Hello!

Codeforces Round #486 (Div. 3) will start on June 1 (Friday) at 14:35 (UTC). It will be the third Div.3 round in the history of Codeforces. You will be offered 6 problems with expected difficulties to compose an interesting competition for participants with ratings up to 1600. Probably, participants from the first division will not be at all interested by this problems. And for 1600-1899 the problems will be too easy. However, all of you who wish to take part and have rating 1600 or higher, can register for the round unofficially.

The round will be hosted by rules of educational rounds (extended ACM-ICPC). Thus, during the round, solutions will be judged on preliminary tests, and after the round it will be a 12-hour phase of open hacks. I tried to make strong tests — just like you will be upset if many solutions fail after the contest is over.

Remember that only the *trusted participants of the third division* will be included in the official standings table. As it is written by link, this is a compulsory measure for combating unsporting behavior. To qualify as a *trusted participants of the third division*, you must:

- take part in at least two rated rounds (and solve at least one problem in each of them),
- do not have a point of 1900 or higher in the rating.

**Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.**

Thanks to MikeMirzayanov for the platform, help with ideas for problems and for coordination of my work. Thanks to Maksim Ne0n25 Mescheryakov and Ivan BledDest Androsov for testing the round.

**UPD**: I also would like to thank step_by_step and eddy1021 for testing the round and help with it preparation!

**UPD2**: You will be given 6 problems and 2 hours to solve them.

**UPD3**: Editorial is published. Thanks to Mikhail PikMike Piklyaev for the help with translation.

**UPD4**:

Congratulations to the winners (official results):

Rank | Competitor | Problems Solved | Penalty |
---|---|---|---|

1 | volamtruyenkyii | 6 | 196 |

2 | IOI2018 | 6 | 238 |

3 | Student_of_Husayn | 6 | 303 |

4 | Omoshiroi_ | 6 | 311 |

5 | Deadpool | 6 | 313 |

6 | AHDPIYKO | 6 | 341 |

Congratulations to the best hackers:

Rank | Competitor | Hack Count |
---|---|---|

1 | jhonber | 70:-1 |

2 | djm03178 | 61:-4 |

3 | applese | 53:-1 |

4 | WAI95 | 41:-3 |

5 | step_by_step | 38:-5 |

6 | greencis | 57:-45 |

530 successful hacks and 401 unsuccessful hacks were made in total!

And finally people who were the first to solve each problem:

Problem | Competitor | Penalty |
---|---|---|

A | AhmedMohamd-BeatMeIFUCAN | 0:02 |

B | S1mplest_gu1 | 0:06 |

C | S1mplest_gu1 | 0:12 |

D | volamtruyenkyii | 0:23 |

E | Ycrpro | 0:19 |

F | MuieEcaterina | 0:49 |

I hope that you will like the problems. If the problems in this round are too easy or too hard, then we will adjust the difficulties in the next Div. 3 rounds.

Good luck!

`It will be the second Div.3 round in the history of Codeforces.`

Actually, this is third Div.3 after rounds #479 and #481.

Thank you so much! Fixed =)

There are not editorial and winners' list?

I will post editorial and winners' list very soon, please wait a bit =)

Also the second one where Vovuh is the problemsetter, thank you

Actually, programmers often count from zero. Just a joke :)

Often?? (surprised)

Always

Finally, Div.3 round :D

Can someone to hack CF rating?? +100 for successful hack. -50 for unsuccessful hack. Just reduce my rating by 1 point. 1600 -> 1599. Because my rating >=1600. So it is unrated for me. But I still want to participate.

Looks like you don't know the CF rules yet.

Nope that's just subtle show-off :p

Yeah you are right :p

Just because it's unrated for you, doesn't mean you can't participate

`If suddenly something does poorly with difficulties of the problems,`

The grammar here is poor. I'm not quite sure what this is trying to convey, but perhaps

`If the difficulties of the problems in this round are poor,`

or even

`If the problems in this round are too easy or too hard,`

would be better.

Thank you for correction, it's fixed now

In fact, I think the difficulties were just about right, nice to have a bit more of a difficulty gradient than what we've seen before

Div.3 ! Round for Newbies like me :)

"if your rating is less than 1600, then the round will be rated for you."

+respect

can anyone explain the difference betwwen div2 and div3.Thank you.

div 2 is rated for > 1600 and the problem is pretty hard and not too easy and for div 3 is rated for under 1600 and the problem pretty easy not too hard (you can check the past contest and learn from it)

for full explanation -> http://codeforces.com/blog/entry/59228 Meow !

Div 2 is rated for all participants less than 2100. There is no lower bound ( as u said >1600).

sorry im misunderstand :p

Div 2 Rated for -infinity < Rating < 2100.

Div 3 Rated for -inifity < Rating <1600

Div 2 got some problems easy, some medium and some hard.

But to help improve the newbies and others learn, CF invented Div3.

Div 3 will contains majorly problems of easy and few medium difficulty.

High Ratings to EveryOne !!

so excited <3 to be Expert, it is just 144 :/

144 is a lot ...

not your cup of tea

Hope to be green lol.

it's just 6 points

Congratulations!!!

Finally a dream come true

nice job mate

If i never join the contest (unrated account), will this contest rated for me?

**I'm sorry for my poor english & question

Yes

thanks :)

May be not.

As it's rated for only Div-3 ** trusted participants**. You must:

take part in at least two rated rounds (and solve at least one problem in each of them),

do not have a point of 1900 or higher in the rating.

N.B.: Read the blog properly.It is rated for all participants with rating below 1600, dude

Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.lol xD you need to read the blog properly.

Scoring?

It's an ICPC style contest. All problems have equal weightage.

Suited for me....

## ?detaR tI sI

Hope that I will become expert after this round. :)

you won't i bet

Let me talk to you after the contest.

Bro you was saying something :)

It might look like I was rude and unprofessional but I did this just to make you really serious for this contest so it goes very nice for you. I commented the same on some other person also, so to give them huge confidence boost. Looks like it worked for you. Good luck for future contest also.

Try to increase your contribution lol.

Thanks for this contest

Wow!! Already 8k. Gonna be challenging.

Is it Div.3 contest if only 3 contestants with rank below 1600 (out of thousands) have solved E and F in first hour??

"If the problems in this round are too easy or too hard, then we will adjust the difficulties in the next Div. 3 rounds."

Also I think that's good. It differentiates contestants more, than everyone solving ABCDEF and rating only by solve times.

Who thinks that 20 points of penalty are too much?

That's standard ACM rules. I think it's fine because it encourages people to write correct code rather than fast code :)

Is it just me or somebody else noticed the mistake in question D!!??

in the output format they have said

"In the second line print m integers — the coordinates of points in the subset you have chosen."

but actually you have to print the numbers itself!! XD

Well the numbers you print represent the coordinates

many times they ask us to print the index so it kind of became confusing!

and again i just asked does anyone else feel the same, just wanted to know how others think! Though no offense to the setters the problem was good!!

Yes, they do mostly ask for the index, I agree. I didn't mean to offend you or something ;)

How to solve " Divisibility by 25 " ?

find lastest 0, 2, 5, 7 in given integer. then we know how many swaps we need to make 00, 25, 50, 75 at the end of integer.

but, you have to consider 0 being at the frontmost number of integer, which is a point i made a mistake.

For example, if we are considering how many swaps we need to make 25 at the end of integer, and if 2 is the frontmost number of integer (before swapping), and if k 0s continue behind 2, (i mean, 200000235234 has five 0s behind 2, k=5), you need to swap k times more. So you can just add k at the answer.

As you said that we only have to deal with the last set of 00,25,50,75 so for the given example we dont have to go for swapping the 0's.We can use the 2 in the 3rd last position and 5 in the 4th last position to get ans=5.I dont understand when will we ever face the problem of getting 0 as the leading digit?

In testcase 8, in which I kept getting WA — 50267

Did you take care of leading zeroes? The result cannot have any. Need to make extra moves to get a nonzero number to the front.

500044444444442

Got it thanks !

Solution for

E: for each case of 00, 25, 50, 75: find the last digit of the case (_0 or _5) in the given string at the rightmost position and swap it to the end (ans += swap distance), then find the second last digit of the case (0_, 2_, 5_ or 7_) in the string at the rightmost position and swap it to the second-last position (ans += swap dist). Then try to swap starting zeroes (if any) with a non-zero digit (which must be at least third-last) (again: ans += swap distance). If the final string is a correct number then consider the 'ans' as a candidate for the shortest answer. Solution with regexes, Perl: 38862670.Can Problem E solved using Recursion "Backtrack" ?

What do you mean? Idk.

I don't know lol.

I tried to Solve it using generating all permutation of the given number using Recursion, but i found a problem in coding it.

My code using Recursion https://ideone.com/9CPZY2, but it didn't give correct answer.

great balance imo

How to solve D?

mis less or equal 3, isn't it?Yup, but I kept getting WA

Yup, the maximum size of subset is either 1, 2 or 3 :)

why cant the max size be more than 3

Because if you have 4 (or more) numbers, the distance between the 1st and 4th will be 3 times a power of two = 3*2^n, which isn't a power of two.

Because 2

^{a}+ 2^{b}= 2^{c}hasa=bandc=a+ 1 as the only integral solutions. Now, If u have 4 numbers then the distance between 1stand 4thwill be 3 * 2^{a}which is not a perfect power of 2.Got it! Thanks

Feels bad when you figure it out right after the contest (:

There are only 3 possible options:

1) Just choose any one to get a set of size 1.

2) Just choose any pair such that

S= {x,x+ 2^{d}} for somexin the array, and somedsuch thatx+ 2^{d}belongs to the array. It works trivially.3) Choose, if possible, for any element

xin the array, the setS= {x,x+ 2^{d},x+ 2^{d + 1}} since it's the only choice to maximize the size (It's not possible to add another one with difference 2^{d+2} since 2^{d+2}-2^{d} is a multiple of 3 and also consider that it also shows that is not optimal to choose different d's as pairwise difference), remember that all those elements need to be in the array.One can get all the information needed by mapping values to indices and just checking for a not null index in each try. Woof!

Got it. Thanks!

Enters Div2 rounds.. Solves A, B & C...

Enters Div3 round.. Solves only A, B...

What ??

I just solved A & C.**T_T**。My B problem failed 8 times.

How did you solve C??

Same here brother!!

Maybe you didn't give the contest all your effort because it was Div3 and you are Specialist.

Focus in the next round.

No.

oh

Was it really a div3 round? Cuz it looked like a div2.

How is this possible that i have TLE when my algorithm calculates in O(n * 30) ?

38858406

Every time you check in map you create a new entry, instead use .find(), it doesn't affect the map size.

thank you

Also use unordered_map so the find complexity is O(1) instead of O(logn)

There are cases when unordered_map gives TLE while map passes. See this.

Idk. I have passed with in 2.5s.

I got WA in same test case than sorted it and got AC till now

The sums were hard sir. Please make it easier next time for rating 1600 and below.

Yes, we know about fast solution to the problem F (with Li-Chao tree). But in hard version this problem is too complicated even for the last problem of the div3 contest. This why we prepared this problem with such constraints.

I thought it was simple dp?

I mean just dp without using any others data structures

Yeah, dp solution have O(n^2) time complexity and this is accepted. But there are exists some other solution with dp and convex hull trick (also known as Li-Chao tree) which have time complexity O(n * log^2 n).

Yes. My bad. I didnt read carefully and though you was complaining about the difficulty of it

Solved it using Li-Chao tree. Learned a new thing. thanks!

Can anyone say why did I get runtime in this code Problem B (Test 7)? Code

I had made 10 submission to get it accepted, which resulted in huge penalty...

I suppose last i is zero. And s[i-1] goes to -1.

Its if(

iand s[i-1].find(s[i])...)C++ sort requires strict weak ordering. for more, https://stackoverflow.com/questions/32263560/errorinvalid-comparator-when-sorting-using-custom-comparison-function

Then, why did this code too get runtime error on test 7 ? Code

Your lambda code still does not satisfy the rule.

This is the right one. without the equal sign.

Thanks bro, I understood my mistake. I should always use '<' equality instead of '<='. But then why did I didn't get runtime in test 1 itself!! Was the output unexpected?

Bcoz, I got penalty of 200 by submitting the code 10 times until it got accepted.

`If the problems in this round are too easy or too hard, then we will adjust the difficulties in the next Div. 3 rounds.`

Again, in the next round :(Will greedy strategy work for 988E - Divisibility by 25?

I mean a number is divisible by 25 if only it ends with 00, 25, 50, 75. So from the end we search for last digit (i.e. 5 of 7**5**) and then for second last digit. If at last there are leading zeros then we make the first non-leading zero number as first.

I got WA on test 23, but I feel I missed corner case!

UPD:Yes, it was a corner case :'(23 was the first one with a 2 digit input (I was using python and got a runtime error when checking for the leading 0 afterwards).

How to prove that in the problem D: 1 <= m <= 3 ???

Assume

x<y<zsatisfy thaty-x= 2^{a},z-y= 2^{b}andz-x= 2^{c}. Then 2^{a}+ 2^{b}= 2^{c}. Notice thatc>max(a,b), so 2^{max(a, b)}divides 2^{c}, and therefore it divides 2^{a}+ 2^{b}, so it must also divide 2^{min(a, b)}, implying thata=b.Then any three number subset satisfying the condition is an arithmetic progression. Therefore if

w<x<y<zhave all differences powers of two they must all be in arithmetic progression. This is impossible because thenz-w= 3(x-w) andz-wisn't a power of two as it's divisible by 3.assume there are four integers a<b<c<d satisfying the condition. let b-a=2^k, c-b=2^l.

c-a=2^k + 2^l must be a power of two, thus k=l.

c-b=2^k.

similary, d-c=2^k

then, d-a=3*2^k, must be a power of two, but is not.

So, 1<=m<=3

try proving with contradiction, what if n==4? (show that it cannot happen by using subtraction of bit pattern of numbers).

Suppose we already have the sequence a[1],a[2],...,a[k]. It is clear that it must be (a[k]-a[k-1]) = (a[k-2] — a[k-3]) = ... = (a[2] — a[1]) = 2^d, d > 0. Try it on 3 numbers, suppose we have a sequence a1,a2,a3 and a2 — a1 = 2^a, a3 — a2 = 2^b, with a <= b. Then a3 — a1 = 2^a + 2^b = 2^(b-a) * (2^a + 1), which is a power of two only in case when a = b. So the sequence should be like k,k+2^d,k+2^d+2^d and so on. Consider an example now with four numbers in sequence, a1,a2,a3 and a4. (a2 — a1) = (a3 — a2) = (a4 — a3) = 2 ^ d Then a4 — a1 = 3*2^d which is not power of 2. The same goes for m > 4.

Missed problem E by just a minute. Contest was over just when I was about to submit problem E :(

In Problem C, my this — code 1 solution got TLE on test 5. I used set_intersection function to find common sum in any two sequences.

Later I wrote this — code 2 in which I have just done linear comparison between two sequences. I couldn't submit it in time.

Can anybody tell if code 2 time complexity will be less than code 1 or not? Also is there any even faster method to do the problem. Thanks!

Why test case 8 in problem E (50267) the answer is 5?

50267 -> 52067 -> 25067 -> 20567 -> 20657 -> 20675

Damnn I too kept getting WA there. I couldn't come up with a test cases like this. The point is in 50267, you want to get the 5 to the right .In doing so your intermediate number is 05267 which has a leading 0 and thus isn't accepted. You need to do another swap — (0,1) which will make all the intermediate numbers not have the leading 0.

Hmmm...

This reminds me of

This statement is false.nice joke, expecting failed on test 11.

Um_nik not just you makes fun :)

In B I think if any code print "YES" for {2,p,pp} then it should receive wrong ans. cause statement says if "reorder" impossible it should print "NO".

Contest of maps and sets

I could only do A, B, C and D. (Before system testing) And I think the difficulty of the questions are absolutely fine !

You end up learning more if there are two questions out of your reach, rather than just 1 or doing all of them. I definitely appreciate the difficulty of these problems. Please don't water it down for the next contest.

My only complaint here is that your previous contest had one question from every topic — DP, graphs, binary search, STL, number theory, etc ... Whereas the other two Div 2 contest didn't cover as many topics.

We got STL and I guess you could call D some kind of number theory where you observe the answer is never more than 3, but still we didn't get a pure number theory one like your previous D — Multiply by 3, Divide by 2 did ! There were no graph problems either.

In short, my feedback is this — The difficulties are great, pre-tests are great, Just add more topics :)

Why are there some guys hacking themselves? lol

because they know they will soon be hacked by reds or etc..

Or they would anyway fail the main tests

A div3 contest with 3 div2 problems for me :(

In D, my mistake was printing a long long int instead of an int. Naturally, long long int wasn't needed. But is this a legitimate reason to give a WA?

That's really strange..

I used printf("%lld", ~) and get AC

Well you not print any solution, look carefully~ Just try it to run locally

That wasn't your mistake — it's just a slightly confusing error message. Your code outputs 0 for that test (as 42 isn't a power of 2) without a second line which causes the error message seen.

You're right. Why was the correct output 1 42, when 42 isn't a power of 2? I can't have that in my subset right? Or can I?

UPD: When the max size was 1, I printed any element. Just got AC. Why is this correct though?

It's given in the problem statement, a subset of size 1 satisfies the given condition.

Didn't read the problem statement carefully. Thanks

please explain solution of C??

Spoiler...store sums instead of items...

While you reading the input get the sum of the numbers from the current array. Every time check if the current sum has been encountered before. Now traverse the array once more and map this value (sum — a[i]) with arrays index, and its position in the array. Check out the code: http://codeforces.com/contest/988/submission/38843860

For each set if elements, first calculate sum of entire set. Then calculate difference matrix by subtracting each element from sum. Use a global map(common for all the sets) for hashing and storing indexes, and check whether each difference element was obtained earlier or not. If yes, you have found common sum producing elements in the 2 sets!

I thought about this approach, but wouldn't creating a difference matrix take O(ni) time for each sequence. and we have k such sequences, so its O(k*ni). Wouldn't it time out?

Also what does global map store exactly? Would be helpful of you to elaborate. Thanks

No it wouldn't time out as k*n<= 2*10^5

Global map stores the difference (between sum and actual element in all the sequences) as key. Lets say I have difference

`D`

obtained on sequence i for a element j, I map the difference`D`

for i and j. The map is of the form map<int,vector>. vector stores for i and j. Later if I find the same difference for seq i1 and element j1 in it, I have stored earlier occurrences of the same difference in map. See my submission for implementation! Hope it helps.D --The size only can be 1,2,3 (but why?) Maybe the "conclusion promble" in div3 is a little bit weird?

Let

A=C+ 2^{a},B=C+ 2^{b},A≥B.According to the problem,

A-B= 2^{a}- 2^{b}= 2^{x}should hold.Divide each side by 2

^{b}. The equation would be 2^{a - b}- 1 = 2^{x - b}.1) If

x-b= 0, then 2^{a - b}- 1 = 1 -> 2^{a - b}= 2 ->a-b= 1 ->a=b+ 12) If

x-b≠ 0, contradiction since left side is odd and right side is even.So there are at most 3 which is

C,C+ 2^{a},C+ 2^{a + 1}IMHO, this kind of problem is little challenging for div3 but worthy at the same time.

The round was cool! I'll wait a new round.

Couldn't debug the runtime error in my solution of B !. Someone help : 38838754 Problem B : RUNTIME ERROR ON TEST 7

Try submitting it using the C++17 Diagnostic compiler, it might help in debugging.

I had the exact same issue. The error is in the comparator.

One of the requirements for a comparator is that

If comp(a,b)==true then comp(b,a)==false

One way to fix it is to return false if a == b before you check if a is in b.

One way to fix it is to return false if a == b before you check if a is in b.I didnt get this, what's the need for that?

http://codeforces.com/blog/entry/59770?#comment-434966

For your case,

return l.length() <= r.length(); — is not right. return l.length() < r.length(); — is right.

When both strings are of equal length, they must be exactly same, for this ques. So, it shouldn't matter whichever comes first. Can you explain why "=" should be removed.?

Because If comp(a,b)==true then comp(b,a)==false is a requirement for the comparator.

If two strings are of the same length, then comp(a, b) and comp(b, a) will both be true, which breaks the requirement.

On the contrary if you remove "=", then comp(a, b) and comp(b, a) will both be false, which doesn't break the requirement.

Thanks :)

bool comp(string l, string r){ return l.length() <= r.length(); } Don't use = int your comparator function.

Can someone tell me what's wrong with my code? 38862572 It is getting a runtime error in testcase 7 But when one of my friends submitted the same code from their account after the contest, they got it ACed I made it a point to compare both codes on an online text comparator and it gave both codes are same.

Try sorting using a strict ordering, i.e, instead of sorting on the basis of '<=' sort on the basis of '<' or '>'.

More info: http://codeforces.com/blog/entry/59770?#comment-434986

Thanks a lot!

What is the complexity of ACd solution for problem D?

O(40*n)

Nlog(10^{9})log(N) passes in time.There are unrated peoplr in the official standing

Hi all, I am new to CF contests. I am not able to locate the hacking option on my dashboard. The 'hacks' tab also has no such option for me (it only shows hacking results involving other hackers and defenders). Can someone help me out with this? Thanks.

Click on any solution (from the standings, the status tab, etc), you should see a hacking option

Thanks, how do I locate my room though? Or I can hack ANY solution?

you can hack ANY solution

You can navigate to a submission from the standings page, and you should see a button that says "hack it" above their solution.

Thanks, how do I locate my room though? Or I can hack ANY solution?

In a typical CF round, you would only be able to hack people in your room, but you may hack anyone in this round.

Thanks a ton for this.

Unexpected verdicts again, hacks #456187 and #456189.

in problem E for sample case 1 :5071 i think answer should be 1:-

5071 to 1075

You can only swap adjacent indices.

ok thanks

Test 27 of problem D is made to fail if unordered_set is used, why is that a thing?

EDIT: It fails with C++14 and passes with C++17, fuck me I guess.

I think it's anti-hashmap test(I guess someone intentionally generated such test).

How could I have known, so basically I can't use unordered_set in a contest? Or I should write my hash function I guess lol.

My solution with the STL map passes.

Welcome to Educational Rounds!

Xor with a random constant before inserting the number into unordered_set also work.

idiot

What?

unordered_set is O(n) // why you use it instad of set log(n)

actually unordered_set is faster than set (some times)

in some problems your solution won't pass unless you use unordered_set

so ,, no one is an idiot

my team's name in ACM is

2 idiotsit is good thing for you :)

thank you for very nice problem set. Had a really great time :)

Is problem E somehow related to Digit DP ??

i dont know about dp solution but i think E can be solved without dp

result is min(case25(), case50(), case75(), case00()). Here case25() is a simulation function that return minimum swap to make the last 2 digit is 25, if it cant make it, return infinity. If result is infinity print -1.

however i'm getting WA because i didn't read the problem statement carefully "your number can't have leading zeroes"...

someone give me hack test for problem E. I want to know if my solution is correct .. thanks :)

I like the idea of harder D, and E(i really think they are harder than the last Ds and Es of Div. 3) so people who just know how to implement fast are ranked lower then the ones who are skilled at getting fast ideas for problems, but implement kinda slower, so those programmers can get rating a bit faster, but also because you don't finish the round in 1 hour. Super Cool problems nonetheless !

i think 2:30-hours Could be better for today

Why will this code return 0xC0000005(Index out of bound) instead of 0?

The input data is as follows.

I guess cmp was a bit not properly

cmp should return whether s1 should be in front of s2 in the result array or not according to your sort rules.

when a=b="aaa",cmp(a,b) and cmp(b,a) both be true.

I don't think so. For example, in integers comparison, we use the cmp function:

Even if a == b, and cmp(a, b) = cmp(b, a) = false, our program will return right result, isn't it?

My fault.

But when cmp(a,b)==cmp(b,a)==true, sort will TLE.

Yes, you are right. So, I am confused to why that code will return 0xC0000005.

Indeed,sort use so called quicksort when n>16

Your cmp may cause out of bounds, so when n>16 it will RE.

UPD:use stable_sort instead?

Chinese version:sort在n>16的时候用通常意义上的快速排序，对于a<b且b>a这种情况(非偏序集?)，指针就跑飞了~

I think that you are also a Chinese, how about we talk in Chinese?[斜眼笑.jpg]

is hacking phase will affect the standing result ?

Yes,for only those who have their solutions hacked,going by the rules of educational rounds.But those who had successful hacks won't be affected.

Why does the contestants who are blue or higher have * marked before their name in status?

They are out of competition participants and the round is not rated for them.

my rating has not been updated after div 3 why please help

same

just wait it takes some time

I'm new to CF contests. Kindly help. Do the final standings that are currently displayed include hack scores too? (+100 for a successful hack and -50 for unsuccessful hack) How do I know if my solution has been hacked? Also, tentatively, by when will the ratings be updated?

If your solution is hacked,you will be able to see it in your submissions.It is informed on the moment whenever your code goes hacked.

The ratings update takes some time..But it is quite sure to be updated within an hour or two.

Do the final standings that are currently displayed include hack scores too? (+100 for a successful hack and -50 for unsuccessful hack)

You may want to use CF predictor extension in Google Chrome.It shows you an assumption of what your ratings increase or decrease will be.It has always given me perfect rating change.

just In the Div2 and Div1 contests you can hack during the contest and get points ,and they show you a window that tells you if u have been hackd . In the educational competitions and div3, you can only hack after the contest is over and the hack results didnt affect your points. For the rating it may take from half an hour to 4 hours Or more to be update.

But if someone's solution is hacked, his solution won't be counted as correct and he won't get full points for it right?

right

though if someone hacks successfully he won't get points for it? what is he hacks incorrectly? will he get negative points?

It is supposed to be that way.Still, there is no update in the standings yet for the one who has attempted a hack.I think the score for the hacking attempts will be rejudged during final system tests.

Actually he will not get any points for it..

he didn't get any point even if he didn't be hacked becuase after the contest his solution will get wrong ansower, the hack help u to debug and update your solution befor the contest is over , and if u lock your problem (you lock a problem if u want to hack someone ) then u cant update your solution .

Hi All, can anyone tell me why the subset size is at most 3 in problem D?

I got it from above comments. Thanks!

Assume points a, b, c are in increasing order, and let d(x, y) be the distance between points x and y. if d(a,b) != d(b,c), they cannot be in the same subset because d(a,c) wouldn't be a power of two. So if a, b, c are in the same subset, d(a, b) must be equal to d(b, c).

Now assume that you want to add another point x (x>c) in the subset. According to what I mentioned above, d(b,c)=d(c,x). But this means d(a,x)=3*d(c,x), which means it's not a power of two. So the subset size cannot be larger than 3.

Edit: Good that you got it already!

Thanks for your effort!

I don't know this is educational round or div3 round :(

Summary of hack for C (test 29): There were a lot of solutions that iterated over an O(K^2) loop. It passed original tests because when K is large enough, there tends to be a lot of YES answers and the solutions would just print it out right away when found.

The largest possible K with the answer NO we can easily think of is when K=20000 and ni=10 for each sequences, where every elements of each sequence are the same. But still some solutions passed through.

I managed to find a test case with K=40000 while keeping the answer as NO, and I think this is actually the worst case. Here's the code of the generator.

not able to submit code and also not able to participate virtually in this.

You can't practice or virtually participate during system testing.BTW system testing is over, you can try now.

thnku :)

can somebody explain why

O(31·N) gets TLE in D? 38851723Worst case of .find() in unordered set is linear i.e O(n) in worst case making your code O(31*n^2).

that is because of collisions but using int reduces the chances of collisions (good hash function of STL), unless the test case was crafted for making TLE.

It's totally possible for someone to add a hack data to make such solutions fail. That's why you shouldn't leave any luck-based things in your code in educational / Div. 3 rounds :)

thanks _l-.-l_ & djm03178 but the problem was pretty stupid, I computed

`v[i]+(1<<bits)`

and`v[i]+(1<<(bits+1))`

again and again, this increased runtime, caching these values passes TL.Never thought this addition can be so evil. 38877122

Haha that's sad, but you still passed TL by a little margin. 38877547 is your code using set instead of unordered set and it's almost 3 times faster.

unordered_set (C++17): 3790ms

unordered_Set (C++14): 2729ms

unordered_Set (C++11): 2995ms

set (C++17): 1263ms

set (C++14): 1341ms

set (C++11): 1263ms

Lesson learnt:DeletedSubmit in set, unordered_set in C++14, but not C++17 for now

That's not the lesson you should learn :)

You need to understand that

`unordered_set`

works based on hashes and therefore, if somebody knows the hashing algorithm used, they can handcraft a test case for which the hashes will have many collisions.One popular technique to avoid this is to add a random seed to your hash function. That way, the hash function is different every time and it is very difficult to generate a bad test.

Have a look at this submission 38889378. It is your code, but with a random element added. It passes in 1153 ms.

thank you for this valuable information, but I think you forgot to take xor again when searching in set. This trick helped me reduce runtime to 998ms. 38890354

In problem D,My submission gets Ac , but I think it is wrong because I thought same numbers can be chosen the same time . However , x — x = 0 , and 0 is not 2^s . it can be simply hacked .just like 3 3 3 3

The coordinates should be distinct.

Oh , I haven't notcied that. Thanks

"There are n

distinctpoints on a coordinate line"So that's not a valid input.

Right. I haven't noticed that.

Bro, the problem D's statement has said that the input contains n

distinctnumbers. :)Yeah. Thanks for telling me that. :)

I am new to this platform.... Just wanted to know.... after how much time rating changes are updated?

I'm still waiting for the ratings to get updated. I don't know why I takes so much time. Hmmmm...

Because it's div3 with ACM ICPC rules and extended hackings (12 hours if I am not mistaken).

For D, solution1 using set passed, but solution2 using unordered_map got TLE. How does solution2 which use unordered_map tend to be slower than solution1 which uses set?

Press Ctrl+F and type "unordered" and you'll find out similar questions and their answers.

Your crafting.oj.uz ratings are updated!

Can anyone explain the approach/solution to problem F?

I'm sorry, I downvoted your comment by mistake.

Editorial??

What is the required complexity for problem C)Equal sums?

In D, why did I have to sort the coordinates array to not get TLE?

TLE submission: 38882910,

AC submission: 38883316.

Thanks in advance.

The problem in your code is not the sorting, but that you use the

`[]`

operator for map look-ups, which creates a new element every time (check the documentation), therefore needlessly increasing your map's size.Use

`find`

or`count`

when you want to do lookups without creating new entries. This submission 38889506 is your TLE code but with this bug fixed and it passes in 1107 ms.As for "why does it work when sorted", it just so happened that, by coincidence, if you sorted your data, you found a solution early on and exited your program before hitting the time limit :)

Oops! Got it. Thank you very much :D

Is O(n^2) not enough for Problem D? n<2*10^5 and time limit is 4 sec.

O(n^2) would be 4*10^10 operations. 1 sec is aproximately 10^8 operations,so order 10^10 is too much.

Can someone please explain intuition / approach behind solving Problem F?

It's optimal to bring at most one umbrella at any time. To get minimum fatigue given the configuration of rains and umbrellas (position, weight) you can use DP 0-1 technique to consider whether to pick current umbrella (to survive the next rain) or not.

maybe I need a div4 contest,because I think div3 is still too difficult for me.

What's the difference between

`if (v.find(x) != v.end())`

and`if (find(v.begin(), v.end(), x) != v.end())`

. (v is a set/vector)The former one got AC but the latter one got TLE? Please help.

in the vector always find O(n)

My vector v is sorted, that's to say, after

`sort(v.begin(), v.end());`

, does that mean it is also O(n) all the time?Yes. How с++ should understand that your vector is sorted. If you want to find the element in the sorted vector, you can use this:

volamtruyenkyii this guy has given his first ever contest and got 1st rank, great. But he has registered just 2 days ago, Well I think he is a high rated coder who has made a new handle just to get rank 1 in Div 3

Not that it really matters, but regardless of his intentions I understand he should not be in the official results because it is stated that "only the

trusted participantsof the third division will be included in the official standings table" and he does not meet the requirements to be such (actually, he is not an official contestant in the "common standings" page, the thing is that he got included in the "official results" table of winners in this post)It was my first time to get tagged in official codeforces round .. Awesome ^_^

Me too : )

I'm glad to be an expect through this round. ~ this is a lovely round. Thanks

hahaha i ac 5 rating from 1500 to 1691!!!!!!!

can someone give me the complete 17th test case of problem F?

can someone tell me why my submission 39007350 is WA at the 17th testcase to problem F?