djm03178 orz....

G.O.D

Ok I commented as per your wish.

So yummy~~

djm03178 (Good problemsetter)

As a participant, I will give you upvote.

As a participant, upvoted and commented :)

// cout << "Hi" << endl;

sir make sure to add some or atleast a dp problem between A-D.

I hope it is a good contest :D

As a tester, I wish I could participate in this round as this round is very well prepared by djm03178. Good luck to all the participants!!

Dude C was easier than B... at least for me. I litterary wasted 2 hours trying to solve B. Props on C though pretest 1 is very helpful to skip a ton of mistakes.

yes... you got my comment as you are problem setter. thanks a lot. this round was awesome

why is this getting downvotes lmao

As a tester, I think he deserves upvotes. He did a great job as a tester.

I think so

Yes

It's my ex-girlfriend's birthday on that day... Should I wish her excitedly, normally or not at all? The only thing I'm excited about is the contest and a little bit about the INDIA-AUSTRALIA cricket match.

Algosexual?

I hope they give out a contest full of "interesting observations" problems.

Please make it like educational round 99 so that we can see where we falter and keep learning new things!

A negative-rating tester whose atcoder rating is more than 2000 :D

Increase in ratings too.

Learnt a new lesson "Don't remind anything if it is already posted on Contest blog". Thank You.

It's already mentioned in bold. Should they just write it in the announcement heading for you?

I'll take that as a No

Why I am getting downvote. what's wrong

I hope to become cyan

How is it a plot twist XD XD

Yeah it's cause he's a lil caught up with planning his "As a setter..." comment for the next contest.

That's how Koreans call Codeforces in Korean communities.

Geez, sad to know that you like seeing such things

I feel great when I see people like you, as some people after reaching expert, stop giving contests in order to uphold their ratings.

Yes, you can cancel it, but remember — no submissions -> contest unrated for you

OK thanks for info

Good luck to you

It is against rules to discuss problems before a contest ends. Please, wait for the end of the round.

The owner's comment makes it seem like the problem in its essence was being discussed. Perhaps that was the plan all along to delete the comments and make it seem like the approaches to the problem were being discussed. On the contrary, statements were made to connote the recent trend in absurd difficulties for Div 2B and there is nothing from the rules here that forbids that. If a specific rule bears that intent, I would humbly suggest that the English version is re-written in a way that the meaning is not lost during translation.

yes and thats making me much sad :.(

This is a Div2 tho. I agree that the second problem may seem harder than usual but not that much. It was an easy to implement solution which needed a very nice observation. I think that it's better for the second problem to have something tricky in it, instead of being very easy and boring. Trust me, this kind of problems are good for beginners too because it teach them to think outside the box :). Also, there are div3 contests which kind of allows the beginners to solve the first 3 problems quick.

Transforming the tries to coin flips, it totally sounds like a common probability problem, which of course it is heh: https://math.stackexchange.com/questions/364038/expected-number-of-coin-tosses-to-get-five-consecutive-heads

I derived expected value for covering different size gaps between successive checkpoints. For $$$g = 1$$$, ie consecutive checkpoints, expected value is $$$2$$$. For $$$g = 2$$$, using expected values, we get $$$6$$$ and then $$$g = 3$$$ we get $$$E(g) = 14$$$, using three equation. and then I thought maybe it is $$$E(g) = 2^{g+1}-2$$$ but dont know the general proof

Edit:

Ok link by robinz62 was useful, thank you

Lol You don't even need to simulate take a look at 1265E - Beautiful Mirrors

first try to find how can you find min value of given array without any changes

The key observation is that changing a number at begin is like removing it from the array.

So we need to find which number, if removed, contributes most.

The contribution per number is the sum of differences of the adjacent numbers, and the number itself.

If you want to see the approach or idea for solution of B:

My approach is to iterate through all elements and guess that if the current element is not there then what will be the answer to make them all equal w/o changing any element.
(It would be the sum of absolute difference of each consecutive element, note: you have removed the current element from the array). This works because you can change the value of the current number to the next/previous number. So, it will reduce the steps to obtain an array with equal value.

So, you can do it in o(n). And you have to see the case for ignoring the first/last element.
it will be easily calculated !!

Note that You need expected 2^(k+1)-2 moves to do 10000...0 where there are k digits.

Number of steps to cross 1 is 2 , 1 0 is 6 , 1 0 0 is 14 , 1 0 0 0 0 is 30 . relation is s = 2*s+2 . Odd number of steps is never possible since to crossing any stage must be even . Now find biggest s which is close to n and keep doing that until n reaches 0 . For example for 20 it will be 14 + 6 i.e 1 0 0 1 0

I guess the expected value converges to an integer when total tries is infinity.

Beating n consecutive levels without a checkpoint is equivalent to getting heads in a coin flip n times in a row. Having that in mind you can read https://www.codechef.com/wiki/tutorial-expectation.

Probability to cross a level is 1/2 . Thus total number of moves require to cross is 2 . Suppose i crossed a level and fell again to it . Then again expected number of moves required to cross it again will be 2 and thus total 4. Hence not only it will be integer but also even.

If you have a range with 1 at the beginning and k-1 0s after, then this range increases the expected value by 2^(k+1)-2. So while remaining length is > 0 you can simply take largest k for remaining length and put 1 and k 0s at the answer.

yes brute force but n*n*40 time compexity so it is same as reading input

triangle space is base*height/2 since they want it multiplied by 2 so it's just base*height so you just need to find the longest line for each number and you can calculate the max space

I got the idea but sadly could not code it in time

If you want to see my approach for this problem, you can see here.

its not strict. You are saving index of each character. NOW, if all 4*10^6 are same character then you complexity will be (4*10^6)^2 (as you are traversing each possible pair of each character) which is definitely not in time limit.