People be like: See a fun comment: upvote; See a comment discussing testers of a contest: downvote; See a red coder write a comment: upvote; See a comment with a lot of downvotes: downvote; See somebody praying to reach a higher level: downvote

People be like: I'm a simple man. I see T1dus, I downvote.

I guess it's 2:30 as per the contests table

I would like to say that some youtube channels provide solution code during contest is running it is totally unfair to most of the participants plz take proper steps to stop this.

To perform better, should i solve the author's previous contests? Please give me some suggestions as i am a newbie.

warm and fuzzy feeling! you mean pain!!

Lol no wonder why a grandmaster gains satisfaction from solving all problems:)

So "exited" that you wrote "exited" instead of excited.

good luck

Points and difficulties have no correlation whatsoever. Difficulty is determined after the contest ends, while point values are there to give us a rough approximation of relative level of these problems. E. g. If two problems both weigh X points that means they are similar in difficulty, but that difficulty could be anything from 800 to 3500 theoretically.

Hacks and unsuccessful submissions are still worth the same number of points though. And if you double every one of them, we will have a 5000, 5500 and 7500 problem.

I think that having 300 points for the first problem would feel a bit better than 250. It's just a more round number. Also it's more than half of 500, rather than exactly half. Not that it really matters, but this might make people feel more comfortable. Especially complete beginners, for whom the first problem might be the only solved problem during the whole contest.

Personally, I have never paid attention to the raw amount of points I got in a contest. I was always focused on my ranking. Maybe I'm in the minority though.

I don't see how using 250 hurts the feelings of div2 participants. Will they be happier if we multiple everything by 10?

There's currently an upper limit of 6000 points per problem in the Codeforces system. You would need almost twice that if you want to keep reasonable geometric progression — which is needed to allow solving problems in a different order, and not to penalize much for failing a medium problem (but solving a hard one instead and getting less points).

Very good point.

I think we should also set everyone's rating to 1500. It might hurt their feelings when their rating is stuck at 1000.

I think I've been advocating starting from 250 for a long time already. It allows for better balancing of harder problems. Even with 250pts the ratio pts/needed time is still the lowest for the easiest problems

This round will have tasks of all levels, and the round itself is "GLOBAL", which also indicates that everyone can participate, so anyone can participate and it will be interesting for him.

So, I am <1200 and I definitely will participate, because I’m absolutely sure that I’ll solve 2 problems at my worst.

The points decreased per minute is proportional to its starting score. A 250 problem ends at 150? I believe (either 3/5 or 2/5 of original), so each minute it's decreasing by 100 / (total_contest_time), which was 150 for that contest. This value is less than 1, so sometimes it'll round up I guess.

Errichto You will not be disappointed :)

Read the announcement :)

I would propose that A will be worth 69 points and B 420 points. That is more fashionable.

There is some problem in the testing of problem E.

For my solution: https://codeforces.me/contest/1552/submission/123763122

It shows colour don't match for 3rd interval in test case 3 despite colour are the same. Please look at this and accept my solution if it is correct.

As a participant i would recommend all to upvote the above comment;)

So much about that lol...

This isn't the place to discuss these topics

When do you think competitive programming will be an official Olympic sport?

One and game over

Congratulations, you've won the bet.

No, its for Div -1 participants

your nickname really tells alot

All questions are too hard for me.

and the last question in the next 6 minutes

First time we are going to see 3800 rating.

numbers a_i and b_i should both be in i-th color. In your case a_3 and b_3 are both in 2nd color

Said an unrated user.

try this 174+(-171) = 250+100.

Gotta admire the confidence!!

How to solve D? Got wa on pretest 2

Sort then check if a[0] is the winner.

The key observation is, that if a>b and b>c, then a>c. So sort and check if first beats all.

Assume the curr_winner is 0th participant. Now, iterate through all the remaining participants. If, the winner is superior to the ith participant, we are fine, otherwise, change winner as ith participant ( because i defeats curr_winner ) and continue checking till the end.

Once done, we have a candidate for winner. We just need to iterate again through all remaining participants and ensure that he defeats everyone ( and there is not a cycle in which case return -1 )

start comparing the first athlete with other athletes if he is superior then all of those who got compared to him are definitely not and if he's not superior to the current athlete being compared then this current athlete could be the one so start comparing from this athlete with the next athletes if you compare him with the previous athletes it will TLE so keep doing that until you have one possible athlete now compare him to all the others if he is superior then answer is that athlete if not there's no answer

Only one overall winner is possible. If an athlete is not superior to any other athlete, he cannot be the winner. Suppose prev[i] is the best choice for first i athletes, then prev[i+1] will be i+1 if i+1 is superior to prev[i] or prev[i+1]=prev[i] if prev[i] is superior to i+1. prev[1]=1. Now you just need to check whether prev[n] is superior to all.

Compare No.(i) person with No.(i+1) person.(1<=i<i+1<=n)

If No.(i) person is lose,it illustrates that No.(i) person is not the champion (since he loses at least 1 round.)and No.(i+1) person will continue to compare with next person and vise versa.

After n-1 round comparation,you can get a person who is the candidate champion and he should compare with other persons to judge whether he is the real champion.

Notice that for any two athletes a and b, either a is superior or b is superior. Since our goal is to have an athlete that is superior to all others, every time we compare any two athletes, we can remove the 'not superior' athlete from our list of candidates.

So we can perform n-1 comparisons and have exactly 1 candidate left. Now, we can check if this candidate is superior to all others. If not, then there are no more candidates so we print -1. Otherwise, this candidate is our answer.

Apply two pointer's approach, take initially l as 0 and r as n-1 now compare between them if l is winner,it means r is eliminated hence decrement r. and vice-versa, finally you will get the ans as l ,check whether l is the real winner or not

My approach: Iterate through all the players and in each iteration compare two-player and eliminate one player, the one with fewer wins. first player 1 vs 2, second, winner of (1 vs 2) vs 3, and so on.

So after a complete iteration, only one player will remain. Now just need to check if he is better than all others or not. So one more iteration through all other players and Check if the candidate won 3 or more previous games with all others or not.

it's the opposite of last round I guess.

It's a typical PQ Tree problem.

https://codeforces.me/contest/243/problem/E

https://onlinejudge.org/contests/278-f2e8ebbf/problemset.pdf Rujia Liu Contest 3 G (which is exactly the same)

Problem M from PreFinals Moscow Workshops 2019 Day 4, Division A: Moscow SU Red Panda Contest.

Sadly yes https://codeforces.me/contest/243/problem/E (and also somewhere else that we still don't know). Of course, we didn't know about it (even though I was a participant in that contest many years ago).

Lmao copypaste PQ tree = free rating (Same problem was in Ptz Winter 2013)

I had also done this problem previously (unfortunately I wasn't able to find it during the contest ...).

Simulate any type of tournament (for example, a well-known tournament tree) and then check that the winner of the tournament (if such exists) is actually the answer (make an additional comparison of him with all of the other participants).

Let's define a directed graph over the players, an edge exists from player $$$u$$$ to player $$$v$$$ if player $$$u$$$ defeats player $$$v$$$. It seems like the problem is similar to finding whether or not the directed graph has a universal sink (in-degree = n — 1, out-degree = 0). To find it, let's assume the first node is the sink, and for each next node, check whether it defeats the current sink. If it does, set it as the new sink, otherwise continue. Finally, there are two cases. Either we have found the sink, or there doesn't exist any. We can check this by testing whether it defeats every other player or not.

Here is the classic problem: Problem 22.1-6

start out with a set of $$$n$$$ possible winners and at each step compare two possible winners, and discard on of them as a possible winner. At the end check if the only candidate is in fact a winner.