It's been a long day without you, Codeforces!
We are glad to invite you to take part in Codeforces Round 969 (Div. 2) and Codeforces Round 969 (Div. 1), which will start on Aug/30/2024 17:35 (Moscow time). You will be given 6 problems and 2 hours and 30 minutes to solve them in both divisions.
The problems are authored by LMydd0225, tzl_Dedicatus545, _FJqwq, _TernaryTree_ and me _istil and mostly prepared by LMydd0225. One of the problems is also prepared by CleinCc.
We would like to thank:
- Artyom123 for his intelligent coordination, and for Russian translation.
- arvindf232, dXqwq, A_G, Gellyfish, N_z__, Tony2_CF, abc864197532, PEIMUDA, hyman00, yuanruiqi, Imdie, Portulaca, superyijin_txdy, SATSKY_2025target_LGM, Error_Yuan, WRKWRK, cfhj, Clan328, cel.in, Misuki, Romy67, misorin, yangchang, Anny, Yoav, turgon314, SF1yer, oloy, Eclipsar, Leonardo_Paes, flakes24, deneribeiro10, fIr0a2t for testing.
- MikeMirzayanov for Codeforces and Polygon platforms.
Hope you will enjoy the problems!
Score Distribution:
- Div. 2: $$$500 - 750 - 1000 - 1500 - 2250 - 2750$$$;
- Div. 1: $$$750 - 1250 - 1500 - 2000 - 2500 - 3000$$$.
badges$$$^\dagger$$$ of the authors:
$$$^\dagger$$$: Began in 2022 at NOI Shanghai, Badge Exchanging has been an activity popular among Chinese CPers. The host school for NOI 2022 decided to print badges containing the avatar of contestants for the contestants, and one can exchange his or her badge with another. The collection of one's badges (of other CPer's avatars) shows the group of CPers that he has met and even made friends offline. It has achieved great success. And since then, it has been a tradition for years.
UPD: Congratulations to the winners!
Div. 1:
- tourist (Updated, and congrats!)
- jiangly
- ecnerwala
- Radewoosh
- Kevin114514
Div. 2:
- mkrukov07
- temp6
- wjq1234567
- bingpao
- AuSquare (It seems that the rank 5 was banned? So let's move on to rank 6!)
The Editorial is out. You can also download Simplified Chinese Editorial in the contest material of the Div. 1 part of the contest.
thank you for no subtasks
Subtasks are good though because if you normally solve n problems with subtasks you you might have a chance to solve n+1 (or n+0.5) problems. And normally I have noticed the first subtasks of problem n is either of the same level or slightly easier than problem n-1.
As the girlfriend of tzl_Dedicatus545, may you good luck & give me contribution!
Love you!
So cute to see <3
pedo....
nothing, nevermind
You two even used almost the same template: this & this
That's interesting :)
Why my girlfriend can't use my template?
Maybe you should use "can't" before "my"...?
hehe, china
hope this contest is good!wish me luck!
Score distribution looks really odd. I think it would go something like this, wouldn't it?:
Btw, the badge custom is really cool, I wish I could have done the same when I was younger T.T
As one of the authors, good luck & have fun!
As a tester, the problems were good and I recommend everyone to participate!
Why are you lying
I found C and D to be good problems.
As a tester, the problems were good and I recommend everyone to participate!(
I tested this round, and I hope you all enjoy it!
As a tester, hope you have fun!
As a tester, when my solution has a different output from the example during testing, I suspect the author first.
Fun Fact: The solution of Div.1 A was initially wrong when I tested the problem.
Fun Fact II: 1A was in the position of 2C before we find out the mistake.
Fun Fact III: There was another 1A before the current 1A was presented, and it was also wrong.
I'm guessing this is referring to the sample with the same number of '1' and '0' leaves + '?' root + odd number of '?' non-root vertices? I feel like there would be a lot of WAs without that sample.
I tried proving that that was the only "edge case" but just gave up and submitted after several hundred submissions AC'd :/
I'm not even sure what you're supposed to do to solve that problem. Just be smart enough to prove it correctly and quickly, or assume that the samples are strong (which to be fair does seem to be a trend in As)?
Not as a tester, when can I obtain your badge _istil
As a tester, I have two badges of one author but none of the other five XD
Time to get a color change in this contest XD
As a tester who hasn't participated a contest formally on Codeforces for nearly an entire year, I hope all of you a positive delta.
As a tester, it's a very good contest and hope u have fun!
The badges are so cute!! Reminds me of trading Pokemon cards :)
I wish I would meet many CPers in real life and exchange ideas too. Also, good luck to all!
hope to reach 1550
As a tester ,wish you have fun and gain positive rating delta.
I hope my downfall will come to an end this week
As a tester, I wish all contestant having fun in this contest.
What cartoon is this.....,,,??.???,???
is there any correlation between being weeb and high rating?
YES
Chinese students are happy to set their profile pictures as anime girls. And so am I, though I don't have high rating lol
you are not Chinese enough
but you are afghan enough
noob sag
So as the photo says Problem D will be a graph problem?
So the C will be need so much attention to be found?
Anyone know the sauce for the top right badge? Also interested in the others too. I think top middle is from "Bocchi the rock"
top left is oshinoko
Yes, top middle is my boy friend's. About top right, _FJqwq himself don't know the source either :(
FreeDurov****
sorry everyone
Were rating constraints for div 2 changed? Pretty sure it used to be 0-2099 yet I can't register(and don't see anyone with 1900+ rating registered).
When Div.1/2 are hold separately, purple would belongs to div.1
Separated Div.1 contests only requires 1900 although they're not very common now.
It has been 4 months since last separated div.1 contest. Interestingly, last time we also have _istil and N_z__ and some other people as authors and testers :D
badges looks nice I hope I can get one someday
All the best everyone. This time +100.
Based Kana Badge <3
Missing Ai (┬┬﹏┬┬)
Hope to solve ABCD for once in contest time. All the time I drop at C or B because of panic WA or TLE or whatever feeling is running in my head...
tourist gonna reach 4K finally!
As a tester, I missed the opportunity to compete officially in a great round! I hope you enjoy it!
The idea of badges is incredibly cool. I believe it should be extended worldwide.
The badges looks so cool!!!
Just curious to know if someone submit the same code in div1 A and div2 C is there a chance to get both his contests skipped.. like is the checker gonna check all codes from both the round together and search plagiarism or check the rounds individually?
you can not give div1 if you are less than 1900, and you cant give div2 if you are rated more than 1900 if both happens at the same time, thus you can't give them together
I'm hopeful this contest will be both fascinating and enjoyable for everyone. Let's aim to boost our ratings and have a great time. Good luck, everyone!
As a tester, the contest is the GREATEST round I have ever seen!
I really want to know why this comment is being downvoted
As a tester,I just realized that it's the one I've tested and my rating-boosting dream was rained(T_T)///GL&HF everyone,not bad probs
As a tester,I tested the round.
As the boyfriend of the girlfriend of the tester yuanruiqi, may you good luck & have fun!
Now everybody knows you have a girlfriend/tuu
Now everybody knows you also have a girlfriend
who asks？
Then where is my boyfriend /ll
here/se
I love u _istil
Good luck & have fun!
My first Div1,let's go!!
Same I hope my rating isn't obliterated.
Wanna VC a div1 in preparation?
my first separate Div.1, let's go!!
Hope it wont be bad for my first div1.
These Badges look awesome. How to get them?
How can one find cool avatars like these for CP accounts? Also, I hope the badge exchange tradition can make its way to international events... Great additional incentive to try reaching them.
Watch more stuff, I think. Anime is a wonderful realm to explore (sorta).
Watch more anime, gotcha
You can also use the photo of yourself (no) You can also find some album covers (yes)
The badges idea is so cool!
Wish to have a good problemset..
As not a tester, I hope to test it live
tourist just signed up... 3947 -> ?
Rooting for Tourist to 4000!
Will tourist have a 4000 rating this time?The fate is coming.
my classes end at 6pm, it'll take me 90-120 minutes to get home, I'm not missing this one, I'll make it!
let get pupil rank with this one! cuyu fighting!
its hard to get 231 points u know :)
hope tourist will reach 4k
...
I watch anime, too. Why can I reach the level as the authors TAT
as a participant, hoping to reach $$$1900$$$.
tourist winning this round is going to feel like Messi finally winning the world cup.
tourist 4K incoming!
Mathforces incoming.
wish to cross 1800
The grandmaster who commented its going to be an extremely chinese round was right. This is basically math olympiad not programming contest.
first time ever I decide to unrate contest after registration by making no submission.
problem A is wow, just wow, what the f**k is that
How to join, where is late registration?
ARE U F KIDDING ME ????? WHAT DIV 2 A ????????
There is an option to unregister, can i unregister now?
I thought once the contest starts no action can be done regarding registration.
There is no "unregister" but if you register and don't make any submissions, it will be as if you did not register
A: Guessforces B: I'm dumb for not reading the statement right C: Pure arithmetic, not really algorithm really
Naming convention for B sucks. l and r is really misleading... should have used l/u for lower/upper bound or something.
I think it was intentionally named like that just so that people would go the wrong path of using segment tree lol.
true I started doing diffrence array method to calculate max then read PS again after 2WA
jiangly has solved all before tourist :( so no 4000 for touristsorry tourist for commenting too soon
.
I start hating cp because of this shit
and we all know that after testing not all cheaters are removed
he did it
tourist solved F!!!
Chinese rounds always shave off a few days of my lifespan, but I still keep participating in them
What a Contest tourist is ahead of jiangly by 5 points only
And even for a period of time jiangly is the only aker!
Nice round. No long queues, no Cloudfares.
Most importantly, we witnessed HISTORY — 4000 rating for tourist!!
tourist has finally become the first person on cf to cross 4000 rating barrier
tourist orz
tourist 4000 rating !!!
Is C really that easy? 5k solves...
Even DuongNeverAlone found it difficult, but a lot of newbie didn't , lol
The fact that I used $$$1$$$ hour in C before using the rest of the time for E, but no use. Come back to specialist soon.
it is really just a normal div2 C but its using arithmetic not algorithms so he might not be good at it
Not really. My performance went down significantly this month and I got my chain of 2 -> 3 problems solved each contest (which is pretty low if I want to maintain 1k7 -> 1k8).
the problem A give away gcd idea. Hint is if you rephrase the problem into doing +-a or +- b for every c[i]. What's the min(max(c)-min(c))?
After rephrase it you realize you can reduce a,b into just 1 number by taking their gcd. so the problem is +- gcd(a,b) now.
Another hint is you should mod every c[i] for gcd(a,b). This is the pigeon intuition something(I dont remember lol, but the main idea is putting n pigeon into m closet. If (m<n) then 1 closet must have 2 pigeon.
Then the rest is brute force over the sorted array O(n) times. Overall complexity is NlogN
How can u do -a or -b?
every c[j]+=a (j!=i) and leave c[i] be. In that way you can see the delta is -a
orz tourist .. the legend has finally made it.. cheers
Please tell me I'm not alone by being bamboozled in this div2A...
Cost 30 minutes for coding gcd and brute force and a lot of time on thinking about optimize just to realized you are being bamboozled (but after this I can solve C in the right path! So maybe not bad at all)
Guessforces ;)
its just skill issue, not guessforces
I used all of my mental strength on ABC and run out of juice for DEF... truly skill issue
The proof is quite obvious though, because we know that the gcd of any two even numbers is not 1.
tourist orz...
tourist orz
Poor Jiangly! Congratulations to the tourist with a rating of over 4000!
I was able to figure out the condition in Div1C/Div2F but couldn't think of how to implement it, Consider the set of elements in the subarray l to r, if we were to take the gcd of the consecutive differences of the elements in the set then the gcd must have a odd prime in its prime factorization, In this Case, the subarray is not brilliant.
Got the idea maybe, We need to take difference of consecutive elements in the array and then use two pointers right?
Basically, but you need to have a way to efficiently recalculate gcd when removing an element (segtree works, sparse table should work as well, there may be other ways), and make sure you don't screw something up if all of the numbers in a subarray are equal.
What?
I reached that for $$$n=2$$$ an array can be brilliant if the difference between the two elements is a power of $$$2$$$
Intuition :
Let the number of elements between $$$a_1,a_2$$$ be $$$Z$$$, I choose the middle element, then half of the remaining elements on the left will be separated from the other half on the right of the chosen average.
So I went from difference $$$Z$$$ to difference $$$(Z-1)/2$$$, and I should always be able do divide $$$Z-1$$$ by $$$2$$$, so $$$Z$$$ must be a power of $$$2$$$ minus one, so the initial difference must be a power of $$$2$$$
$$$2^n$$$ does not have any odd prime O_O
I think he meant that the array is brilliant iff the gcd of differences doesn't have an odd prime factor (which is true for all subarrays that have at least 2 different elements; all numbers being the same is a special case).
Yep
tourist orz...
Solve div2D at 02:27:30... This contest is really tough for me.
tourist orz
Yet another "Found bug 30 seconds after the end" contest for me... The problems are great though.
Same, I found my error for Div2E 2 minutes after contest ended. If I had found my error 2 minutes earlier I would've likely gotten CM :(
can someone explain why this solution does not work for div2 problem D
https://mirror.codeforces.com/contest/2007/submission/278834900
isn't that all we need to do is determine if the root has a number? if it has, we only try to give numbers opposite to root's, else we can determine weather iris can give her turn to the opponent
If I understood your code correctly, your way of checking whether Iris can pass the turn (the (qm&1) bit) uses the number of all question marks instead of using only the number of question marks in vertices that aren't leaves or the root (playing in other question marks doesn't pass the turn).
yes you understood correctly but before this operation i put qm -= lq + 1 i subtract question marked leaves and the root
Whst if the root is not question marked?
If the root doesn't have a question mark, qm isn't used at all (in fact, this change only occurs if the root has a question mark), so it doesn't matter.
Oops, my bad. I'm pretty sure I actually caught the mistake now: ln + (lq/2) + (qm & 1) should actually be ln + ((lq+(qm & 1))/2). When you have an even number of leaves with question marks, it doesn't matter if Iris passes until Dora changes the root or if Iris changes the root herself, she can always change the same amount of leaves with question marks (lq/2).
ahh, i got ac with that
when 6 seconds left i submitted with cout << ln + (lq/2) + (qm % 2 && lq != 0) << '\n'; but it didn't work. i guess i noticed the problem a bit late. thanks.
What was your thought process for realizing that Iris can "win" an extra leaf by "passing" turn to opponent sometimes?
actually one of the testcases forced me to realize that, i see if we play first, Dora can immediately give the same number.
Ah yeah, I got that pretest wrong and my mind blanked... I was so convinced that Iris could only score 1 even after drawing it on paper.
I guess it is important to observe that the first move is "useless," so by moving first, you are actually at a disadvantage.
how to solve d1C?
I couldn't submit solution because of a bug, but it looks like the condition is $$$GCD(|a[l + 1] - a[l]|, |a[l + 2] - a[l]|, ..., |a[r] - a[l]|)$$$ must be a power of $$$2$$$ (or $$$0$$$). It is the same as $$$GCD(| a[l + 1] - a[l] |, | a[l + 2] - a[l + 1] |, ..., | a[r] - a[r - 1] |)$$$ and then you need a sparse table to find these values fast for each left position $$$l$$$.
wow, thats cool! can you plz explain why it works?
Consider all possible values of $$$A[j] - A[i]$$$ on a segment and find their GCD $$$g$$$. If $$$g$$$ is $$$0$$$: all elements are the same, segment is good. If $$$g$$$ is a power of two you can just fill all the gaps between the numbers so that each of them is in the set. If $$$g$$$ has any prime divisor $$$p$$$ rather than $$$2$$$ -- all elements will have the same remainder modulo $$$p$$$. Since $$$p$$$ is odd, that means for all adjacent numbers in the set $$$x + y$$$ will never be divisible by $$$2$$$.
And then you can just prove that GCD of all pair-wise differences can be found as the GCDs I've mentioned above.
thanks!
You don't want to have $$$a[l]$$$ in there: if your subarray was $$$[2, 8]$$$, this would give you gcd $$$2$$$, which is a power of $$$2$$$, yet the subarray isn't brilliant. Otherwise, yeah (also, you can use a segtree instead of a sparse table, but that doesn't matter).
Thanks, good catch! Fixed the formulas.
great problems! sad that limitations in E are so high :( didn't have time to figure out how to improve $$$O(n \log^2 n)$$$ to $$$O(n \log n)$$$.
Same (i figured out while impelementing nlog&2 in last 15mins but no time to switch then? ), but nlog^2 passed in 1s
We had no way to make $$$\mathcal O(n\log^2 n)$$$ TLE under the 4s constraint, but IMO it is harder than the intended $$$\mathcal O(n\log n)$$$, so I think it is unnecessary.
And it seems that $$$\mathcal O(n\log^2 n)$$$ solutions are just with 2 logs, it cannot be improved to 1log...? I don't know because I am not very good at data structures )
it can be improved though? I can keep a frequency array of dimension log instead of a segtree
I think if you can't tle something (my code wasnt even that optimized and it passes in 1s), consider letting it pass freely. IMO its more fair this way
Also if you have a template of segtree, nlog^2 is easier (atleast to me). nlogn needs additional ideas.
I had centroid + std::sets. TLE test 9. But I submitted at the last minute, maybe I would be able to optimize it if I had more time.
Mine was super quick log^2
Just a for loop + iterative segtree
I guess I can replace an std::set with an iterative segtree too... Probably should've done that.
Intergalactic Grandmaster incoming?
GOD tourist
Was anyone able to pass HLD in Div1B?
The intended solution wasn't HLD but simple segtree manipulation apparently :( I also tried HLD initially.
I don't think it's segtree, my sol was O(n)
How is tourist that invincible?
how to solve C?
the funny thing is that it has 5k AC
it's not this easy to get all of this but thx to cheaters actually
It's about average div 2 C though :/
Maybe get better?
bro, u haven't seen the number of submissions in the last 30 minutes for C & D
and for sure I already solved it in the first hour, but seeing this is too annoying
I agree, I just think it was nothing unusual compared to other contests. It's, like, the stable amount of cheaters, The Horde of Disgrace.
First let's imagine you have only A, not A and B. Then you can fix the difference between values if it's not smaller than A. e.g. A = 5, values = 5, 10, you can make diff = 0. But if it's 5, 11, the lowest diff is 1.
Second, now we have A and B. If you have A = 5, B = 10, then B is useless, because A * 2 = B... or, actually, because GCD(A, B) = A! So now we have a clue. If GCD is 1 (e.g. A = 5, B = 9), we can ALWAYS make a difference of 0!!
The idea: calc gcd(A, B), then for every number get it's Value % GCD(A, B). Maximum of this array — minimum of this array is your answer... UNLESS...
Let's say GCD is 7, and you have values 1, 6. But 1 + 7 = 8, and 8-6=2 is better than 6-1=5. So now you need to actually iterate the array of values % gcd(a,b) you got before to see if you can optimize the solution by adding to the values[i] your gcd. Though to understand this there were steps in between that I have missed in the explanation.
If no fst.Tourist will get the first 4000 rating. Congratulations.
The 4-th sample of D1A reduced it's difficulty a lot. Otherwise it's problem rating could increase by 100.
Yes! I debugged using that case. Maybe authors want to make A a bit friendly (?)
I've never experienced so much skill issue solving a div2A q.q
Experience with GCD problems was required for the A, I think... and for the C as well... GCDforces.
Funny thing is that I've immediately noticed bezout's identity on problem C but couldn't figure out that any 3 consecutive numbers, with the first one being odd, are coprimes (fro problem div2A) :/
How do you solve C with bezout's identity. I just know it means ax+by=gcd(a,b). For C I realized that eventually the difference between possible linear combinations of a, b with different positive x and y integers converged on gcd(a,b) if you sort all the unique values that were generated. So you get a difference array that will eventually [....,gcd(a,b),gcd(a,b),gcd(a,b),...,gcd(a,b)], just be gcd(a,b) at the end.
You actually used the theorem. It state's that gcd(a,b) is the minimum positive value we can get from a*x+b*y with x and y as integers. From what I understood of your comment it seems you just did not fully understand the theorem, and treated this as an unrelated observation. Sorry in advance for any misunderstanding.
I will explain to you my solution for the div2C problem. The first step is to understand that, by using bezout's identity, we can add or subtract (by adding it to every element except one) a multiple of
G := gcd(A, B)to the elements of array C. That means that we can shift backwards all elements of C so thatC_i := C_i mod G. Sorting all elements (C_i mod G) and keeping them distict, it's a matter of asking if there is a prefix that is useful to shift forward by G (for example, if I had the elements 1 mod 5 and 4 mod 5, it would be useful to see the 1 mod 5 as a 6 mod 5, making the solution 6-4=2 instead of 4-1=3), which means finding the smallest[ C_i + G - C_(i + 1 mod N) ] mod Gin O(N).bro just please tell how to get intuition of shift back thing or minus thing it's like the operation is plus how did you know that minus won't make any differ this is very hard to observe for me
Honestly, when I first read the problem I assumed that you could subtract, only after the contest ended I noticed you could only use addition lol. You could get the intuition for using subtraction by noticing that it would really be useful if you could subtract A and B, unlocking bezout’s identity, it’s just a matter of justifying why is it possible.
I got rekt in A just to have a better start at C, so maybe it's fair trade in div 2 after all
lost a ton of rating but atleast it went to good cause
Finally 4000 gonna be breached
Some very cool problems, thanks a lot to the organizers! :)
Passed the example of D 5min after contest ends...
f**k me and my understanding for B , dammnn
The little boy from Belarus on behalf of every little boy wearing his shirt. Tourist on a million backs, Tourist for a million flashbulbs.
Gennady Korotkevich aka Tourist becomes first ever competitive programmer to cross 4000 rating mark on Codeforces.
Greatest programmer to ever exist!
A and C were uh, peculiar problems, but very well written and thought out.
hardest Div2A I've ever seen. 40min of thinking with 0 progress .. euler totient? integer sieve? how it is possible
(count of odd numbers)/2
I spent 30 minutes on brute force and gcd and random stuff... until 31st min just print odd_count//2 and submit
just brute force
that was my first approach and obviously didn't worked. take first one, remove, find first one that is coprime, remove, find last one that is coprime to first and second.
orz tourist
Hints for Div2 C?
this
Can someone explain why I see this?
It's Chinese editorial. English editorial will be published soon (probably since the translations are done).
I had the logic for D but my bad implementation trolled me :(
Yeah, kinda sad there's no in between in codeforces (it's not like it's really possible to implement such system in a good way though), I'd imagine a half-grade for the idea... but when you have one and can't implement it (because there's some case you are missing despite the fact it's working), you get nothing, it's heartbreaking.
i could implement it but 1 error in a part of the code cost me the wa in pretest 2
I edited my answer, I've meant exactly this. have this case:
GOAT tourist ！！Can someone tell me the rule that the problem score in div1 decreases with time?
Look at this blog
I've spent a significant amount of time trying to solve Problem B , then I realized the problem I am trying to solve is not B.
How bad a problem statement can be? => This bad.
Same lol
As a competition on Codeforces, I believe there are too many data structure problmes in this game. Both D1D and D1E require some data structures to maintain, and most of D1C's practices also require the use of data structures. These questions are not difficult in terms of thinking, but they require a lot of time to write code. I don't think this is in line with the style of most Codeforces competitions. At the same time, D1E allows the time complexity $$$O(n\log^2 n)$$$ approach to pass when there is an $$$O(n\log n)$$$ approach, as long as you write it well, but in reality, this approach is not difficult to think about, only requiring some template code to be combined and slightly modified. I think this game is more focused on testing code implementation ability rather than short-term thinking ability. If it is played under the IOI format, it may be a qualified game, but it is not suitable for Codeforces. Or, in other words, it pioneered a new style of codeforces div1 competition ;)
nvm
D1D doesn't require a data structure other than prefix sum. My code for C and D was shorter than for A and B, with some thinking you don't need to use any data structures and the implementation is quite short for both C and D.
Releasing Chinese editorial before the English one??
Maybe only me got FST on Div1, Ofast + unroll-loops maybe harmed me >< 278842172 278846221
During the round I got PT passed and PT=ST but I got WA-FST, very strange...
Thank you, tourist, it was the best and most epic live broadcast of my life!
He was live? where? I mean on which platform?
I meant my regular updating of the results table :D
Too much math to be honest.
mfw I'm 5 minutes away from AK div 2 ToT
Aug 30th: World tourist's day.
nice
Congrats!
Congrats tourist for making history 4000+ ! UPD: Congrats on reaching Tourist.
tourist you are a living legend. Congratulations!
nice, only one with this color username
Congrats , Orz
Congrats for new title "Tourist"
nice
Congrats on the achievement!
Will you be participating in rating contests now?
Tourist thats Crazy
Orz lord tourist
congrats!!!
Man, the joy I had was one of the highlights of my cp journey. Congrats.
it's cool to see tourist becoming Tourist.
let's see can anyone reach Tourist...
Tourist, in a league of his own (quite literally)
Congrats!
I think people are more excited on tourist reaching 4000, than tourist himself
For me it is, even if I drop 41 points in this contest.
MANNNNNN I got cooked so badly :( All because of Div. 2 A
The worst Div2 in the whole summer
My dream is lgm ✖️ My dream is Tourist ✔️
The Tourist has passed Codeforces; he needs Codeforces 2☠️
The contests was epic. I haven't seen such good problems in a very long time.
Maybe it's better to update the post so that we can see his new color.
Why do Chinese authors like multiple test cases so much? It almost reminds me of the days when I worked on HDU problems all day
It has nothing to do with Chinese authors, pretty much every CF rounds uses them a lot. Making strong test cases without the multiple test cases per file can often be very hard, so people use $$$10^4$$$ test cases per file to put every small case in one file to catch WAs a lot more easily.
Currently, it is a rule of CF that you must use multi-tests unless it affects the time complexity.
Thanks for your explanation, that's my ignorance.
Guess difficulty
Div2A — 800
Div2B — 1000
Div2C — 1400
Div2D/Div1A — 1800
Div2E/Div1B — 2200