).

For every potential problem set (which can be conviniently expressed as bit mask) we need to check if it satisfies all needed criteria. We can simply find the sum of problem complexities and also the difference between the most difficult and the easiest problems in linear time, iterating over the problems that we included in our current set/bitmask. If this problem set can be used, we increase the answer by one.

Complexity of this solution is *O*(2^{n}·*n*).

## 550C - Divisibility by Eight

This problem can be solved with at least two different approaches.

The first one is based on the "school" property of the divisibility by eight — number can be divided by eight if and only if its last three digits form a number that can be divided by eight. Thus, it is enough to test only numbers that can be obtained from the original one by crossing out and that contain at most three digits (so we check only all one-digit, two-digit and three-digit numbers). This can be done in *O*(*len*^{3}) with three nested loops (here *len* is the length of the original number).

Second approach uses dynamic programming. Let's calculate *dp*[*i*][*j*], 1 ≤ *i* ≤ *n*, 0 ≤ *j* < 8. The value of dp is true if we can cross out some ditits from the prefix of length *i* such that the remaining number gives *j* modulo eight, and false otherwise.

This dp can be calculated in the following way: let *a*_{i} be *i*th digit of the given number. Then *dp*[*i*][*a*_{i}*mod*8] = 1 (just this number). For all 0 ≤ *j* < 8 *dp*[*i*][(*j* * 10 + *a*_{i})*mod*8] = *max*(*dp*[*i*][(*j* * 10 + *a*_{i})*mod*8], *dp*[*i* - 1][*j*]) (we add current digit to some previous result), *dp*[*i*][(*j* * 10)*mod*8] = *max*(*dp*[*i*][(*j* * 10)*mod*8], *dp*[*i* - 1][*j*]) (we cross out current digit).

Answer is "YES" if *dp*[*i*][0] = 1 for some position *i*. For restoring the answer we need to keep additional array *prev*[*i*][*j*], which will say from where current value was calculated. Complexity of such solution is *O*(8·*len*) (again *len* is the length of the original number).

## 550D - Regular Bridge

Let's prove that there is no solution for even *k*.

Suppose our graph contains some bridges, *k* = 2*s* (even), all degrees are *k*. Then there always exists strongly connected component that is connected to other part of the graph with exactly one bridge.

Consider this component. Let's remove bridge that connects it to the remaining graph. Then it has one vertex with degree *k* - 1 = 2*s* - 1 and some vertices with degrees *k* = 2*s*. But then the graph consisting of this component will contain only one vertex with odd degree, which is impossible by Handshaking Lemma.

Let's construct the answer for odd *k*. Let *k* = 2*s* - 1.

For *k* = 1 graph consisting of two nodes connected by edge works.

For *k* ≥ 3 let's construct graph with 2*k* + 4 nodes. Let it consist of two strongly connected components connected by bridge. Enumerate nodes of first component from 1 to *k* + 2, second component will be the same as the first one.

Let vertex 1 be connected to the second component by bridge. Also connect it with *k* - 1 edges to vertices 2, 3, ..., *k*. Connect vertices 2, 3, ..., *k* to each other (add all possible edges between them), and then remove edges between every neighbouring pair, for example edges 2 - 3, 4 - 5, ..., (*k* - 1) - *k*.

Then we connect vertices 2, 3, ..., *k* with vertices *k* + 1 and *k* + 2. And finally add an edge between nodes *k* + 1 and *k* + 2.

Build the second component in the similar manner, and add a bridge between components. Constructed graph has one bridge, all degrees of *k* and consists of *O*(*k*) nodes and *O*(*k*^{2}) edges.

Complexity of the solution — *O*(*k*^{2}).

## 550E - Brackets in Implications

Let input consists of , *a*_{i} is 0 or 1 for all *i*.

Let's show that there is no solution in only two cases:

1) *a*_{n} = 1.

, for all *x*, and no parentheses can change last 1 to 0.

2) Input has the form or its suffix with at least two arguments.

This can be proven by induction. For input there is no solution, for longer inputs any attempt to put parentheses will decrease the number of 1s in the beginning by one, or will introduce 1 in the last position (which will lead to case one).

Let's construct solution for all other cases.

1) For input 0 we don't need to do anything.

2) For input of the form we don't need any parentheses, the value of this expression is always

3) Expression in the form (where second missed part consists of ones only). Then .

Complexity of the solution is *O*(*n*).

please correct the format of problem C editorial,, it seems print the date

Thank you!

Problem C is very easy. Don't need Dp

yes, it has explained in this editorial

Oh. yes. I'm sorry :D

Can someone tell me why my A got TLE? : http://codeforces.com/contest/550/submission/11420305

I used Regexp in Ruby.

https://swtch.com/~rsc/regexp/regexp1.html. I think Regexp has a exponential running time as mentioned in this post. Also from the post :- "Perl, PCRE, Python, and Ruby are all using recursive backtracking". Hope it helps.

Do you mean bi-connected component by strongly connected component? If not, what is a strongly connected component in an undirected graph, I tried to google it and found nothing :)

Me too, I don't understand why there is a strongly connected component in undirected graph.

no bi connected components are different . Suppose you have a undirected tree then it is strongly connected but not biconnected. Biconnected components means there are always atleast two edge disjoint path between any pair of vertices in the component. It means there is no bridge in that component.

Good to know about biconnected. So what does strongly connected mean?

in the bottom of the explanation for C there is a O(8·len). but i think that in the big O notation you shouldn't write constants.

Yes, you are right. It's written in such a way to show that if you have the same problem with some d instead of 8, you can solve it in

O(d·len).Hey!

For problem C, can you please explain the method to restore the answer using prev[i][j] in its DP solution?

You should calculate it when you are finding DP: for example,

When you are printing answer, you may do it recursively, like

Thank you for your reply. I have one more confusion though.

When we cross out the 'i'th digit (i.e when we do not consider it), why is the DP equation : dp[i][(j * 10) % 8] ?

I mean, why are we doing (j*10)%8 ? Why not just leave it as it is, because we are crossing out the ith digit?

Should it not be like below.?

if (dp[i — 1][j] == 1 && dp[i][j] == 0){ dp[i][j] = 1; }

Yes, you are right, it should be

jinstead ofjmod10 in dp and prev. Thanks!ADJA, could you please fix the editorial accordingly?

Excuse me, could you explain to me the first if? :

How can we calculate

when we know

I really don't understand although read editorial 3 times.

Thank you very much indeed.

Actually O(8*len) = O(len). It's not necessary to write the constant, but you can.

Seems Codeforces has new feature: autocomment about post changes. Please ignore this comment.

The idea was not to edit such auto-comments so that everybody ignore them =) I guess we'll remove the option of editing auto-comments.

In problem C, I think the first approach can be better than O(len^3), O(125*len) can bee achieved, check 11430246

hey can someone explain how people solved B without using recursion ?

You can solve it by Bitmasking. You can Refer [my solution].(http://codeforces.com/contest/550/submission/11425633)

Here is my solution (it doesn't use recursion): 11424247

Each set of problems corresponds to a binary number. For example, if we had 5 problems total, then the number 10101 would represent a set containing problems 1, 3, and 5.

Because of this, we can just count from 0 to 2

^{n}in binary and test every possible subset.Use the first n bits of an int mask to represent a subset. If the i-th bit is on, then it means the i-th element is in the the subset. Just iterate the mask from 1 to 2^n — 1.

Here it is !! http://codeforces.com/contest/550/submission/11424876

Read this tutorial on how to find the power set of a set. POWER SET

http://codeforces.com/contest/550/submission/11421622

In problem D, how did you come up with the magic figure of 2k+4?? Can you provide any intuition for that?

Actually it is the minimum needed number of nodes (we thought about requiring it, but agreed it would be too hard).

I or Timur will post the proof later, if you want it and won't come up with it yourself :)

Hi

Please post explanation if possible.

Hi. Actually I just noticed that this proof is available in the Russian version of editorial :)

Here is the translation:

For

k≥ 3 let's find minimum number of nodes needed. The graph will contain at least one bridge and at least two strongly connected components, that are connected with other part of the graph with exactly one bridge. Each of these components will contain at leastk+ 2 nodes (one node that is connected to the bridge is connected withk- 1 edges with other nodes in the component. Other nodes in the component will have degreek— odd number, so the total number of them should be even, to make component right graph by its own [see handshaking lemma]. So this min number of other nodes isk+ 1. Therefore total number of nodes in the component is minimumk+ 2 ---- one incident to bridge,k+ 1 not incident).There are at least two components with at least

k+ 2 nodes in each, so minimum is 2k+ 4.When we want find a bridge side in th graph . so we can think of two grpah has only one vertex's degree is

`k-1`

,other's degree is k. And then we connect the graph and another.If we want to get the grpah which has only one vertex's degree is`k-1`

,k+2 vertexs can get the graph.... And maybe solving other constructive algorithms can help you.i didnt, i went for 4*k-2. its pretty easy actually. first consider the bridge and name the vertices u and v. then add k-1 other vertices to the right side and connect all of them to v and them add another k-1 vertices to the right side and connect all of them to the previous k-1 nodes so now those k-1 nodes + v have degree k but the new vertices have all degree k-1 since k is odd then k-1 is even and you can connect every two adjacent pairs of the new k-1 vertices to each other so they now all have the degree k and now do the same thing for u and you are done.

For me its more like 2 * (k + 1) + 2, which is equal to his formula.

To get weight k in one component we need exactly k + 1 nodes. We have two components, so its 2 * (k + 1). Also we have a bridge, which is 2 more.

Shouldn't the degree of EVERY node be equal to k? The degree of the (k+1) nodes in each component is k, but what is the degree of the 2 nodes which form the bridge? Isn't it just one?

Yes it should. And it would :)

My algorithm: Generate graph of k + 1 nodes, each connected to each.

Remove k-1 edgesfrom there and connect these nodesto the bridge one. After that, you'll have k+1 nodes with k degree, and bridge node with k-1 degree. Duplicate this graph, so you get two of them. Connect bridge nodes. Solved.UPD:Forget to tell, that we should pick edges for removing carefully — each of them should have unique nodes connected, so no duplicated allowed (because if any dup k-degreee would be violated)Got it. Thanks very much.

The solution for E is not clear to me Please help if possible

You must be clear fist on the fact that

(x->1 = 1)This means that the last number in the input must be zero.

So now our sequence takes the form

"xxxxxx...0"Now we want to make the second to last number equal to 1 in order to have

(1->0 = 0)Now we have two cases:

"xxxxxx..10"You can just leave the sequence as it is ... all the numbers to the left of the 1 will just have no effect .. and then the final value will be

(1->0 = 0)"xxxxxx..00"We need to turn this new 0 into a 1 somehow in order to be like case 1.

In order to get rid of this zero we have to match it with some other number.

From the rules, we have

(1->0 = 0)and(0->0 = 1), so obviously we will try to match this 0 with another 0.So we will search for the first 0 to the left of this 0.

So our sequence now takes the form

"xxxx...0111..1100"We can get rid of this group of 1s by grouping

(1->1->1->1->....->0 = 0)So now our sequence takes the form

"xxxx...0(111..110)0"="xxxx...000"Now you will just have to group these two 0s together to change them into 1

"xxxx...(0(111..110))0"="xxxx...(00)0"="xxxx...10"="0"Great explanation , i just wish codeforces editorials were written like this .

awsome explaination....thnks a lot.

I was unable to complete my code of D during the round. I did it differently and it was more intuitive.

My submission : CodeCould have done better by creating just 2 such bipartite graphs and removing (

k- 1) / 2 edges from each one of them.I suppose you meant edges not nodes.

Yes, sorry for the typo.

Could have done even better by taking a completeGraph(k)[or K(k)]. Refer: http://codeforces.com/contest/550/submission/11437241

EDIT: LOL, just realised the editorial does the exact same thing.

Problem B can be solved with partial sums?

Well, you can find all permutations of problems and test each segment of each permutation using prefix sum, but the complexity would be O(n!*n^2) and you would get a TLE. I don't know if we can do better than this (using partial sum, of course).

anyone did Problem B using recursion? I did using bit masking . Just curious !!

I've solved it with recursion (Backtracking) 11423350

Ok! Thanks for explanation!

In the recurrence of problem C,

I think when we are crossing out current digit,

recurrence should be:

dp[i][j] = max(dp[i][j], dp[i - 1][j])

Correct me If I am getting it wrong?

You're right. The editorial solution is appending a zero to j.

Can you explain why need to append a zero to j? thx

There's no need for it. In fact, it's wrong as grayhathacker pointed out.

For problem D, if k is even, then Eulerian cycle exists, thus there are no bridges.

We can solve problem C in O(1000 * len) without using dp. First find all numbers which are less than 1000 and divisible by 8. Convert them into strings. Now check if any one of them is a subsequence of the given string. This can be done in O(len).

Seems wrong...

E.g. string: 214

The answer (YES, 24) is not a subsequence of the original string.

UPD:Or maybe you meant the solution like this one?24 is a subsequence of 214. Check the definition of subsequence here en.wikipedia.org/wiki/Subsequence

Yea, sorry, thought about it like "substring"

I think this would be 124 not 1000. Because there are only 124 numbers less than 1000 (Not including 1000) that are divisible by 8.

125

E easier than D :)

I also think that E is easier than D but that is subjective. There are [to the date] 363 accepted solutions for E which is lower than half those for D (754 accepted solutions).

What is the optimal solution for question B? Thank You

I can't get what the editorial says about the dynamic solution for problem C. Can someone tell me how is it working with a c++ code?

can anyone provide dp solution for problem A? As dp was a tag in problem a

almost 2000 user tag edit access... now it has "chinese remainder theorem" XD . can anyone provide chinese remainder theorem solution for problem A?

I almost used dp to solve it :P

11422078

haha cool enclosed is my dp solution http://paste.ubuntu.com/11587440/

You don't even need to process the 3 digit numbers in 550C - Divisibility by Eight , It can be done by adding Two Zeros as the Prefix of that number , which would transform a single , two and three digit number to three , four , an five .. which can easily processed by the O(N^3) solution.

UPD : Just don't print the leading zerosIn problem C I generated first 125 numbers which are divisible by 8, then convert each of them to string and find if any of them appears in the given string as a substring. This can be done in O(125*len)

so many brute questions , not sure is codeforces or bruteforces.

I really liked D! Theoretical graph problem is much interesting than a algorithmic graph problem.

Somebody should fix the tags for Div 2 Prob A. Seems like someone went crazy and added every possible tag to the problem.

in 550C editorial what is meant by "crossing out"

Yes, you're perfectly right.

11447158 Is it solution for C in O(n)?

You could solve C in linear time ( O( (1000/8) * len ), actually ): just check for each multiple of 8 less than 1000 if it is subsequence of the input.

Seems that.But if n is large and mod is prime, that means is ok?

In Problem E, when I used the inbuilt 'string' in C++, I got a TLE at case:8. When I replaced the inbuilt 'string' with the character array char[], it got AC.

Here is my code using inbuilt string which got TLE : http://codeforces.com/contest/550/submission/11455328

Here is my code using character array which got AC : http://codeforces.com/contest/550/submission/11455369

Any reasons for that?

Anyway, array of characters is a slightly faster than the

`string`

class- but it should not make sense and should not give an advantage to who uses array of characters instead -but someone may use the

`string`

to produce a much slower running code inadvertently, How is it possible ?Every appending operation uses

`+`

has a complexity`O(n)`

where`n`

equals to the length of the string,- since`C++`

does a resize operation -you may use the built-in

`push_back`

function instead or even initialize your string to have such a maximum length and append elements by a direct access using index . look here is your first submission, just swap`+`

by`push_back`

and got AC in`139 ms`

hey ... can u tell me why we are taking the factors of 8 upto 125 in number only in C problem...? need help plz...?

Can someone explain the DP solution from the editorial ? I dont understand this part-> dp[i][(j * 10 + ai)mod8] = max(dp[i][(j * 10 + ai)mod8], dp[i - 1][j]) , what is this supposed to do ?

Is there a reason that you choose to use 2k+4 nodes in 550D?

For problem B I see many solutions have this statement if(i&(l<<j)){...} in the 2^n loop, what's this mean?

(i&(1<<j)) is basically used to check whether the j-th bit in bit representation of i is set or not.

As an example, consider i=5(101). (i&(1<<j)) will be true for j=0 and j=2.

Thank you very much. I made a mistake. I were thought that is 'L'<<j

550C- Which data type should I choose for the integer n <= 10^100 in C++? I am using long long, but with it test# 23 gives incorrect answer probably because of limit of long long.

`std::string`

will fit just well. :)Can somebody please explain why we do we have always this general notation of for(int j=0; j < x ; j++ ) dp[( 10*j + added digit ) %x ] +=dp[j] . I mugged it up . but now i want to know reason behind this

I didn't get the idea behind the DP solution for the problem, "Divisibility by Eight." Could someone kindly explain the solution in greater detail?

Let's first define DP state as dp(

index,mod) = {0,1}, which means using the digits between the leftmost digit and digit atindex, is it possible to make a numberxsuch that x%8 ==modWe loop through all digit (i.e. loop through all

index) for each digit, we have two choices naturally:For the first choice, dp(

index,mod) |= dp(index-1,mod) for allmodFor the second choice, it becomes tricker.

Consider dp(

index-1,mod) for allmodIf it is true, then if we append current digit,

the modular at

indexwill become (mod*10 + digit(index))%8So we just update this way: dp(

index, (mod*10 + digit(index))%8) |= dp(index-1,mod)Here's my code using this idea: 12143232

Hey kirakira amazing explanation! But i would like to point out that even though your solution got accepted, the way you handled the dp table for filling the store table .. distorted the dp table filling base logic and even confused me in the beginning.

You just cannot assume

`dp[0][0]++`

even though you have added

and further in the final ans checked`store[0][0]=" "`

This way it distorts the dp table filling logic and even confused me in handling the base case. This can be managed just by adding

The rest is same .. this is my accepted solution 16596112 . Still better explanation than the editorial :) It helped me. Thank you!

Can you tell why you are performing the bitwise ORing in the recurrence formula.

why we consider dp(index-1,mod) for all mod

What on Earth has happened for the test on question A? My code is returning "NO" for the input "ABA", but when I submit it, is says that "YES" is being returned for the string "ABA".. What?!

Need

O(n*8)solution for problem550CCould not understand dp solution for C. Can you explain it more clearly with code? ADJA

Hi... .

Could you please help me with my code for problem 550A - Two Substrings?

submission: 21399858

Corrected... . submission:22328773

Please explain me the dp solution of problem C . thanks in advance :)

awesome solution to div2 C !

if same ques with 9..then can we use 2nd approch?

The test cases of problem C are quite weak. My program 32731408 passes the test cases, while it fails on 8460.

550A - Two Substrings Test case 17 is : ABAXXXAB with Jury answer "YES"

I think it does not contain two non-overlapping string "AB" and "BA".

A**BA**XXX**AB**

hey someone help me plz @madhur4127 , suraj021 ? why we are taking the the numbers divisible by 8 below 1000 only in Div-2 C problem ?

Because a number is divisible by 8 if the number obtained from its last three digits is divisible by 8.

See this caption for generalization.