Hello, Codeforces!
It's been a long journey, and I'm finally pleased to invite you to participate in our Codeforces Round 955 (Div. 2, with prizes from NEAR!), which will take place on Jun/25/2024 17:35 (Moscow time).
This round will be rated for all participants with a rating is below 2100. Participants with higher ratings can participate out of competition.
During the round, you will need to solve 6 problems. You will have 2 hours to solve them.
The problems for the round were prepared by nik1998, egor4444ik, iamdimonis and me.
We sincerely thank everyone who provided invaluable assistance in preparing this round:
- Our insightful coordinator 74TrAkToR for excellent communication, helpful advice and assistance in problem preparation!
- Kolychestiy for improving one of the problems.
- EJIC_B_KEDAX, A_G, zwezdinv, Dominater069, culver0412, TheScrasse for red testing.
- Kolychestiy, CSQ31, dope, sadness, harsh__h, meowcneil for yellow testing.
- onlytrall, veleboks, zarubin, robotolev, TimVen74, Vladosiya for purple testing.
- Alenochka, LightSky, Chugunov_Alexander, slash0t, PoDuReMaN, katzi, Vamperox, redpanda for blue testing.
- Matrosk1n, bessarab, dariasaligina, GRIVA_, viteli, dazlersan1 for cyan testing.
- Kirillar2, _supersonic_, dan00ile for green testing.
- venlesN for gray testing.
- MikeMirzayanov for the wonderful Codeforces and Polygon systems.
We are pleased to announce that NEAR has supported the round!
NEAR was founded in 2017 by Illia Polosukhin, one of the creators of Transformers, and Alex Skidanov as an attempt to build an artificial system capable of solving competitive programming problems. You can read more about that attempt here.
Ultimately, NEAR pivoted into building a blockchain protocol, which it launched in 2020.
This year, NEAR started NEAR.AI, a new lab with a mission to build AI systems that are open and available to everyone, instead of being controlled by a few mega-corporations.
One of the areas of focus is making models capable of reasoning reliably, and for that, competitive programming problems provide a great environment. To help us build it, we invite all the Russian-speaking members of the Codeforces community with a rating of Specialist or higher (1400+) to help us annotate step-by-step explanations of solutions to competitive programming problems. We want to annotate a very large set of problems of all difficulty levels, and pay relatively high rewards in NEAR per annotation. Don't speak Russian but speak English? Stay tuned here, we will be launching the same project for English-speaking people very soon, and will likely sponsor another round when it happens.
The round also features prizes in NEAR. Participants who rank in the top 16 will receive Ⓝ 16 each. The next 32 participants in the overall ranking (including unofficial participants) will receive Ⓝ 8 each. The following 64 participants will receive Ⓝ 4 each; the subsequent 128 participants will receive Ⓝ 2 each, and finally, the next 256 participants will receive Ⓝ 1 each.
Additionally, 64 random participants from the top 4096 in the overall ranking will each receive Ⓝ 4.
Score distribution: $$$500\,—\,1000\,—\,1000\,—\,1750\,—\,2500\,—\,3000$$$
Wishing everyone good luck and high ratings!
UPD: Let's continue the series of announcements with a photo of the authors :)
UPD 2: Editorial!
UPD 3: Congratulations to the winners!
Div 1:
Div 2:
UPD 4: Wonderful video analysis of A-E tasks. Thank you, Shayan!
Just curious, are the prizes based on full ranking or just official/Div.2 ranking only?
Prizes will be distributed among all participants of the round (including unofficial participants)
Why not just wait for a combined round instead of awarding div1 participants for solving div2 problems.
Even if div1 participants were excluded from prize list, they would have made an alt and took the prizes anyways so ig it doesn’t matter
You could just say "official participants" to reduce the number of alts but that's not my issue with this. If you're going to give prizes to div1 the contest should be at least a bit challenging for them. It's their money I guess.
Could someone please explain what's Ⓝ in the prizes?
It's NEAR's own cryptocurrency:
You can read more about it here
Currently 1 Ⓝ = 5.35 USD
I am just going to ask the same question.Thank you!
You are welcome bro
Is anyone going to pull a rainboy?
Hello everyone, hope to reach CM in this round
u mean expert ig
No I mean it
i like it
bro is already expert lol
Hope to reach expert in this round
As a tester vodacbaoan is genius
December 2023: "AlphaCode 2 is worth a gazillion specialists, competitive programming is done for!"
Fast-forward to present day six months later: "Sup nerds, wanna annotate some problems?"
I am 2100 and I can’t register officially :(
Oops, let's fix the blog :)
Hey Authors
I'm curious to know all your ages/experience? since it's great to find people with 2-3 years of experience building contests!!
Hello :)
We are 20-year-old students. Contests are done only under the dedicated guidance of our senior friends)
you guys look 10 at best
As a tester problems are really interesting and challenging!
As a tester, I literally tested just now because the coordinator invited me to test a few hours ago.
I will come back after a long time not participating in contests
Get ready ;after a while you can't participate like past
OMG! Another FelixArg round.
The expression on the face of boys explain how dangerous their mind is.
What is meant by blue testing, purple testing, etc...?
they mean problem color as per rating (eg 1200,1300 green problems) if i am not wrong.
no, they "purple testing" means testers are purple. "blue testing" means testers are blue. and so on.
i remember you guys removed 24-word seed restore in the default web wallet and my near coins were gone, very funny. then someone told me i have to paste it in the url to restore it. please be careful about backwards-compatibility, would hate for the new ai to suddenly vanish. looking forward to nearcrowd round 2!
weird score distribution, I've never seen 1000 point div2c
thinking speedforce, unless upsolve D of course.
64 random participants from the top 4096 in the overall ranking will each receive Ⓝ 4, that means we have a 1.5625% chances of becoming rich! All the Best!!
$20 is rich? 🤡
$20 richer!!
it is equal to 20 meals here .__.
Rich enough to:
Buy live contest solutions :)
interesting score ditribution
Can someone help me ? How am i supposed to change my compiler from c++14 to c++20 (in my own laptop) ?
You have to download and setup a newer version of GCC. Here is one of the places you can download it, under Release versions. Then you have to do the usual, adding it to system path.
is there any penalty?
Me after solving D : (^_^) Checking standings afterwards : (o_o)
nice tasks, 2k submissions on D wow
thx to cheaters actually :)
I'd say it was not that hard of a problem, albeit a bit scary looking. Once you realize that you only need to get the difference of zeroes and ones in each kxk matrix to get the impact of an operation then it becomes a fairly simple number theory problem
B > C
Animated Video editorial for problems [A-C]:
https://youtu.be/-btzjr-u4dM
problem D must have something to do with 2D prefix and maybe some gcd yes? I believe
yes
C<<<B
how to solve C? i tried queue, greedy, pretest 2 fall
discard all a_i > r
your right pointer is the index you are currently at. you have left pointer lp. you keep the sum of subarray [lp, i]. if this sum is greater than r, keep moving lp one to the right and if it holds <= r for its sum at some moment immediately take all the cards at [lp, i]. if it holds < l, you have to proceed to the next index i + 1. 267353736
indeed spent 1 hours to solve B and C under 10min
Can someone tell how to solve D?
My thought process for D : Find net difference between two types of mountains, this will give the value to increase. Now in each kxk submatrix, find difference between the number of two types of mountains : this will give the amount we can increase or decrease. And then I couldn't proceed further.
That's correct. Then you can use GCD to figure out whether you can sum the list of kxk difference into mountain_diff.
The difference should be a linear combination of the amount you can increase by each of the k*k matrices. Hence, it should be divisible by their gcd.
same, my approach was to first calculate sum of ones and zeros and get difference between them , now create prefix 2d array to store count/(difference between them) of ones and zeros in a matrix then for all possible matrix if dimension kxk save difference between one,zero in an array so now my problem boils down to classic Coin Change DP(with infinite coins of each arr[i]),but it takes O(sizeofArray*Sum) probably TLE
any hints please ??
Suppose the difference of the numbers of each type of mountain is $$$k_1,\dots,k_\ell$$$. Recall from number theory that the only numbers we can form as a $$$\mathbb{Z}$$$-linear combination of integers $$$k_1,\dots,k_\ell$$$ (i.e. $$$c_1k_1+\cdots+c_\ell k_\ell$$$ for integers $$$c_i$$$) are multiples of $$$\gcd(k_1,\dots,k_\ell)$$$. Therefore, if the difference is some multiple of the $$$\gcd$$$ of all the $$$k_1,\dots,k_\ell$$$, then there is some series of moves we can do to get the difference to zero (plus some edge cases where some/all of the $$$k_i$$$'s are zero).
I was knowing ax + by = c has solutions only if c is divisible by gcd(a,b) but was not knowing this is true for any number of variables.
If you have $$$a_1x_1+a_2x_2=c$$$ iff $$$\gcd(a_1,a_2) \mid c$$$, then you can write (for some $$$v$$$)
and you can continue by induction :)
I didn't know this. If it's so simple to know if a sum can be obtained by some combination, then why don't we use this property in subset sum problem or coin change problem?
It's simple because every item can be used for any integer number of times, but it's not the case for subset sum problem.
This will only hold when the coefficients can be any integer (including negative) ?
Yes. For example, with $$$2,3$$$ you can form $$$1$$$ with negative coefficients as $$$1(3)+(-1)2$$$ but clearly not with positive (the result would be greater or equal to 2!)
It would hold if the coefficients can be any integer. But the reverse direction may not be true (i.e. it's possible to get all multiple of $$$gcd(a_1, a_2, \ldots, a_n)$$$ even if some of the value can't be used for any integer number of times).
Is there a name for that theorem or is it just a common concept?
It's a generalization of Bezout's identity.
This theorem also holds for all Euclidean domains (in particular, for polynomials in $$$k[x]$$$ for any field $$$k$$$!).
You can find the net difference in all $$$K \times K$$$ submatrices and let gcd of these differences be $$$G$$$. It can be proven that you can only change the total height difference by some multiple of $$$G$$$, therefore you only need to check if the height difference is divisible by $$$G$$$.
How to do B?? Spent more than an hour and couldn't solve it.
After a while, the value of x just starts to loop in between 1 and y-1 . So just find out the position at the end making use of some %
Tried something similar to that.
I tried to convert x to 1. If it was possible then I gave x + k%(y-1) as answer and if not then returned x only.
Submission : 267370934
At First , we need to ensure that x<y ,only after that x can become 1.
So you've really decided to use the $$$3 \cdot n + 1$$$ problem in a contest? I'm really salty right now if you didn't notice.
even though i couldn't solve it, i feel that C is much easier than B. From the One's I've read, A-C. I can say that the problems were very interesting and I think i would enjoying upsolving them.
The trick to B is to try to simulate the operations and output the x values as it changes, then it becomes apparent that it starts looping between 1 and y-1 when x reaches a value equal to y or less. And to get to that point you just perform as many operations as you need to get x to the next multiple of y but you don't actually need to add 1 every time just find the next multiple of y and do all the ops at once
B was very hard for people who aren't that good at math. :(
u basically simulate the process, don't need to be at math
i thought it only needed memoization. but there's a math trick here?
yeah you can basically simulate it until x < y, and in that case x will go up to y and then down to 1 in a loop forever.
Why are all the A's getting hacked..?
Thanks for the contest. I got my lesson. I wont give any contest from now on which has the word TrAkToR in its description.
Why so many hacks on A? (observing those on page 1 of standings)
A: If the sign of (x-y) changed, then (x-y) must be 0 at some time, so answer is NO. Otherwise, the answer is YES.
B: If x<y, then the value of x will change among {1, 2, ..., y-1} periodically, so the answer will depend on k%(y-1). Otherwise, after (y-x%y) operations x will become (x+y-x%y)/y, which is not greater than (x/y)+1. Because y>=2 this can happen not more than log(x) times until x<y, so we can simulate the process in O(log(x)).
C: Let dp[i] be the anwser for first to i-th cards. We can binary search for the largest j such that sum(j+1, i)>=l, if sum(j+1, i)<=r we have dp[i]=max(dp[i-1], dp[j]+1), otherwise dp[i]=dp[i-1].
D: Let the balance of heights B=(sum of snowy heights)-(sum of not-snowy heights). If B==0 initially the answer is YES. If we let c[i][j]=1 when (i, j) is snowy, c[i][j]=-1 otherwise, then when we do operation on a k*k square, the change of balance will be sum(i,j)(c[i][j]) where (i,j) is in the square. So we can calculate the sum of c[i][j] over all k*k submatrices using 2D prefix-sum, and by the bezout's theorem, the change of balance can be any multiple of GCD of these sums. In order to make B=0, the initial value of B must be the multiple of GCD.
E: First think about case for n=2**d. If k>=d the answer is n*(n+1)/2, and if k==d-1 the answer is (n-1)*n/2, let's assume k<=d-2. We can see the answer is sum(1<=r<=k+1)((2**r-1)*(2**r)/2 * C(d-r-1,k+1-r)). Denote this as f(d, k).
We need to find all maximum intervals [L, R) such that (L<=x<R) --> (popcount(x)<=k)). So we have popcount(R-1)<=k and popcount(R)>k. Because popcount(R)-popcount(R-1)<=1 we have popcount(R-1)==k and popcount(R)==k+1. Assume R has r trailing-ones. We can see for R-2**r<x<R we have popcount(x)<=k, and popcount(R-2**r)>k, so L=R-2**r+1. Then the highest d-r-1 bits of L will have k+1-r one-bits, so there will be C(d-r-1, k+1-r) maximum ranges [L, R) where R has r trailing-ones, and each range contribute (2**r-1)*(2**r)/2 to the answer.
Now think about arbitrary n. We can split [0, n) into several intervals with length of 2**j, where n has the j-th bit. If we let L=floor(n/2**(j+1))*(2**(j+1)), R=L+2**j, the answer on interval [L, R) will be f(j, k-popcount(L)). If popcount(R-1)>k, there will be no valid interval containing both R-1 and R, so we can add this to the answer and consider for [R, n) seperately. Otherwise, when popcount(R-1)<=k, we can see that for any R<=x<n we have popcount(x)<=k, so we can add (n-L)*(n-L+1)/2 to the answer.
F: Let L be the minimum i such that a[i]>a[i+1], R be the maximum i such that a[i-1]>a[i]. Then at least we need to sort subarray [L, R]. Let X=min(L<=i<=R)(a[i]), Y=max(L<=i<=R)(a[i]), if there are any L1<L such that a[L1]>X or any R1>R such that a[R1]<Y, then we also need to sort them. To find L,R we can maintain a set S={1<=i<n: a[i]>a[i-1]} which can be modified in O(log(n)) each operation. To find X,Y we can use a segment tree, and to find L1, R1 we can use binary search on a[i].
how to solve F?
C is greedy $$$O(n)$$$, you don't need a binary search.
yes but bs solution is easier to understand why it works.
also easy to implement without considering different cases. check his solution 267340021
The greedy isn't too hard to prove either. At any step, you chose a valid interval with minimum right side index. That is correct because you end up with more possibilities for the rest of the problem than with any other valid leftmost interval (since its right index would be larger).
Who can prove D?
Bézout's identity
Wow, first time hearing this. Thanks! Very counterintuitive for me.
I just thought about Linear Diophantine equation, and the condition of it to be solvable is in CP-algorithm :v.
I am just not studying number theory, haha. I just thought about knapsack.
If d[i][j] is the positive difference between the number of snowy caps and bare caps in the k*k submatrix whose top left corner is (i,j), then you can change the total difference in sum of heights, by c * gcd(d[i][j]), where c is any integer. (bezouts identity).
So just compute this gcd and check if it divides the positive difference between the sum of snowy heights and the sum of bare heights.
Can obviously be done in linear time by precomputing the rectangles.
Wrote a overcomplicated DP + Segment Tree solution for C. D was a good problem.
In fact, I think that question C is much less difficult than question B, and I was thwarted by question B for forty minutes but passed C in ten minutes.
Can anyone teach me how to past B problem?
When $$$x$$$ will become less than $$$y$$$, there will be a cycle: $$$x$$$, $$$x+1$$$, ..., $$$y-1$$$, $$$1$$$, $$$2$$$, ..., $$$x-1$$$, $$$x$$$, which cost $$$y-1$$$ moves, so you just take $$$k$$$ modulo $$$y-1$$$ after $$$x$$$ will become less than $$$y$$$.
Lol, i just read F and isn't it just much much easier, than D and E? It took me like five minutes to come up with the solution.
lol it appears to me that it's harder than D and E, im just being stupid...
Can anybody please help me understand the problem with my code for problem C(Boring days) ~~~~~
signed main() { ll t; cin >> t; while(t--) { ll n,l,r; cin >> n >> l >> r;
vector<ll> a; for(ll i=0;i<n;i++) { ll x; cin >> x; a.push_back(x); } ll i = 0; ll j = 0; ll sum = 0; ll count = 0; while(j<n){ sum+=a[j]; if(sum<l){ j++; } else if(sum<=r and sum>=l){ count+=1; i = j+1; j++; sum = 0; } else if(sum>r){ while(sum>r){ sum-=a[i]; i++; } if(sum<=r and sum>=l){ count+=1; i = j+1; j++; sum = 0; } else{ i = j+1; j++; sum = 0; } } } cout << count << endl;}
}
~~~~~
bro, your final else block is the problem, its the same as if block
include <bits/stdc++.h>
using namespace std;
typedef long long ll;
signed main() { ll t; cin >> t; while(t--) { ll n,l,r; cin >> n >> l >> r;
vector a; for(ll i=0;i<n;i++) { ll x; cin >> x; a.push_back(x); }
ll i = 0; ll j = 0; ll sum = 0; ll count = 0;
while(j<n){ sum+=a[j]; if(sum<l){ j++; } else if(sum<=r and sum>=l){ count+=1; i = j+1; j++; sum = 0; } else if(sum>r){ while(sum>r){ sum-=a[i]; i++; } if(sum<=r and sum>=l){ count+=1; i = j+1; j++; sum = 0; } else{ j++; //only change } }
}
cout << count << endl; }
}
267411889 check this, cuz the copy paste was horrible
Thanks , but can you explain why that was wrong in which test case it will fail
well you else block is basically (sum < l) right? and in that condition you need to j++, i.e., only move the front pointer as the sum is already smaller than l.
For example, 6 9 10 1 3 3 5 4 8 the original code will fail this test case and give 0, instead of 1. because it will not go to the pair i=5,j=4(which sums up to l) but would rather go to i=4,j=4
267409135 Can anyone tell me why is this not working? Like is my logic completely wrong or is there some edge case that the code misses?
Hello everyone, I need somebody to hack my friend's code: https://mirror.codeforces.com/contest/1982/submission/267413232
I am so sure that His code will overflow since he just used 32bits integer (int / i32). Hurry up! Don't let him discover it! Just hack the submission!
PS: His computer's battery has run out and I submitted his code after contest. But why it AC!!!!! I cannot accept it!!!!!
Thanks!
Update:
Sorry, his code is correct. Just sum up all the "-1" and "+1" will not exceed i32. My bad! Actually, he cannot submit his code since the battery problem, I want to easy him that his code maybe cannot pass the test (So that he will not lose this big chance to increase his rating). Okay...... Now he will be definitely very sad and regretful.
Hacking phase was open during contest. Its over now
What I mean is uphacking. QwQ
really? I thought $$$500^2\times 48$$$ won't overflow! did I miss something?
upd: he even changed the char
'1'into int 1! it seems more unlikely to overflow now.You are right, thanks! I have just noticed that!
Is filling "TON wallet" field (and "Secret Code") with account id from mynearwallet in settings enough for getting reward in NEAR cryptocurrency for current round? Or what should I do to get it?
I did surprisingly well in this round. This was a really fun one, questions were not very straightforward, I liked it a lot.
My rating is 1110.I participated in this contest. But my rating not changed.Why?
Till now i was happy that i became specialist but all of a sudden ratings changes is disappeared, can i know what's the reason behind it?
They are probably running plagiarism checks and they will need to recalculate the ratings so it's kind of messy right now
But why they released earlier and now making such messy but i don't think that this happened before
This happens sometimes but usually the plagiarism checks are ran for multiple contests at once (And then you have the message that "the ratings for a few contests are temporarily rolled back") but I assume they wanted to do a single contest plagiarism check here because of the contest having prizes
In that case then they have check plagiarism for both today's contest and recent div 3 contest.
Maybe, not sure how exactly it works but I'm assuming that the plagiarism check has to be ran manually and now they are doing it for this contest because of the prizes
Subject: Issue with Judgement in Problem C
Hi,
I encountered a possible error in the judgement of my submission for Problem C during Codeforces Round 955 (Div. 2). It seems there was a bug during the judging process.
Details:
My Submission During Contest: Submission 267358679 My Submission After Contest: Submission 267429867 During the contest, in the 2nd pretest, 6th test case, my submission was judged incorrectly. The actual output should be 0, but it was judged as 1, resulting in a wrong submission.
After the contest, I resubmitted the same solution, and it was judged correctly with the output being 0.
I would appreciate it if the issue could be reviewed and corrected as it affected my contest standing unfairly.
Thank you for your attention.
Best regards, Deshik conan45
Bruh, it's not even the same code, I don't know what you are trying to achieve
I have one question for the cf community, why to give the score distribution beforehand, people can initially guess the difficulty of the contest, isn't it non exciting it would be more fun if we are not known that and may be we can keep it constant in every contest
FelixArg , My solution for D shouldn't have passed the test cases, but it did.
I knew the current difference between type1Sum, and type0Sum. ( let's call it RequiredDifference )
I knew for each of the sub-rectangle of size k*k, what are achievable differences. Lets call them (d1, d2, d3 ... dk).
I knew that I had to do a1 * d1 + a2 * d2 + a3 * d3 + ... ak * dk = RequiredDifference
I didn't know how that gcd ( d1 , d2 ... dk ) | requiredDiff .
But I knew, for ax + by = z , then gcd(x, y) | z. So I used this two variable equation property and passed test cases. Refer to my solution.
The test cases were weak for D. Ideally my solution shouldn't pass. It was later hacked by djm03178 .
bro i wrote the same solution as yours and mine was also hacked lmao.
So many people here are saying that B was a lot difficult than C. I did B in 20 mins, but it took me almost a day and more than 10 attempts to finally get AC on C. I had actually figured it in some time but it was giving TLE again and again. Don't know if I am too dumb or too smart lol!!
you are too smart bro
Exposing Cheating in Competitive Programming Cheating undermines the spirit of competitive programming. The Telegram channel selling Codeforces solutions, with Gautam_Himanshu implicated in using these solutions to unfairly boost their ranking. This not only disadvantages honest participants but also erodes the platform's credibility. The community must stay vigilant, report suspicious activities, and promote ethical practices. By addressing such issues, we can ensure a fair and rewarding environment for all. Let's work together to uphold the values of integrity and fair play in competitive programming. Report any evidence of cheating to maintain our community's standards.
Creating: .cfuser.near.
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Can you please help me?
Even though I ate negative delta the problems were very good, I enjoyed the round so much.
Hey folks! Here's an attempt of an explanation of problems A to D of this context with Tourist's codes https://www.youtube.com/watch?v=3fPvRAowuqU. Would be glad if you find it informative!
How do I redeem the 1N prize? Sorry, this is the first time I've won something in an online contest, so I don't know how to redeem stuff. Thanks in advance for the reply!
I have the same question. And how do I determine if I have won a prize（because of unofficial participation）? Should it be a message, or other forms?
I saw your rank, I think you are also eligible for the reward. They said unofficial participators will also be able to claim rewards. Ranking under 496 (including 496th rank) should be enough. I saw something in the settings->Wallet section (but it is only visible in Russian language). But I still don't know how to claim it. If you find out, please let me know too :D
OK,thank you, maybe you have noticed, the form for filling out the award information has been sent out by the system about 5 days ago.(You can see it in "talks" page, which is near the position of your profile)
I wrote this in case that you missed it. Please note that the DDL is 7.8 24:00 :)
Dear Codeforces Team,
I recently received a notification regarding a significant coincidence between my solution for problem 1982B (solution ID: 267384934) and the solutions submitted by Muhammad_Tawfiq (solution ID: 267384934) and Antartic_Nafeez (solution ID: 267386858). I take such matters very seriously and would like to address this situation promptly.
Firstly, I would like to clarify that my intention has always been to adhere to the rules and maintain the integrity of the competition. I assure you that I have not intentionally shared my code with anyone nor copied from other contestants.
To provide more context:
Code Review: I have thoroughly reviewed my code and confirmed that it is my own work. However, I understand that similar solutions can sometimes arise due to common approaches to solving the problem.
Use of Public Platforms: I have not used public platforms like ideone.com with default (public) settings to share my code. All my code is developed locally or on private, secure platforms.
Potential Common Source: If there was any unintentional similarity due to a common source or widely known algorithm published before the competition, I am more than willing to provide details of my references and thought process.
I request a thorough review of this case and am ready to cooperate fully by providing any additional information or evidence required. My goal is to clear up any misunderstandings and continue participating in Codeforces competitions fairly.
Thank you for your attention to this matter.
Best regards,
[Muhammad_Tawfiq]
Almost the same code,stop joking bro.
exactly the same code, he's definitely joking
Hello Codeforces,This is in regard to plagiarism being put on one of my solutions. I want assure you that no such incident has been done by me the only common issue can be a code template i use for some questions .You can check all my previous contests and submissions.This was the first time i used this template seeing many other use a template of their own.If unintenionally I have made a mistake I apologize and assure that i wont be using templates as well from now on
seems everyone has received plagiarism message
is that the reason why rating change was rolled back?
is rating rolled back for everyone?
yeah, i didn't get the plagiarism message but my rating is also rolled back.
Cheaters are even copying their excuse messages. Kind of funny.
How long do the rollbacks take? I need to submit my resume with a screenshot of rating as proof in 4 hours:(
it's finished
Can anyone please!!! tell me what did I do wrong here in question D. please!!! I am disgusted
It is showing "Amount to claim: 0." in the prize link, what's the problem? How can I resolve the issue?
Update: Fixed
Same problem
Can I know how to claim the prize because I got 4 NEAR but when I enter the form link and make cf near account nothing happens can someone explain how to claim the reward please?
Why I tried to connect NEAR account via here and clicking "Connect your account with NEAR ecosystem", only to get a text which reads, "500: Internal Server Error" in the url "https://acade.studio/cf/[account name]/[some UUIDs]"?
Also, I tried other guy's account on my computer, which can normally connects NEAR.
The link works either if you have rating 1400+ or if you won prizes during the round.
If you did win prizes, send me a DM, I will help debug.
All
Near Prizes-
I got 4 Near as a prize and I did sign up on Acade Studio as well as claimed the prize. However it is not showing up on the Acade Studio built-in wallet balance, and on MyNearWallet, it says it requires a deposit with NEAR to be able to do transactions. What can I do, or will the prize be given after some time? I did claim the prize almost a week back.
Do you remember what account did you withdraw the prize to?
Also can you confirm that the amount to claim on the CF page shows up as 0?