Hello Codeforces!

On Nov/22/2021 12:35 (Moscow time) Educational Codeforces Round 117 (Rated for Div. 2) will start.

**The problems are based on the Southern Russia and Volga Cup 2021, which is a regional contest of ICPC. If you have participated in it or are planning to play it as a virtual contest, please refrain from taking part in the Educational Round.**

Series of Educational Rounds continue being held as Harbour.Space University initiative! You can read the details about the cooperation between Harbour.Space University and Codeforces in the blog post.

This round will be **rated for the participants with rating lower than 2100**. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.

You will be given **6 or 7 problems** and **2 hours** to solve them.

The problems were invented and prepared by Adilbek adedalic Dalabaev, Vladimir vovuh Petrov, Ivan BledDest Androsov, Maksim Neon Mescheryakov, Roman Roms Glazov and me. Also huge thanks to Mike MikeMirzayanov Mirzayanov for great systems Polygon and Codeforces.

Good luck to all the participants!

**UPD:** Editorial is out.

NOTE THE UNUSUAL TIMING

eagerly waiting for the contest.

Why Unusual timing again? :(

Now that's usual:(

Finally a cf contest after so long .....All the best to the fellow programmers !!

contest after long time in unusual time , why in the last period the contests in unusual time :(

On behalf of all Russian schoolers, I protest against the unusual timings on weekdays.

Isn't that just a good reason to skip school?)

On behalf of myself I strongly support the unusual timing, it gives an opportunity for competitors around the world to participate :)

Will try to perform better this time. let's hope for the best. Good luck everyone!

Good luck everyone!

Last week one contest this week five contest I am not complaining, but it would have been better if contests evenly spaced them in these two weeks. Lately, there has been a lot of inconsistency in contests either there are very few, or there is a lot in a 7 to 10 days period. Anyway, today is contest day, so best of luck to everyone.

Can you change the timing i have meeting from 4:00 to 4:30 pm , please

I need this much confidence

lol,it was sarcasm ,xd

this comment made my day xddd

Will the problems be arranged in the correct difficulty order? Or will it be shuffled as we see in icpc?

## Mathforces

How to solve D?

How to solve E?

i did it using ternary search on the number of messages to be pinned , and then selected the books in a greedy way, got tle on tc 149 , but after one optimization it ran comfortably. Update: the solution was wrong

thx, got it

Your function isn't parabola or convex. When T(len of answer) is [1; 20], ur function gets chaotic values. When T is [21; 2e5] fucntion decreases on all segment

No wonder it got hacked

Your ternary search solution has been hacked :)

best contest i've ever done!Does anyone know the proof for why it's never optimal to select more than 20 pins in 1612E - Messages??

If $$$s_i$$$ is the sum of all $$$k_i$$$ with $$$m_i = i$$$, then the expected return you get from including $$$i$$$ is equal to $$$\frac{s_i}{t}$$$ if $$$t \geq 20$$$. Note that you greedily want to select the largest $$$t$$$ values of this for the messages, and you can show that this value is greatest when $$$t = 20$$$ (since averages decrease since you're taking lower and lower sums).

I mean many people made the guess for first 20 but I wasn't smart enough ig.

If you take 20 best messages out of $$$m$$$ taken in terms of expected value contribution and leave just them, their expected value will increase proportionally to $$$\frac{m}{20}$$$. But since they were maximal total EV will not decrease.

Alright thanks this is pretty helpful!

Suppose we have selected the $$$y$$$ messages with the max $$$x$$$, where $$$y$$$ $$$\geq$$$ $$$20$$$, and $$$\sum{k_i}$$$ $$$=$$$ $$$x$$$ for all $$$i$$$ where $$$i$$$ is one of the chosen messages. If we select the $$$(y+1)_{th}$$$ message $$$msg$$$ where $$$\sum{k_j}$$$ $$$=$$$ $$$z$$$ for all $$$j$$$ where $$$m_j$$$ = $$$msg$$$, the first $$$y$$$ messages are still the best, and their total expected value decreased from $$$\frac{x}{y}$$$ to $$$\frac{x}{y+1}$$$, that is, decreased by $$$\frac{x/y}{y+1}$$$. However, $$$msg$$$ only introduced $$$\frac{z}{y+1}$$$, and $$$z$$$ $$$\leq$$$ $$$\frac{x}{y}$$$, otherwise it would be one of the best $$$y$$$ messages.

idk how many guys pass G, I want to say that F is much easier than G

When I began solving G, it had 17 solves while F had like 3.

Most people who reached F/G didn't have time to solve both, so it was probably just an observer effect here.

Hi, can you please give some hints on how to solve F? For me G seemed easier than F.

You can just optimize bfs.

The queue contains tuples $$$(a, b, d)$$$ ($$$d$$$ is the number of moves to reach $$$(a, b)$$$). Let $$$(a, b, d)$$$ be a useless tuple if there exists a tuple $$$(a', b', d)$$$ in the queue, with $$$a' > a$$$ and $$$b' > b$$$. If you remove useless tuples from the queue (e.g. by rebuilding the whole queue if its size is $$$> 2 \cdot 10^5$$$), the bfs is fast enough.

How to solve D ? I called gcd(difference, maximum number % difference), but if the difference is larger, then it gets executed huge number of times...

Basically, the numbers you can reach are everything reached by the Euclidean algorithm for a and b.

I guess resultant number was eventually max-min*k such that max-min*k>=min after that max and min will change so you just have to check whether x%min==max%min otherwise pass for min,max%min

Hmm... I think G is easier than E. E was difficult to translate.

nice code

nitin_05 136448698

I dont know how total accepted submissions of problem D went from around 1200 to 1700 in last 20 minutes ;_;

Maybe it's just cheating again...

traditional -100 for me kekw

EDIT: ok i got the mistake.

i should've been prepared more before joining this round

Can Anyone provide the intuition behind D, it ate up my whole hour, still couldn't find a logic.

When assignment operations are only based on substraction of two no.s . Think GCD .

i tried GCDs, but idk how to actually solve

Go through the same procedure as GCD , and before recurring for GCD(b%a,a) check that whether (b%a)+(j*a) is equal to x or not , j is from 0->r such that (b%a)+((r+1)*a) is greater than b . For clearity refer this https://codeforces.com/contest/1612/submission/136448350

Can you please explain this part? I would be very thankful to you.

CodeAs I said before if x's value is b%a+(j*a) where j is 0 to r , then x is achievable . So x=(b%a)+(j*a) => x-(b%a)=j*a => (x-(b%a))%a=0 , as j*a%a=0

Should E really pass in $$$O(nlogn*20)$$$? I did write a solution which passes but with a high execution time. Can someone hack it?

Did anyone observed precision handling error in python in question 3

Yes — do use

`from decimal import Decimal`

to handle floating point numbers with higher precision in Python.What's the approach for E?

Observation 1: if you're going to pick X messages, it's obviously optimal to choose the mi such that sum(min(ki,X)) is maximised (why? consider how you would write the expectation calculation as an equation). So the first thing we want to do is sort the messages by the sum of ki values assigned to that value of mi.

Observation 2: it's never optimal to go beyond 20 messages (or, more specifically, max(ki) messages). Why? Suppose we have exactly max(ki) messages. Then our expected value is E20 = sum(ki)/20 for the users whose messages are in this set of 20. If we add another message, the expected value E21 = sum(ki_prev)/21 + sum(ki_21)/21 = E20*(20/21) + sum(ki_21)/21. Suppose E21 > E20. Then we have sum(ki_21)/21 > E20*(1/21) = sum(ki_prev)/(20*21), and hence sum(ki_21) > sum(ki_prev)/20. This means that the 21st message has a greater sum of ki values than the mean of the first 20. But this is impossible because the 21st message has a sum of ki values less than or equal to the first 20 values.

So it's proven we will never go beyond 20 messages. Now we just try all possible values from 1 to 20, and find which gives the greatest expectation. There's a tricky nuance (which I believe I got wrong in contest) where you have to sort each time according to sum(min(ki,X)), rather than just sorting according to sum(ki).

Code

And you don't even have to make the second observation, I, for one, completely missed it. You can just sort all E20 highest to lowest and check all prefixes of this order.

I Was looking for this proof precisely, Thank You for the clear explanation !

How to solve C?

Regarding this submission: https://codeforces.com/contest/1612/submission/136484148

If I'm not wrong, it's complexity is clearly $$$O(t * 10^6)$$$ for cases which results in

`-1 -1`

. So, it must exceed the given time limit easily (as $$$t \leq 3000$$$), but on Custom Invocation I have tried and it runs in $$$\sim 1500$$$ ms.Can anyone kindly explain me why it is so?

Because 3e9 operations can be fast enough

How to solve F?

Can someone explain their approach to Problem D

What's the hack case on E?

Here are the video Solutions to the first 5 Problems In case you are interested.

Thanks for the video editorial found it very helpful, especially for D!!

Great Explanation

Thanks for the good sample tests to avoid overflow in problem C.

Here's a heuristic solution for problem F: https://codeforces.com/contest/1612/submission/136557294

The idea is that it's generally much better to replace the smaller number to increase our power quickly (which greatly overpowers any gain from synergies), but for small values this might not be the best option. We take all possible choices for the first 6 moves, and then apply the following:

This passes all test cases but I highly doubt it is correct, and would welcome a valid hack.

To not keep you waiting, the ratings are updated preliminarily. In a few hours/days, I will remove cheaters and update the ratings again!

I panicked in this contest and lost a bunch of ratings. My biggest drop yet (-100). I couldn't even solve A. Is there anything I can do to get a better grip of my nerves in a contest and not panic when I see an implementation heavy or math problem in the contest?

Keep participating in contests

do not make submission and you will not lose rating

164 pretests in E and still they do not have tc that fails the most obvious wrong solution)

Now you can still HACK others, right?

Yeah, purple+ can uphack for 7 days after round end iirc

I don know that!Thank you.

What kind of time of sending tutorial is normal? Looking forward to the solution.QWQ

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).I received a System mail saying that my solution 136459207 matches with idkmyname_ submission 136471505. I believe this is because we both have used nor's publically available BigInt library

I had also put this link as a comment in my code, as one should do while copying third-party code. MikeMirzayanov please look into this

I'll revert the punishment

I got this email from Codeforces which says that my solution for problem C coincides with too many people.

It goes like- Attention!

Your solution 136470651 for the problem 1612C significantly coincides with solutions bihari_bandar/136430343, greyman/136431111, going-too-far/136431248, SnigdhaReddy27/136438434, Jee_none/136440961, loser53/136448455, Pratyush_tripathi/136449209, manishchembeti/136452512, yash112124/136453654, misra99/136456090, ashutoshtr/136458047, TheThinker_08/136458129, venom0912/136460586, Secret_Superstar12/136461961, pratik2912/136462123, shivam.yadav98939294/136462732, Karan10100/136462833, 123Alpha/136465897, crosscheck371/136466967, darkcoder_347/136470103, codeprakhar/136470651, ramanand_rv/136470925, noob_tusharb/136471502. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

Now, I don't understand how my solution can so perfectly coincide with so many people even when I don't know who they are! I even have never used Ideone.com.

I was previously trying to solve this question using simple maths(quadratic equation solution) but that didn't work so I thought of another method that involved simple binary search. It worked.

I understand that the code of binary search is freely available at almost any programming website which leads to this coincidence.

Codeforces community... How can I get this fixed?

Please look into this matter MikeMirzayanov

why does this kind of coincidences always happens to indians?

At least compare my solution with others before demeaning India or its coders.

yes

you just changed the names of the variables

just another dirty indian still lying after being caught

Yes Sir It was my bad to argue with you. You are absolutely correct! :)

Though your mind is still filthy like your DP.

ohno i smell dead bodies!!

did you took a bath in the Genzies river today?

Haha Nothing like that exists

Don't even know the names of rivers. Poor boy

SOMEBODY HELP! https://codeforces.com/blog/entry/97183

I want to ask question F, if I just think about q=0, does it have the same number of answers as it has the right answer? How long does it take for the Fibonacci number to form Max of m,n