Привет, Codeforces! Давно не виделись :D
Codeforces Round #461 состоится в среду в 20:15 по московскому времени. Обратите внимание, что раунд чуть позже, чем обычно.
Раунд будет рейтинговым для участников из второго дивизиона. Первый дивизион тоже приветствуется, но вне конкурса :)
Спасибо Коле (KAN) за координацию раунда, Грише (vintage_Vlad_Makeev), Олегу (xen), АмирРезе (Arpa) и Сене (craborac) за тестирование и, конечно, Майку (MikeMirzayanov) за Codeforces и Polygon. Отдельная благодарность отправляется Диме (Dmitriy.Belichenko) и Камилю (pseuda) за идею одной из задач.
В этом раунде вы будете помогать игривому монстрику Импу. Не пугайтесь, условия будут короткими!
Задач будет шесть со следующей разбалловкой:
500 — 1000 — 1250 — 1500 — 2000 — 2750
Удачи!
Обратите внимание, раунд перенесен на 40 минут, чтобы избежать пересечения с раундом на CSAcademy.
UPD. Контест закончен.
UPD. Системное тестирование окончено! Разбор.
Поздравляем победителей!
Div. 2:
Div. 1 (unofficial):
UPD. Выложены разборы D/F.
Really loved your last round but hopefully this time, there will be a better difficulty distribution in the problems!
I tried to make it a bit more balanced this time :)
I am still trying to solve your C up to this day.
Hey! This might help. Spoilers for problem C : Cave Painting
Instead of checking whether the remainder left by dividing n by 1 to k are different or not, lets find the max K such that when n is divided by 1 to n, it gives different remainder all the times. For e.g. whenever n is even, K = 1
Note that all these 1 to K will give remainders as 0,1,2,...K-1 as they are different and they don't have other choice.
So just find K for the given n and check if it is greater than equal to k.
Hope I am clear.
Возможно, там опечатка. Грише*, а не Гришы. Извините, что докопался :)
UPD: исправлено.
Когда в ближайшем будущем не завезли раундов в удобное время...
Это еще цветочки...
В 9 вечера — наоборот удобно.
codeforces contest is always great and amazing , my bad , may be i will not be able to attend this contest due to travelling , missing it so much :( :( :(
codeforces contests are always great and amazing , i always try to attend the contests , but my bad , may be i will not be able to participate in this contest due to some travelling stuffs , and it is much painful for me , competitive programming is my passion
Ohh god CM problem setter again.
How to download Test #15 of 650C Table Compression?
In input :
1000 1000
1002 1002 1002 1002..... (1000 times)
In output :
1000 1000 1000 1000..... (1000 times)
Coz you were asking so many times.
the other numbers must be n*m(1000*1000). And also if there is only 1002(1000*1000 times) the answer must be 1(1000*1000 times).
oh but downloading the full test cases is not possible
Moreover it is highly unlikely that you will be able to debug your code with such a large input /output file
The names of the problems will be about colors of grapes?
Is there a relation between the name of the problems and some colors of grapes? :O :D :)
Is it rated?
Are You Retarded?
it's not my comment. but i still want to apologize
Obligatory comment: "Hope the problem statements are as short as the announcement"
Have seen you being quite responsive in the last contest. Keep up the good work, and well, wish a quality and balanced contest ;)
P/s: Yup, I still can't figure out the most optimal income when using Perun's ult xD
Thnx :D
xen
Хоть кто-то живет в 2к17 году.
PS. Если что я говорю о GreenGrape.
PSS. Если вы еще не поняли нажмите на xen
Для меня он навсегда останется xen-ом!
Thanks a lot for this timing
Oh... This contest is even later than others T_T
Too bad I can't enter it because of... time zone problem...
Wish you guys good luck and high ratings anyway ^_^
When it starts in your country?
00:35……
its not a friendly time for UTC+8
It was supposed to start at 01:35 AM here, and it even became 02:15 :)
Этот раунд рейтенговый?
Переведя шутку на русский, не заставишь ее звучать по новому. Это так не работает.
I hope my rating will be more than 1900. Then this contest is unrated for me. I don't like rated contest. My rating will down if I don't do well in this contest. CONTEST, CONTEST and CONTEST ... RATING, RATING and RATING ... TRAINING, TRAINING and TRAINING ... ......
if you cheat in this contest, this contest is unrated for you too, and if nobody discover it, you can have a high rating:)
Thank you guys, who create codeforces contests. They have really interesting problems. Live long and prosper. :)
Good luck to all
let's see, can i be a Candidate master at the end of this round or not ...
wish me luck... :)
Good Luck!) Hope you will be a Candidate
why people doesnt like this post whats wrong about it
He didn't get cm :\
sorry,what is cm?
Candidate Master
you're right, i was far away from my ideal performance last night ...
too upset for it... :(
well ...
now i'm :)
congrats
So sad that magic is over, can't troll with different colored grapes anymore :(
I wondered what is different from participant division 2 and division1
-the purple color and all the colors above are for Div1 contestants.
-Div1 guys can participate in Div2 Rounds unofficially.
really awesome, wait for some blossom
A prime number contest!
Nice observation!
not u again
This one is for you
→ Literally JoJo video
→ Inserts "reference meme"
→ Literally JoJo video
→ Obligatory weeb comment
Look on the bright side, man, this time there are no pupil users as problemsetters :D
"Yet the statements are guaranteed to be short :)" this is the best announcement so far.
Yet the statements are guaranteed to be short :)
Thank you very much:)
"Не пугайтесь, условия будут короткими!"- то чувство, когда раунд не от adamanta...
Unusual early scoring!
"Codeforces и Polygon"
Learning some russian.
Learning everywhere. :)
Learning so much that you start to understand Croatian. (Source: COCI 2016/17 Round #6 solution, p. 4, line 8)
Hope another great problem set from this amazing head. Big chance to rise up. CF is too good to stay away from. :)
Can anyone help me with one problem? I have good answer on test 2 on my computer, but on codeforces I get WA no matter what I do. Obviously compiler settings or something isnt same. Just visit my problem and check last submissions, problem C from few rounds before. 34983468
Test ur program on custom invocation and u will undrestand why got WA.
It's quite strange no one noted that this round coincides with CSA round #68 tomorrow! Can't you just delay it 30 more mins? GreenGrape
UPD: the contest is delay 40 mins!
More like, CSA coincides with Codeforces Round 461.
linguistically, is there a difference?!
I mean, CSA consistently hosts contests only in that time slot.
I guess it went under the radar since they start 30mins earlier until recently.
CF round moved by 40 minutes.
Glad to hear this: Yet the statements are guaranteed to be short :) Looking forward to it !!!
How to download Test #15 of 650C Table Compression?
Okay, this should be the last time I see this anywhere.
You cannot do that. At all.
(Well, except if the problem setter provide the full package publicly).
For convenience, all test cases are compressed in case they are too big, so large input/output files are near to complete omittance.
Usually, if you get stuck with a big tests, it can only mean a few things:
You ran out of memory (MLE verdict).
You ran out of time / your algorithm was too complex (TLE verdict).
You handled the undefined behaviors poorly (WA/RE verdict).
These tests are mostly used to tests how a code works against large input streams, and not to be corner cases, so please check your own source code for places that can be optimized, not just spamming the blogs all the time.
Normal guy's Imp.
Codeforces member's Imp.
Is it legal?
I registered him manually, just a prank ^^
How is registering someone in a contest a prank?
It’s just weird so a little funny. Pseudo will be like “wtf, dude” .. hence.
Seems someone also wanted to prank you and registered you manually :P See you on leaderboard :D
This will be the first time someone solve all problems in 1 minute :D
I am gonna solve A, B and C today !
Best of luck !
you will solve nothing we will come newbie
o yeah ! I have solved A, B and C !
only pretest passed
only pretest passed**
Let's see after system testing !
i hope we solve nothinh
i hope we solve nothing
Problem — C Time limit exceeded on test 60 ! :'(
be peptimistic
Wow :O
And I thought heaven didn't exist.
Two line summary of the announcement :
1. The statements are guaranteed to be short :)
2. There will be six problems with the following scoring: 500 — 1000 — 1250 — 1500 — 2000 — 2750
The time is a little late for me,but I am still glad to take part in the round.
It's really a good chance to train myself. :)
I just want to say it's too late for Asian contestants . Hope I will stay alive for being such late......
One hour earlier, but yeah, I feel you, dude...
12:15 AM in UTC+7 (Hanoi, Vietnam). Uh-oh...
Hope I can become Expert!
Happy!
Congrats :)
Sanwli saloni, adaayein manmohni Tere jaisi beauty kisi ki vi ni honi Thande ki botal main tera opener Tujhe ghatt ghatt main pee lon Coca Cola tu ROUND GONNA BE AMAZING !
You Should Behave On website like codeforces , though fun is necessary , but people seach for good comments over here . I hope you understand it ! Good Luck for contest .
i think i will be green this time,thank you GreenGrape.
Pessimism level: Notali.haidar
Тык.
Тык.
В этот раз бы сложности задач более сбалансированные что ли.
is it rated??
i think you love negative contribution.
it increases my ego and confidence , yes.
if you really get annoyed because of some online comments you need help:)
may be this site is not just a random website for many people
who knows :D
I want a 'HaHa' button(as in fb) in codeforces, people have become too funny :v
can't wait.....
Again a round with near 7000 participation :)
I hope the long queue is fixed before the contest starts
I hope the int vector is fixed setprecision(10) after the contest end ;!
MikeMirzayanov
What's wrong? I thought you can lock the problem as long as you have the "Pretest passed" verdict, irrespective of whether you get hacked or not. Correct me if I am wrong.
Nope, you can't lock after you got hacked. But I think it happened because times are very close.
Oh okay, that's interesting. This probably happened because of the long hack queues today.
very long queue in hacks
HackerRank will be proud of this contest
I think this contest are more "Successful hacking attempt" messages than "Accepted"
A round full of hacks. Even one of my friends hacked me, nice contest
Hackforces in a nutshell...
Anyway, to be serious, maybe it was just me, but I think the scoring is still a little bit underrated. This should be 500-1000-1500-1750-2250-3000. C is sufficient to be called a Div2C, D is somehow quite twisted (got 4 WAs verdict on the 4th problems, all on pretest 4).
Not really sure about my opinion about E and F, coz' I have just read it for a short while.
After all, it was fun actually. And these Imps are just cute ^^
UPD: I took my words about D back. Somehow I wish I could think sharper...
Even you hacked me. This is first time in my life when I got 3 hacks on A
Everyone has their own bad days...
I also faced the same situation as you, once at Codecraft-18 and Codeforces Round 458 (Div. 1 + Div. 2, combined).
Can someone please provide explanation for third testcase of problem E.
I think you are confused just like I was — summoning a bird doesn't increase your mana, it increases you maximum mana.
Ohh Thanks.Same mistake
Take 1 bird from the first nest (you don't have enough mana to take two birds), take 10 birds from second nest (your mana capacity is 17 and you have 15 mana so you can take them all).
Problem D is solvable with radix sort?
I just used normal sorting with a comparing function that returns
numS[a] * numH[b] > numS[b] * numH[a]Hell man! :'( I tried to do it with normal comparison too but that condition didn't pop up to my mind! How can you prove this works ?
Given a string s, denote f(s) to be the number of subsequences 'sh'.
Now given strings s1, s2, ... sn, it is easy to prove that if we denote xi, yi as the number of 's' and 'h' in string si, that
.
To maximize our desired value, all we need to look is the value
.
For n = 2, it's trivial that the comparing function wewark described works.
We will induct on n — say the comparing function works for n and we prove it works for n + 1.
Clearly,
, as after deciding which sn + 1 to choose, we are left with optimizing the value with n strings. (induction hypothesis is used here)
Now, all we need to do is prove that we need to put sn + 1 as the string with the highest
value. This is simply greedy, easy to prove.
The 1st problem made for the hack..
Idea behind C ?
For small Ks (probably <= 1000) you will find two exact values, so just implement
just do it with brute force :)
Ok. Limits are very large so how do you guarantee fast running time ?
it will find a pair soon
Thats what my questions is. How to prove that ?
It is known that (a very important consequence/equivalence of prime number theorem)
.
For n, k to be good, (n + 1) must be a multiple of
. So n must be like ek.
This should be good enough, as running time is
.
Or, say primes >10 will multiply, so it be over 10^18 fast...
Bruteforce works. :)
This was my approach.
gcd(1,2,...,k) = 1Therefore,
lcm(1,2,...,k) = 1*2*3*..*kAlso, since the remainders have to be distinct, one observation is that for any i between 1 to k, n%i should be equal to i-1.
This means:
n = i*C + i-1 ==> (n+1)%i = 0.Therefore n+1 should divide all the numbers from 1 to k. This also means that n+1 should be equal to the least common multiple of 1,2,..,k (in this case its just the product.) or any multiple of this lcm.
Looping from 1 to k and stopping the loop whenever product crosses
n+1. Product crosses n+1 easily by the time we reach i=50 in the loop. We return "No" in this case.If k! is equal to n+1 then we return "Yes"
try this
bool f;
int top=0;
for(int n=1;n<=100000;n++){
for(int k=1;k<n;k++){
f=true;
for(int i=1;i<k&&f;i++){
for(int j=i+1;j<=k&&f;j++){
if(n%i==n%j)f=false;
}}
if(!f){
top=max(top,k);
break; }}}
printf("%d\n",top);
just look at the output and you will find out why it is possible to do it in brute force
People have this idea:
Since you need to have all unique values till k. They should be obtained serially from 1,2,3,..k like this:
0,1,2,3,4,5...k-1so,
nshould be of the form:2x+1, 3x+2, 4x+3, 5x+4, 6x+5, 7x+6,...all at the same time. But if something is of the form5x+4.. it cannot be of the form2x+1.. (coz one is odd and other is even, for even x, similarly chose a pair for odd x)Edit:
The conclusion seems flawed. Maybe we can continue the logic like this:
Number has to be of form:
2x+1, so3,5,7,9..3x+2, so5,8,11,14..... ....It kind of creating a sieve of eratosthenes, which generates a prime number.
I don't what that gets us :)
Locking after being hacked
how to solve D?
sorting : )
merge or bubble
consider to string a and b
let the number of 's' in a be a.S and the number of 'h' be a.H
if you arrange a and b be ab
the you will get a.S*b.H pairs
so we know that if a.S*b.H>b.S*a.H then a should be in front of b
What was the hack for A?
if you donnot have a copy,you cannot get more copy
Most common one: x 1 (while x is a positive even number).
Expected answer: "NO" (as there are no initial copies to generate new ones)
0 0
0 1
1 0
3 0
2 1
2 1
output
NO
something!=0, y=0 and something!=0, y=1
Problem A just burn my room
What is wrong this this solution for E?
Let dp[i][j] be the max mana we have at tree i with j summoned in the trees before us. The max capacity is always w + j·b for a fixed j so we don't have to store that. I used long long int here and for dp table so no overflow. The answer is the largest i such that dp[n][i] is valid. I kept getting WA on pretest 5.
it's correct
Did you use sliding window maximum?
problem is that max summoned birds numbercan be ci * n = 103 * 104 = 107, so your array will have size dp[103][107].
EDIT: I didn't read task carefully :{
How to solve B? Seems easy cause lot of people solved it, but I only solved C and D.
Wow.. I'm a fool :) Thanks for your reply! (And all the others too)
can you please tell me how to solve C?
Try 1~50
n mod 1 , ... , n mod k has to be different. but you can easily notice that n mod 1 is always 0. So, n mod 2 must be equal to 1, because n mod 2 is 0 or 1, but it cannot be 0. ... n mod i must be equal to i-1, which means that n+1 is divisible by 1, 2, ..., k.
For a <= n, b <= n, let c = a ^ b, if all other conditions hold, then increment answer
a^b^c = 0 implies a^b = c. So bruteforce for a, b, and check if the 3 numbers can form sides of a triangle.
You can iterate through all possible cases of a and b. c can be calculated by bitwise xor:
As
a xor b xor c = 0, we can agree thatc = a xor b.If c is not lower than a or b, and not greater than n, and a,b,c forms a non-degenerate triangle (simply check that if all 3 inequalities of triangles are proved right), then you can add 1 to the answer.
Simply iterate over all a, b values
. Check a + b > c, a + c > b and b + c > a
I actually passed it with brute forcing which is O(N^3). It gave me 780ms even though it doesn't seem to be the intended solution.
Your condition ( c < a + b ) in the third loop in your code makes it about O(n^2) i belive
it is still O(n^3) with very small constant.
Test for hack C:
1 2Isn't it "yes" ?
Yes. Many people (me too) have writen:
if( n < k ) cout << "No";Bad luck, man
fortunately I had written n<=k so it failed pretest and I was able to correct it. and now i'm purple.
pretty sure this is the case that got me
How to solve F? I tried taking all N numbers initially and consecutively removing the one which will decrease f(I) the most but such that f(I) ≥ k will remain true. Just plain intuition, didn't work.
I bruteforced n<=17, implemented it for larger n, it worked well for all local inputs. Still kept getting TLE on the 7th pretest. I just wonder whether it was my complexity, I/O, special countertest designed to make this kind of solution fail...
Explanation — Problem F
Man, them noise and echo are simply unbearable which makes your speech unrecognizable.
Is it so interesting to post one picture several times?
In question D, what is the pretest 4?
I think there should be a limit on the number of hacks for each problem... Yeah hacks are a part of CF, but when a hacker with ABC has more points than a non-hacker with ABCD, I think there is a problem.
I think so too.
On today's A,same hacker hacks same participant many times…
For example,Hack 0 1 -> Hack 1 0 -> Hack 0 0…
At least,some fix is need to this system…
what is the output for test case 0 1???Isn't it "YES"??
Yes
For 0 0 and 1 0???
NO NO
Everything is ok. but how/why my code is hacked??
for 10001 1?
should be No
found "NO"
Is case sensitivity a problem?
NOvsNo?3 1 -> NO…?
found "NO"
4 1 should be NO
Can anyone give me some idea for 922F - Divisibility
will this work for very small k?
can someone help me find why i was hacked in problem C? 35018691
n=1 k=2 answer:yes
Short statements not means a lot of hacks.
This was the biggest slaughter since I started to participate at contests
omg, it was great. hope i'll see you contests later. GreenGrape, live long and prosper.
When will the system testing start ?
Update: System testing started :)
When someone asks how life is going on?
While everybody is discussing today's contest, I just found my love <3
Thanks Mike :)
Isn't it obvious that current hacking system sucks? It's a brilliant idea, but really bad realization. Hacking should give you points (maybe even more than 100) if you find hidden not obvious bug in someone's code. But if you just take n = 0 corner case in div2A and it gives you 1500 points that's so unfair: you get that much points not because you are smart or your debugging skills are better than average, but because you have gray guys in your room who always do this mistake.
That's why I think that hacking points must be not a constant, but a function, which depends on how much hacks there are on the problem.
if problem difficulty is 500 points so on one hack your receive 100 is 20% but if problem difficulty is 2500 points so on one hack your receive 100 is only 4%
til i am a grey
Agreed. I hacked 18 solutions (Problem A) using only two test cases And guess what? My own solution missed one corner case and I got WA in system test. But 1800 points gave me an unfair advantage.
Many among you would be disappointed because of all the hacks, but I feel that the problems were super exciting and I enjoyed all the brainstorming while getting hacked.
+1 to GreenGrape
When you find one of the testcases for A
I think that there is a problem with D testcases. This is my AC submission.
The problem with my solution is that the comparison is made entirely on integers, and therefore it could overflow.
I'll try to generate the testcase, I think that:
is enough.
EDIT: oh yes, I think it was enough. My solution gave 50000, but when reversing gave 2499950000. I don't think I'm the only one who failed for this.
That's actually quite strange since tests 11 and 12 are made exactly to counter this.
I would like to remember that overflow is an undefined behaviour, so anything could happen, even getting an incorrect AC (A guess for why this happens is that the compiler may do some simplifications like x*y/x = y — avoiding the calculation that gets the overflow — maybe something in that line is what happened)
I tested it using custom invocation, and it gives the incorrect output there (50k instead of 2499..etc).
Oh, I think I know what's happening here. My submissions treats different the strings consisting of only 'h' or only 's', so that's why my submission passed.
lucky you didn't loose 100 points on tc 11.:)
35024915 this solution gets tle on your test, if i am not mistaken. if so, D should be reevaluated
Looks fine to me. Total characters are 200000.
yeah, i realized that i tested it wrong right now, sorry
long long contest!
#define int long longaccept my C and D :)Can't believe my STUPID code for C got pretests passed even and finally failed on test 76th!!! I used k instead of i in my loop.
Failed.
vs
Accepted.
Couldn't proof the solution for C during the contest. But now, I think I got a proof.Correct me if I'm wrong plz.
If the answer for n is "yes", then for all i from the set [1,k] will have distinct remainders.
We can easily see that for all i [1,k]
n mod i == i — 1,
so, we can say (n + 1) mod i == i == 0 ;
That means all i [1,k] must divide (n+1) simultaneously.
Which means (n+1) must be divisible by the lcm(1,2,3,........,k).
Now the lcm(1,2,3,.........,k) would be > 1e18 when k is > 41(or something) .
That's why the brute-force solution worked.
ohuenniy round
ploxo materit'sa
who is sorry-haghani ?!
Haghani ...
I think he is Deemo !
it's possible ... :)
Can anyone explain? :( 35036774 35011492
never use >= or <= on a cmp function. this is your accepted code 35038659
<= got Accepted 35037198
could you please explain why should we avoid <= or >= in compare function?
you can read why here
Thanks :)
Oh thanks:(
don't know why but writing this line in cmp function gives AC 35038641
Solving the B in O(n^3) no problem... i would say a O(n^2) solution was intended but anyway: http://mirror.codeforces.com/contest/922/submission/35039399
lol. Does it pass with int as well?
yes... :p http://mirror.codeforces.com/contest/922/submission/35054737
Ironically,GreenGrape has never been green in his/her entire Codeforces history.
Why is my code giving runtime error on test 5 in D : Link
This was one of the best and most interesting contests on codeforces that i have given. So much of hacks and good difficulty distribution i should say ,awesome GreenGrape
раунд очень годный был
GreenGrape has never been Green on Codeforces.
Actually he was during new year magic
I don't understand A number problem. please, anyone, help me to understand this problem. problem link : http://mirror.codeforces.com/problemset/problem/922/A thanks
you start with
o. Then you do magic copy to get additionaloandc.Now you have :
o,o,cNext you do magic copy on
cand get additionalcandc, so a total of 3c,c,c.Likewise.
http://mirror.codeforces.com/contest/922/submission/35048494 why i am facing runtime error thanx in advance
try swapping these statement in your comp function if (!h1) return true; if (!h2) return false;
But what is logic behind this swap statement
If 2 equal objects are passed it should always return false. before it could return true.
Problem B, O(n^3) without pragma 35154154....using pargma only make it a little faster..35129118
In Problem B, why (1,2,3)is not a valid set?
Because it's a degenerate triangle (1+2 = 3) the