Hope to reach expert this round! (Tho there's no way an unrated jumps all the way to expert)

Weak TestCase Contest

Can you speak Chinese

You missed an opportunity to name him Rudeus (Greyrat) and have back-to-back anime rounds :'(

In the problem B, why this case is valid?

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in the problem statement it is mentioned that

Rudolf has a 3×3 field — the result of the completed game.

how can this test be result of the Game where one player is making all moves?

Also, All the sample tests are maded in that way that one player is making move after another.

If it is valid test then I think it was better not to introduce this problem as 3 player Tic-Tac-Toe game.

E2 hard, have no idea

I liked the problem, but something odd happened in the evaluation. I passed E1 and E2 during the contest, but in the post-validation got TLE in E1. It's such an odd thing to pass E2 and fail E1, and it is due to the fact that TLE strictness vary during and after the contest. Of course, if I had noticed this TLE, it would have been easy to submit the same program that had succeeded for E2...

are you a time traveler?

Sorry bro , but this is div 3. You are at wrong place.

Can relate

lol, ur hopes got shattered

bro wrong blog this is upcoming div3 :skull: and copying code from online is allowed if it was there before contest

Congrats

is it hard or something?

First time to participate in unrated Div3

yes

what is the minimum number of problem you need to solve to reach expert from you cur rating in this round?Just curious to know.

Me too...

You can use this extension to predict delta even during contest, it is pretty accurate

Actually it was my first contest...can you please tell me when will the ratings be updated?

have they all been accepted? nothing hacked or failed in system testing?

I used quadratic formula to solve for potential cubes and then precomputed possible powers until 10.

Didn't Solve it.. But i think we don't need to do a binary search over k's powers we can check all of them "aka. brute force " as maximum possible power to have is p = 63 for k = 2 and p decreases with any increasement in k .

So the goal of the problem is to find some k and l such that k^0 + k^1 .. k^l = n and l>=2. Notice that l can be at most around 63 ( when k is 2 ), and therefore, you can iterate on l and binary search on k, yielding a log(n) solution with a moderately large constant factor.

No, It was a singular B.S. problem. There are at max 60 possible powers of k.

I.M.O. you can't B.S. on possible powers of k as you can't be sure if Mid value fails whether you should use more power of k or less power of k. as base number is not constant

You can guess the number whose exponents we are adding should be moreOrLess though. as k is kept constant

i ,myself am not sure if this explanation is sufficient, you can look at this submission for more info : https://mirror.codeforces.com/contest/1846/submission/212833690

2^64 ~ 3^41 ~ 10^19 ~ 250^8 ~ 7000^5 ...

I brute forced the solutions for the first 1e5 values. For values above 1e5, I did 2 things.

First: Fix the length of the formula between length 3 to 5.

Second: After assuming a certain length is correct, lets say 4, you know the formula should look like this: k^0 + k^1 + k^2 + k^3 = n

Since n is known, you can binary search for a k.

you have wa1 you can fix it

nvm i was using setprecision wrong. i will get negative delta for sure

You can use cout << fixed << setprecision(10); and make sure you use long double.

set precision to the answer.

for c++, add this before printing answer "cout << setprecision(8) << fixed;"

same

You misunderstood problem D. It wants you to be precise to the 6th decimal place. If you google "C++ float precision" or "C++ double precision" then you find out that both datatypes are accurate enough. Float is precise to the 7th decimal while double is precise to the 15th decimal.

If you ever find yourself in the need of bigger datatypes, you may want to check out this: https://cp-algorithms.com/algebra/big-integer.html

optimization of E that I overlooked during the contest but turned out to be valid: The only possible candidate for $$$k$$$ is $$$\lfloor (n)^{(1/r)} \rfloor$$$. This is due to the fact that $$$x^r < (x^{(r+1)}-1)/(x-1) < (x+1)^r$$$. Look at the binomial expansions to find out why.

how to solve G with dp?

In problem G, is it true that every medicine can be used multiple times? If not, how do you prevent multiple uses?

I implement F the same idea as yours but getting WA. Any idea why?

212701178

wait why is E $$$O(log^3(n))$$$ not $$$O(log^2(n))$$$?

what did you like about these problems? G is very obvious and E2 is just stupid math.

the amount of high-rank newbies is insane. There is no chance to get rating for honest newbies like me. I am very disappointed with that. Every time I think I do better there are tons of other people solving the same problems. In this way is really impossible to increase rating and this is very demotivating. Finally there is no chance all the cheaters will be caught, so that's the story.

What do you think Codeforces can do? It happens every contest in Leetcode as well.

This is more like an Indian problem that spreads everywhere in my opinion. There's nothing Codeforces or Leetcode can do if most of the Indian users lack integrity and consideration for other people.

TLDR: Indians lack skills and talents so they cheat (low rating overall). Blame India for producing top cheaters instead of top geniuses in algorithm

never mind,we solve this tasks for increasing our skill,not just for the points

I'm not experienced in Java, sorry, but I'm wondering if it's because hashmaps can't store multiple copies of the same object. Some participants can have the same scores and penalty.

You are adding the same penalty multiple times.

time+=(ar[j]+time); should be time+=sum