Educational Codeforces Round 169 [рейтинговый для Div. 2]

4 месяца назад, # |

hopefully expert after this one

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4 месяца назад, # ^ |

good luck mate

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4 месяца назад, # ^ |

good luck

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DalChawal

4 месяца назад, # ^ |

why is this comment at -12???

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4 месяца назад, # ^ |

idk either lmao, it is what it is

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Kolyanchick

4 месяца назад, # ^ |

because this is cf

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mayank.nichlani22

4 месяца назад, # ^ |

coz he's probably a cheater,see his submissions

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4 месяца назад, # ^ |

if you accuse me of cheating, I suppose you have evidence?

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mayank.nichlani22

4 месяца назад, # ^ |

← Rev. 3 →

-17

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4 месяца назад, # ^ |

Look through every single one of my submissions, and you can see I ALWAYS write long long int. Any single time I use 64-bit integer, I always type long long int. You just picked a random submission and the most random reason and decided to go with it? cope and seethe newbie, maybe focus on actually solving some problems instead of throwing baseless accusations.

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mayank.nichlani22

4 месяца назад, # ^ |

-7

So maybe you debug every question using ai,who knows!Also I am actually pupil,I didn't know multiple accounts are not allowed on cf,Hence kept this one only for practicing.

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4 месяца назад, # ^ |

yea sure, I also use ai to type my code when I practice as well right? you're a complete clown, stop embarrassing yourself. oh and since when is AI considered cheating?

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4 месяца назад, # ^ |

← Rev. 2 →

I used to write long long int every time myself before I started using a template that had long long. "clearly ai generated" bro is confident too. maybe try to come up with better proofs next time, at the very least you could avoid so many wrong submissions Starting to blame everyone you see for cheating using such petty claims is making things worse

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kalpitdon7

4 месяца назад, # |

← Rev. 2 →

Independence day Contest !

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panku8433

4 месяца назад, # |

← Rev. 2 →

All the best to all. Hope you all get +100.

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Mohamed_Saad62

4 месяца назад, # ^ |

I need 130 :) but 100 still good and will make me happy

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4 месяца назад, # ^ |

thanks

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.XiaoXia

4 месяца назад, # |

How to gain rating in EDU Round? I've droped the rating at almost every the EDU round.

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Manish_Jangir

4 месяца назад, # ^ |

Same here

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123gjweq2

4 месяца назад, # ^ |

+45

during the contest, if you see that your position is too low to gain rating, it is best to solve another problem

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Nick9

4 месяца назад, # ^ |

based

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bhardwajsatyam

4 месяца назад, # ^ |

What do you mean?

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pink_floyd25

4 месяца назад, # |

Noob doubt: What's the difference between Education rounds vs Div 3 vs Div 4 ?

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Azuken

4 месяца назад, # ^ |

educ ≈ div2 and div4 < div3 < div2

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Adityaxv

4 месяца назад, # |

I'm glad that there are no tester comments.

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Abdo_Hussain

4 месяца назад, # |

i love edu rounds <3

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lucas.morrison

4 месяца назад, # |

Pupil rank happens TOMORROW

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lucas.morrison

4 месяца назад, # ^ |

Looking forward to weeks of therapy for failing to do C over my misinterpretation of what the k variable meant...

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nvvy

4 месяца назад, # |

← Rev. 2 →

How to Hack someone's solution?? Shayan bro i think you should make a video on this topic!! I tried many times to hack someone.. but always ( Unsuccessful Attempt )

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https://codeforces.net/contest/1999/hacks/1066931

4 месяца назад, # ^ |

← Rev. 2 →

You should give test with high strength. Here is an example: 274841071

If you can't understand this code, it doesn't matter. You just need to know what he is doing. In this code, he's using Substring all the times which is $$$O(|s|)$$$ in time. It will get a TLE by a test with high strength. Thus, I give a test like this: 1, and then 200000 ?s, and then an a.

Unaccidentally, he is hacked by me.

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nvvy

4 месяца назад, # ^ |

You just used generator to generate worst case for s right?? But how did you figure out it'll give TLE..

How to practice for this?? I am very bad to figure out the time and space complexities

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4 месяца назад, # ^ |

You can search the submissions in “Status”. And sort it by execution time at the end of the page. And get to the last page. The submissions there are almost TLE so that maybe TLE after you hack them.

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4 месяца назад, # ^ |

What generator do you use?

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4 месяца назад, # ^ |

Usually I use C++’s output. You can use what you familiar with.

if the test isn’t that long, just input the test by text is enough.

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4 месяца назад, # ^ |

There you've wrote string s of length 2*10^5, how did you do that, and how to generate those big test cases?

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4 месяца назад, # ^ |

← Rev. 2 →

There is a “Generate Input” Botton when you hack. Press it and paste your code there.

Example:

#include<bits/stdc++.h>
using namespace std;
int main(){

    cout<<“1\n”;
    for(int i=0;i<=200000;i++)                 cout<<“?”;
    cout<<endl<<“a\n”;
}

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4 месяца назад, # ^ |

Okay, this way! Thanks.

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nvvy

4 месяца назад, # ^ |

I tried this multiple times but never got successful Attempt

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4 месяца назад, # ^ |

Try more, you’ll gain from it.

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4 месяца назад, # ^ |

What’s more , see which problem is hacked most. It may mean that it is easier to hack than other problem.

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zer0_123

4 месяца назад, # |

← Rev. 2 →

Sorry I’m a beginner here, how are edu rounds different from normal contests ?

PS: Got it, thank you!

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NOTHING_2_LOSE

4 месяца назад, # ^ |

← Rev. 2 →

The difficulty level is usually the middle ground of div2 and div3 constests.

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4 месяца назад, # ^ |

You can't hack during the contest, but you can EVERYONE hack after contest.

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tobyAcodes

4 месяца назад, # ^ |

I am a total beginner and have no idea what hacking means in this context. Like changing the answer or sth? Where should I start from

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https://codeforces.net/help

4 месяца назад, # ^ |

Read about hacks.

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https://codeforces.net/blog/entry/456

4 месяца назад, # ^ |

← Rev. 5 →

In some contests like Div. 2 Round, there is a Codeforces Format .

Maybe in most websites, it will give you the result at once when you submit. But if Codeforces, there is only a set of tests called Pretests.

It just means that this is just the preliminary test which shows you if the code can run normally. If you passed pretests, you can lock this problem and hack participants who pasts pretests in your room, by giving tests with high strength to make them failed passing this problem.

P.S. If you lock the problem, you can't resubmit any more until the contest ends.

To get more information, you can read this blog

In Div. 3 , Div. 4, and EDU Round, it is EX-ICPC Format. There will be a 12-hour-long phase at which you can hack everyone after the contest.

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4 месяца назад, # ^ |

← Rev. 3 →

Different from Div.1 , Div.2 , Div.1+2 contests, it is an EX-ICPC Format.

Similar to Div.3 and Div.4 contests. You can see participants rated and unrated separately in the standings after system testing. You can view last Div.3 contest as an example: Codeforces Round 966 (Div. 3)

And view this round to see the probable difficulty of each problem. Educational Codeforces Round 168 (Rated for Div. 2)

standings: View rated participants rated separately by pressing the Both divisions v botton at the rightmost of your screen. Then cancel the tick of Show unofficial one.

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_regular_show

4 месяца назад, # |

Thanks to all who contributes to this contest. Hope I can go back to specialist after this one!

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Enatsu__8

4 месяца назад, # ^ |

hoping for the same!

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4 месяца назад, # |

Hope for 1300+ rating

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4 месяца назад, # ^ |

good luck (you did well in the last contest congrats)

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4 месяца назад, # ^ |

thanks, you too

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truth

4 месяца назад, # |

Let's go specialist in this one

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3rian_jr

4 месяца назад, # |

I want to go back to Newbie...

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Nahodkin

4 месяца назад, # |

Ждём каждый educational раунд!

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21Cash

4 месяца назад, # |

My first contest as specialist. Good luck everyone.

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Md-Merazul-Islam

4 месяца назад, # |

Best of luck everyone. Hopefully, we all increase our rating. And I will back specialist.

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Md-Merazul-Islam

4 месяца назад, # |

Best of luck to all. Hopefully, we will increase our rating in this contest =)

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4 месяца назад, # |

+15

Bet it or not, I'm going to reach Master this round.

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4 месяца назад, # ^ |

Good luck to you. Good luck to everyone.

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4 месяца назад, # ^ |

Not happening, ahahaha

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4 месяца назад, # ^ |

← Rev. 3 →

I didn't say I bet it. haha.... ha... Failure is common.

Nevermind. I'll be still there next round.

How can you reply me just minutes before the contest, are you watching me? Thank you for your watching stuttering

Maybe I'm going to change my handle into Lmao_Fox if this happens too many times.

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Ami_Nul

4 месяца назад, # |

hopefully expert after this one

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ArcticFox

4 месяца назад, # |

← Rev. 2 →

best of luck guys

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Ahmed_Yosry

4 месяца назад, # |

any interactive problems?

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ankitchauhan

4 месяца назад, # |

I want to regain my pupil status

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ankitchauhan

4 месяца назад, # |

I want to regain my pupil status.

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jatrophalouvre

4 месяца назад, # |

+40

If I don’t get CM I’m gay.

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4 месяца назад, # ^ |

+25

Congratulations! At least it's legal in Singapore since recently.

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Enatsu__8

4 месяца назад, # ^ |

nice way of coming out of the closet

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CodeKira

4 месяца назад, # |

hope i become pupil , pls be easy for me

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CodeKira

4 месяца назад, # ^ |

and it happened nice

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Enatsu__8

4 месяца назад, # |

Hoping for a positive delta!

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VS34

4 месяца назад, # |

In Educational Round dose speed of solving question matters or only number of questions solved during the contest matters and does not depend even on the rating of the question solved??

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4 месяца назад, # ^ |

← Rev. 2 →

Speed matters. Rank is sorted first by number of solved problems, then time (10 minutes penalty for 1 wrong submission)

That means solving A and solving G are the same. It just calculates the total number of your solved problems and then the time.

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flexyw1be

4 месяца назад, # |

I don't care what rating I will have I just solve tasks for fun

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Maria_Sultana

4 месяца назад, # |

-9

InShaAllah, I will get +100 today

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modassirfaruquee

4 месяца назад, # ^ |

I also prayed to Allah to make me capable of solving at least 4 problems but i guess it wasn't better for me

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SW7777

4 месяца назад, # |

Hope to solve four problems!

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Steven-_-AbuAlkhair

4 месяца назад, # |

c < b

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deepak_221

4 месяца назад, # ^ |

https://codeforces.net/contest/2004/submission/276681600 why its wrong

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Steven-_-AbuAlkhair

4 месяца назад, # ^ |

se is changed after se+=min(k,fi-se) so you need to use temp like t = in(k,fi-se) , se+=t

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deepak_221

4 месяца назад, # ^ |

i was dumb thanks for solution

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Karatinos

4 месяца назад, # |

-13

worst contest((

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donpaul2718

4 месяца назад, # |

← Rev. 2 →

+26

Great D but C < B.

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123gjweq2

4 месяца назад, # ^ |

was C really that easy?

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https://codeforces.net/contest/2004/submission/276691814

4 месяца назад, # ^ |

it was easy I still got 6 WA's cause i didnt read "total" increase <=K

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123gjweq2

4 месяца назад, # ^ |

ahh, that changes a thing or two

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aviralarpan3301

4 месяца назад, # ^ |

or three

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jigyashugupta

4 месяца назад, # ^ |

OOOHHHHH MMYYYYYYY GOODDDDDDDD!!!!!! YES!! YESS !! YESS!! YESS!! NO !! NO!! NO!! NO!!

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lmn207

4 месяца назад, # ^ |

gotta be honest the difficulties of problems from A->D is too low, compared to normal div2 rounds or just edu rounds. I choked this one but generally speaking, i think this contest was not so good

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Baby_Bird

4 месяца назад, # ^ |

D offered a little bit of implementation fuss despite being not that hard conceptually.

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Jerrasterix

4 месяца назад, # ^ |

← Rev. 2 →

Mind pointing out where my mistake was ??

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phsads

4 месяца назад, # |

← Rev. 2 →

+60

so what the hell are you supposed to do for E (nims and games are kinda like the opposite of my skillset lol)

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UP84

4 месяца назад, # ^ |

Find grundy numbers using SPF.

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kwangg

4 месяца назад, # ^ |

+13

I don't know why it was named "not a nim problem". I solved it as a nim problem. Case analysis on small numbers showed that the nimber of a pile is basically the "rank" of its smallest prime factor (where a prime having rank k means it's the kth prime number, except that the rank of 2 is 0.)

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adaptatron

4 месяца назад, # |

← Rev. 2 →

Video editorial for D: Colored Portals

Here's a very fancy implementation of D: Colored Portals using Bitmask tricks that was introduced in D: Cases from a recent CF round. I also have a video editorial for the old problem.

Let's represent all portals as bitmasks of length 4. Then, a path between 2 nodes is possible if and only if the bitwise AND of both the numbers is not zero.

Assume $$$x < y$$$. On a number line, we know that the shortest path between 2 points is the straight line connecting them. So, if $$$a[x] \& a[y] \neq 0$$$, then, the answer is $$$|x - y|$$$. Otherwise, we need to jump to a portal that differs in exactly one color (i,e, the bitwise AND should contain exactly 1 set bit). From there, we can jump to $$$y$$$ directly.

Define $$$ldp[i]$$$ to be the nearest portal to the left that differs in exactly one color. You can populate it from left to right by maintaining a map of the latest bitmasks seen so far. Similarly, you can populate $$$rdp$$$.

There are only 3 possible paths. We take the best one amongst them.

$$$x \rightarrow y$$$
$$$x \rightarrow ldp[x] \rightarrow y$$$
$$$x \rightarrow rdp[x] \rightarrow y$$$

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4 месяца назад, # ^ |

i got the solution down immediately, but i could not implement it at all, fk me

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4 месяца назад, # ^ |

I also got the solution, but my implement is not as good as the method above so... I don't know where I went wrong (I use binary search btw)

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4 месяца назад, # ^ |

i tried using a map and then i also tried bs and some other stuff and i just couldnt get it. unlucky :(

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Saivishwas

4 месяца назад, # ^ |

276681877

You can look at my case work. I am laughing at my own code.

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4 месяца назад, # ^ |

i shouldve just quit trying to do it nicely and just do that. good job ngl

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aditya0610

4 месяца назад, # ^ |

Nice work

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4 месяца назад, # ^ |

← Rev. 6 →

oh well I notice where I went wrong, need to reverse the array and position to binary search to get the correct answer. Not searching the y+1 or x+1 or (x+y)//2 will do

UPD: after fix the bs bug, my solution got TLE in test 6

In the hindsight, just maintain ldp and rdp will do me a favor more than bs

UPD: Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)

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Astroflexx

4 месяца назад, # |

+20

And the trend of 4k+ solves on D in Edu Rounds continues...

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FuyuMai12

4 месяца назад, # ^ |

This D is only of Div2C level at most.

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_inferno_

4 месяца назад, # |

whoops

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Wandoka

4 месяца назад, # |

+170

Why would you ever call a nim problem — $$$\texttt{Not a Nim Problem}$$$

Like I really don't understand, because it is funny?

When I read the name, I though that it is a bad idea to include "Nim" in the name of the problem, because when you search for nim tasks, this one would show up.

When though of the solution, I became even more confused. It was a nim problem. So why the author is telling me it is not true?

I thought that the name was part of the statement. I started to doubt myself that I got something wrong. Turns out everything is fine, nim works. Downvoted the contest when I saw "Accepted" verdict.

And also this statement:

It is not allowed to take from the pile a number of stones y such that the greatest common divisor of $$$x$$$ and $$$y$$$ is not equal to $$$1$$$.

Why you say "not not A" instead of just " A "

We can rewrite it like this:

It is only allowed to take from the pile a number of stones y such that the greatest common divisor of $$$x$$$ and $$$y$$$ is equal to $$$1$$$.

This sentence is so confusing that even an announcement was made, where it was stated that

In other words, gcd(x,y) must be 1.

If it is so confusing, maybe the statement should be fixed? I don't understand why the announcement was made, and this sentence was not changed.

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4 месяца назад, # ^ |

+26

Lol, bro, relax

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san9blin

4 месяца назад, # ^ |

Agree with the title, just why?

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beyondpluto

4 месяца назад, # ^ |

Oh bro, is it a nim problem?
I immediately got idea for single pile, thought of using nim for n independent piles, but then I saw the problem name and thought its going to be some special case where nim doesn't work. And spent whole contest trying to solve without nim lol. I should understand nim better.

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CrackInTheSky

4 месяца назад, # ^ |

+12

this nim thing is unfamiliar to me. any suggested resources ?

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.-__-.

4 месяца назад, # ^ |

I really think you do not need to know that at specialist rank.

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JasamritRahala

4 месяца назад, # ^ |

Hey, I was confused too. This article seems good :) https://codeforces.net/blog/entry/66040

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CrackInTheSky

4 месяца назад, # ^ |

thanks

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The5threich

4 месяца назад, # ^ |

You can read the wikipedia article on sprague-grundy theorem and Nim Sum. I think they are pretty well written.

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Nishu_2005

4 месяца назад, # ^ |

Agreed, read it wrong due to the not not thing and thus couldn't do it in time.

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4 месяца назад, # |

bruh C question due to "total" K wasted my 1 hour and 5 WA anyone else also faced same issue?

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UP84

4 месяца назад, # |

Every page from problem to submit takes so much time to load. It's infuriating.

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4 месяца назад, # |

← Rev. 2 →

ARE THESE GRUNDY NUMBERS CORRECT ?

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mr_swashbuckler

4 месяца назад, # ^ |

Yes

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4 месяца назад, # ^ |

I could generate grundy numbers, but didn't know how to extend them to 10^7. Small hint please ?

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fonmagnus

4 месяца назад, # ^ |

← Rev. 3 →

+38

For even numbers, grundy is always 0. For 2, it is obvious because you can only take 1 and the grundy of 1 is 1. For the rest, because player is not allowed to take 2, then $$$g(2)=0$$$ would always be the mex of all options

For prime numbers, the grundy is $$$k$$$ where $$$k$$$ is the order of prime numbers (e.g. $$$g(3) = 2$$$, $$$g(5) = 3$$$ etc). This is because all prime numbers $$$p$$$ can have options of $$$1,2,3,\ldots,p-1$$$. Therefore the mex is always $$$k$$$ because it hasn't existed yet

For the rest, the grundy would be smallest prime factor of that number. Because it would be the smallest number that doesn't exist for all its transition (hence, it similar to the nature of mex). The other number would exist because if the smallest prime factor is $$$p$$$, then all numbers between $$$1, 2, \ldots, p-1$$$ can be taken of that pile. Hence, all numbers between $$$0, 1, 2, \ldots, g(p-1)$$$ can be exist as transition moves and won't appear as mex.

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4 месяца назад, # ^ |

← Rev. 2 →

Thank you very much for explanation.

I could see the first and second observations from the brute-force that I had written.

But I couldn't deduce the third observation :( . otherwise I would have solved this.

Anyone who is interested in how to generate grundy numbers here, please refer to this.

c++ code to generate grundy number

~~~~~~

grundy[0] = 0; grundy[1] = 1;

    grundy[2] = 0;

    repl(i,3,100) {
       set<int> grudySet;
       repl(j,1,i-1) {
         if(gcd(i,j) == 1) {
          grudySet.I(grundy[i-j]);
         }
       }

       int mex = 0;
       while(grudySet.find(mex) != grudySet.EN) {
         mex++;
       }

       grundy[i] = mex;

    }

    repl(i,1,100) {


       cout << " for the i " <<  i << " its grundy number is " << grundy[i] << endl << endl;


    }

~~~~~~~

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4 месяца назад, # ^ |

Don't worry it took me 90 minutes to get it haha. F was much more straightforward today, so it's a good thing I wasn't afraid to skip.

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4 месяца назад, # ^ |

small doubt, can we generate all the primes till 10^7 within the given time-limit ? ( never generated primes till 10^7, so asking ).

We need to find the what is the order of the given prime number, and we can't do that unless we have all the prime numbers within 10^7.

Do we need to implement SegmentedSieve ? or normal sieve is sufficient ?

fonmagnus

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fonmagnus

4 месяца назад, # ^ |

I think if we skip the even numbers, then we can do it for $$$5 \times 10^6$$$ numbers and should be fine (it's kinda close to TL but I guess that's one way to prune it)

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4 месяца назад, # ^ |

Sure, will try later.

I wish, the pile sizes were within 10^5 range.

then I would have solved this question in 25-30 minutes :P .

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4 месяца назад, # ^ |

yes we can. my submission ran in 406ms and i am running sieve till 1e7

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october_rain

4 месяца назад, # ^ |

can we generate all the prime still $$$10^7$$$ within the given time-limit?

yes, use linear sieve

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4 месяца назад, # ^ |

Here, you forgot to include a link.

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4 месяца назад, # ^ |

Normal sieve, if implemented correctly, runs in O(n*log(log(n))), which is sufficient (my sol passed with it).

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indian_rounds_LULW

4 месяца назад, # |

+68

Don't use " NOT NOT " in your statements ever again please

And yes, I googled E, sue me (I suspect many others did or have seen this problem before anyway)

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MVP_Harry

4 месяца назад, # ^ |

This makes me wonder how many people thought of E during the contest by themselves instead of just googling. Because I couldn't think of anything after more than an hour (sad noises**)

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indian_rounds_LULW

4 месяца назад, # ^ |

← Rev. 2 →

I have spent most of the time going back and forth: "This is coprime nim. No, this isn't coprime nim. What if I take only divisors? But I can take any non coprime number. Wait, this is coprime nim, gcd should be equal to 1"

In regards to coprime nim (and not divisor nim, non-coprime nim and other variations) I've come up only with the most obvious stuff on my own:

0 is losing, 1 is wining and even numbers are always losing (since you are always forced to reduce the pile to some odd number, and from and odd number you can always go back to an even one)
for a prime number any move except for the number itself is possible

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Fck_cheaters

4 месяца назад, # |

say my name

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aqua4

4 месяца назад, # ^ |

Fck_cheaters

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SuperMo

4 месяца назад, # |

D is maybe just next and previous tables but in order for me to implement it I would need maybe 400 line looking forward to see how people did it nicely

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jzcrq

4 месяца назад, # |

I'm curious if E is a classic problem of SG function. I don't know how to do it, but I have a vague feeling that it can do it.

→ Ответить

Samarth_Rajj

4 месяца назад, # ^ |

what is SG function?

→ Ответить

Chiral

4 месяца назад, # ^ |

Sprague-Grundy function, super useful for nim-like problems.

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jzcrq

4 месяца назад, # ^ |

Consider treating the game process as a DAG, taking the SG function value of a node as the mex of the SG function values of its child nodes.

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ExpectoPatronum

4 месяца назад, # ^ |

← Rev. 3 →

I just generated all values of SG and guessed, then I realized that $$$g(x)$$$ is equal to the smallest prime factor of $$$x$$$ ($$$g(x) = 0$$$ if $$$x$$$ is even))

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mathtsai

4 месяца назад, # |

I come back to pupil again, thanks to problem B : )

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tickbird

4 месяца назад, # |

← Rev. 2 →

C < A < D < B

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JakobZ

4 месяца назад, # |

+40

Maybe I'm salty for not solving E, but is it ok to put a somewhat well known problem that apparently has a solution on the internet to a contest?

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jmyszka

4 месяца назад, # ^ |

← Rev. 2 →

+13

I am not a big fan of E myself but its edu round so the authors are somehow justified

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Nishu_2005

4 месяца назад, # ^ |

Should such a contest be rated for anyone at all? C,D,E all were kind off not div2 level. If someone had a day where they did B and D fast enough (I couldn't, maybe thats why i am putting this out), E was a walkover.

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JasonMendoza2008

4 месяца назад, # |

Lovely problems thanks, problem B and D were a bit implementation heavy unless I just missed the obvious solution and did something unnecessarily long.

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SuperMo

4 месяца назад, # ^ |

I think that D need a trick like bitmasks or else it would be so much pain

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JasonMendoza2008

4 месяца назад, # ^ |

My solution was a few hundred lines long without bitmasks haha

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4 месяца назад, # ^ |

I got WA test 2 in D until the contest is over... unable to fix the code in time

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i3mr125

4 месяца назад, # ^ |

binary search the indices would be enough no need for bitmasks

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Dinh_Van_Tung

4 месяца назад, # ^ |

OMG! same avt. best of luck

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i3mr125

4 месяца назад, # ^ |

yeppiee, u too

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4 месяца назад, # ^ |

look at mine 276671836

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vermish77

4 месяца назад, # |

I dont get my code is O(qlogn) time complexity in D but still getting tle on test case 6 , can someone please help where it I am wrong , this is my solution link [submission:276660215][My submission link](http://codeforces.net/contest/2004/submission/276660215)

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4 месяца назад, # |

← Rev. 2 →

Why this code (https://codeforces.net/contest/2004/submission/276666440) shows strange behavior when all the array elements are equal, for example:
5 4
3 3 3 3 3
Shouldn't the answer be 3 here?
Also i tried printing the final array after all the operations done, and in this case it is 3 3 3 1 0, but according to code none of the elements should change. Where i am going wrong?

Update: There was problem with my keyboard language setting due to which a special character was in the input instead of 3. But the solution is still wrong i think.

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LiuXingJian

4 месяца назад, # ^ |

The variable ind will still bea -1 after your operations on nums, adding an another verification may help.(Though the solution is still wrong as you don't need to adjust all the numbers)

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4 месяца назад, # ^ |

Okay, let me try once again, thanks.

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4 месяца назад, # |

Wasted so much time on B lol, though D was very enjoyable!

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o.khalifa

4 месяца назад, # |

I enjoyed these problems

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LJY--2002

4 месяца назад, # |

How B ?

→ Ответить

sandbag

4 месяца назад, # |

+14

If you ever feel stupid, I got +7 penalties on D despite solving it in 20 minutes due to writing this function:

Code

bool overlap(string&a, string&b) {
    if(a[0] == b[0] || a[1] == b[1]) return true;
    return false;
}

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4 месяца назад, # ^ |

small piece of advice.next time create character set. and put characters of both the strings inside the set.

If the set size is less than (a.size() + b.size()) , means there is an overlap <3 .

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MohammadHijaz

4 месяца назад, # ^ |

I had the same issue and couldn't figure it out in contest lol. 276656805

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ZyadFva

4 месяца назад, # ^ |

← Rev. 2 →

Say less

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4 месяца назад, # ^ |

i got also got penalty on D, because of this: wrong submission: 276659462 correct submission: 276671836

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anuragmishra0521

4 месяца назад, # |

I'm a newbie. Any suggestions for me on how to learn from these contests and ace these problems?

→ Ответить

4 месяца назад, # |

← Rev. 2 →

can any one tell me what is the wrong in my code in (d) or if i missed some cases may be 276679658

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34z12000

4 месяца назад, # |

← Rev. 3 →

+14

My sol for D looks like abomination 276680579

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4 месяца назад, # ^ |

at least you solve it

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4 месяца назад, # ^ |

R u blind? He didn't.

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4 месяца назад, # ^ |

oh my god i was think he solve it

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san9blin

4 месяца назад, # ^ |

He did later.

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4 месяца назад, # ^ |

Lol, u r also blind xD since he didn't

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san9blin

4 месяца назад, # ^ |

Not during contest, but he did :)

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4 месяца назад, # ^ |

Doesn't count :D

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4 месяца назад, # ^ |

thank you

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JasonMendoza2008

4 месяца назад, # ^ |

It clearly says Runtime error on test 2 so they didn't.

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34z12000

4 месяца назад, # ^ |

← Rev. 2 →

sorry, wrong submission

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Astroflexx

4 месяца назад, # ^ |

Holy mother of God what is that code

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4 месяца назад, # ^ |

Bro what, 800 python lines...

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blablabla1803

4 месяца назад, # ^ |

← Rev. 2 →

276679322 Mine is not too far off xD , just that mine TLE'd

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4 месяца назад, # |

Lmao i just brute forced the first 100 grundy numbers and stared at the screen for 45 mins until i observed pattern

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4 месяца назад, # ^ |

what is the pattern in this problem?

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4 месяца назад, # ^ |

for even its 0 and for odd it is same as the grundy number of its smallest prime factor

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4 месяца назад, # ^ |

Oh god, I've just realized that I misread the statement completely LOL. There was two "not" in the statement @@.

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apex

4 месяца назад, # |

How do you solve D??

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4 месяца назад, # ^ |

if x1 and x2 dont share a colour, then you want to find the closest city to the right of x1 and the closest city to the left of x1 that share only 1 colour, then you calculate which of the 2 paths is better. im so salty i couldnt implement this, but this is the solution i think

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4 месяца назад, # ^ |

i was think if you find the closest city to minimum one between of them you can find the answer and i think this do same idea you mention but i got wrong 2

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adaptatron

4 месяца назад, # ^ |

I added a 13 minute video editorial. Hope this helps.

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differentialplantain

4 месяца назад, # ^ |

← Rev. 2 →

For each pair query, there are three cases you have to consider:

Case 1: The two portals are not opposites (i.e. they share a color)

In this case, you can just directly teleport with a cost of their distance in the array

Case 2: The two portals are opposite (they do not share a color), but between them there is a portal color combo which is different from both

In this case, our cost is also just their distance in the array b/c you can just teleport to the element between them to get from one to the other

Case 3: The two portals are opposite (they do not share a color) and there is no element between them which has a different color combo than either one

In this case, we have to calculate the shortest distance from the left of the leftmost query point to an element different from it which is not its opposite and likewise for the rightmost query point to the right instead. We then take the minimum of these distances we call mindist which naturally gives the total cost for the query to be 2*mindist + the distance between the query points. If there is no such minimum distance (i.e. there is no way to get between the query points at all) then we instead output -1.

One can implement the third case by iterating forward and backward through the array for each color combo to fill an array of pairs calculating the index of the nearest element that is not equal and not opposite to its color combo both to the left and right.

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4 месяца назад, # |

can you tell me what's wrong in this solution for C? https://codeforces.net/contest/2004/submission/276677983

my code for C could pass the sample test cases but i kept getting WA after submission :(

i wasted 1 hr on it lmao, was C really so easy?

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Sylvester_

4 месяца назад, # ^ |

try using long

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4 месяца назад, # ^ |

just tried using long long, still WA

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Sylvester_

4 месяца назад, # ^ |

Look at my code if it help

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denk

4 месяца назад, # ^ |

The problem is that if the array is initially of odd length, then you create a phantom element equal to 0 and further consider it as the original one and, accordingly, you can add values to it, which cannot be done, in fact, it does not exist.

Test when the code is not working correctly:

3 1

1 5 5

Program output: 0

Correct answer: 1

According to the logic of your program, element 0 is created and the array turns from [1, 5, 5] into [0, 1, 5, 5]. After that, in order to get the optimal answer, your code adds 1 to 0 and you get an array [1, 1, 5, 5]. Thus, the result will be equal to 0.

It turned out that we optimized the response using a non-existent element.

In a situation where the array is of odd length, you need to use another method — immediately add the minimum value in the array to the answer and then remove it from the array, thereby making it an even length

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4 месяца назад, # ^ |

wow i wasn't expecting an answer so well explained lol. thanks a ton, i understand it now.

honestly i feel i could easily be cyan but i always mess up small things like this, for example related to the range of inputs, such as writing >x instead of >=x for some loop, etc. and realizing that after wasting 25 minutes on that. since you're blue do you have any suggestions for me?

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denk

4 месяца назад, # ^ |

The best practice will be to solve implementation tasks outside the contest and set a goal to pass the task the first attempt. That is, spend a lot of time checking all the key points of the code and be 99% sure that all the places are implemented correctly.

In the contest itself, you can:

1) Write a naive solution for small values and compare the answers received with the code that you are going to send

2) Manually check the boundary values, experiment with the boundaries (try to change > to >= or even to <, try to change the data type (if an overflow occurs))

3) Check each part of the code and intermediate values separately (whether the array was built correctly, whether the answer is updated correctly at each step, whether the necessary variables are used everywhere, etc.)

Usually these steps are enough, if not immediately to send a solution without bugs, then after an incorrect attempt to quickly find where it can be the error and fix it

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4 месяца назад, # ^ |

makes sense, thanks. i'll try solving a bunch of div. 2 Bs and Cs and try to get them right on the first try

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sv1shan

4 месяца назад, # |

Is there a $$$ O(n^2) $$$ solution for F?

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sstrong

4 месяца назад, # ^ |

just use hashmap

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https://codeforces.net/contest/2004/submission/276639528

4 месяца назад, # |

← Rev. 2 →

too tight constraints for D. This is a QlogN and not passing? [EDIT SOLVED . I was declaring the vector again]

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4 месяца назад, # ^ |

O(N+Q) is possible (via previous/next tables for all six pairs, for example), so this may be a TLE intentionally

→ Ответить

https://codeforces.net/contest/2004/submission/276683926

4 месяца назад, # ^ |

But there will be only 4 pairs possible

with vectors+lower bound tle.

Do you mean to say the complexity is not O(qlogn) but O( (n+q)logn) ?

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4 месяца назад, # ^ |

I meant to say that an O(n+q) solutions exist, so I'm not sure if O(qlogn) is intended to pass. As @rahulmysuru7 pointed out, your code runs even slower, so it's not relevant to you, I was just guessing what might be the reason

→ Ответить

4 месяца назад, # ^ |

Ohh I see. I will think of N+Q solution too.

Thanks for the response <3

→ Ответить

rahulmysuru7

4 месяца назад, # ^ |

← Rev. 2 →

@mahendraakshansh ~~~~~ for(int i=0; i<4 ;i++){ set ss = mp2[vv[i]]; .... ~~~~~

yeah ,you fked up here. you are declaring and extracting the entire set inside here. making your code O(qlog(n) * n)

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https://codeforces.net/contest/2004/submission/276681345

4 месяца назад, # ^ |

fck me thank you so much.. I am a noob

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Vitaly503

4 месяца назад, # |

+44

E is too educational even for educational round

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newStartNotSure

4 месяца назад, # |

not sure why this is TLE?

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vermish77

4 месяца назад, # |

I dont get my solution is of O(qlogn) but I am still getting TLE on sixth test case, tried so many things but still got tle on sixth, please tell where I am wrong my submission, My submission Link

→ Ответить

umar.aligarh

4 месяца назад, # ^ |

you are copying the whole vector everytime in binsearch, this could be the issue vector vec=mpp[s];

→ Ответить

vermish77

4 месяца назад, # ^ |

And, when I thought its better to be good at code writing , I fcked up. Thnx man got accepted

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RoomTemperatureIQ

4 месяца назад, # |

+13

B is the hardest out of the first 5 tasks for me 😭 Also, how tf does E have like 1400 ACs? Do so many people know sprague grundy now?

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4 месяца назад, # ^ |

apparently its a well known problem with solution on the internet, im just surprised so many ppl managed to actually implement D.

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sandbag

4 месяца назад, # ^ |

B was also hardest for me. I wasted 20 minutes on casework before just (correctly) guessing the pattern. E was googlable.

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4 месяца назад, # ^ |

I remember seeing a cf blog about game theory a few weeks ago that explained how to solve Nim problems, so that might be why

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LJY--2002

4 месяца назад, # ^ |

so, How B? Help me please !

→ Ответить

RoomTemperatureIQ

4 месяца назад, # ^ |

I just bruted for all bridges [i, i+1] upto 100 in the end

→ Ответить

4 месяца назад, # ^ |

← Rev. 2 →

Break the problem in several cases.
ref : (https://codeforces.net/contest/2004/submission/276649235)
Case 1: When there is no overlap ----> only blocking one gate will work.
Case 2: When there is an overlap at exactly one point ----> Block two gates.
Case 3: When both the intervals are same ---> Answer will be difference between them.
Case 4: When all the 4 points given are distinct ---> answer will be diff b/w overlap region + 2
Case 5: When only 3 of them(l, r, L, R) are distinct ----> answer will be diff b/w overlap region + 1

Analyse all the cases in order on pen & paper you will get the idea.

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isjaya

4 месяца назад, # ^ |

Oh I missed case 2, completely skipped my mind

→ Ответить

4 месяца назад, # ^ |

Same here, took time to figure it out.

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pmmitkovv

4 месяца назад, # ^ |

You can just consider the intersection of the two intervals(you need to place a door between every two cells). Then if there is anything to the left or right of the intersection you need to add 1 door for each case. In case the intersection is empty the answer is 1 as you pointed.

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LJY--2002

3 месяца назад, # ^ |

thanks everyone

→ Ответить

4 месяца назад, # ^ |

should I feel proud that the problem I found easiest was hardest for Candidate Master (even thought I could only solve till C) :///

→ Ответить

_inferno_

4 месяца назад, # ^ |

← Rev. 2 →

no, as they have a room temperature iq

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Karabutsa

4 месяца назад, # |

+61

Not a Nim problem

looks inside

a Nim problem

Why? Just, why?

→ Ответить

Vassu05

4 месяца назад, # ^ |

← Rev. 3 →

+17

Fr I was not even thinking around the lines of NIM after looking at the title. 1400+ accepted submissions for that is crazy

→ Ответить

JasamritRahala

4 месяца назад, # |

+17

One small step for man, a giant fall for my rating.

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comebackking

4 месяца назад, # |

+12

Don't know why I hate L and R problems which have O(1) solution. Just hate the casework! P.S Talking about problem B

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PaciukZvichainyi

4 месяца назад, # |

← Rev. 2 →

Damn, my programming skill is so bad that I couldn't implement C. The first test cases in problems are weak and doesn't help to find even some edge cases. E has no explanation for how Bob wins in the third case. Next year will be our year.

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4 месяца назад, # |

+10

if u ever feel stupid then just know that I exist : wasted my whole contest on C only to realize it was total increase<=k, not k for every element

→ Ответить

Au.

4 месяца назад, # |

The question of double negative sentences is very unfriendly to foreign contestants, which seriously affects the questioning, and I feel very angry about this

→ Ответить

Aman_Garg

4 месяца назад, # |

Can someone pl look what is wrong in my implementation of D.

Link

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teoet6

4 месяца назад, # ^ |

if it_low==pos.end() you should try finding the last position <= x

→ Ответить

sandbag

4 месяца назад, # |

D was actually easier to implement than I thought after I realized you could iterate over the colors and only check when you found an overlap for both endpoints. Then just binary search on the nearest index.

Code

bool overlap(string&a, string&b) {
    if(a[0] == b[0] || a[1] == b[1] || a[0] == b[1] || a[1] == b[0]) return true;
    return false;
}

void solve(){
    int n, q; 
    cin >> n >> q;
    vector<string> v(n);
    for (int i = 0; i < n; i++) cin >> v[i];
    map<string, vector<int>> mp;
    vector<string> cols = {"BG", "BR", "BY", "GR", "GY", "RY"};
    for (int i = 0; i < n; i++) {
        mp[v[i]].push_back(i);
    }
    for (int i = 0; i < q; i++) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        if(overlap(v[a], v[b])) cout << abs(a-b) << endl;
        else {
            int ans = 1e9;
            for (auto j : cols) {
                if(overlap(j, v[b]) && overlap(j, v[a])) {
                    if(mp[j].size()==0) continue;
                    int l = -1, r = -1;
                    int it = (lower_bound(mp[j].begin(), mp[j].end(), a)-mp[j].begin());
                    if(it-1 >= 0) l = it-1;
                    r = it;
                    int lid = (l != -1) ? mp[j][l] : -1;
                    int rid = r<mp[j].size() ? mp[j][r] : -1;
                    if(l != -1 && lid != -1) ans = min(ans, abs(a-lid) + abs(lid-b));
                    if(r != -1 && rid != -1) ans = min(ans, abs(a-rid) + abs(rid-b));
                }
            }
            // not sure if both loops are necessary, just did to be safe
            for (auto j : cols) {
                if(overlap(j, v[b]) && overlap(j, v[a])) {
                    if(mp[j].size()==0) continue;
                    int l = -1, r = -1;
                    int it = (lower_bound(mp[j].begin(), mp[j].end(), b)-mp[j].begin());
                    if(it-1 >= 0) l = it-1;
                    r = it;
                    int lid = (l != -1) ? mp[j][l] : -1;
                    int rid = r<mp[j].size() ? mp[j][r] : -1;
                    if(l != -1 && lid != -1) ans = min(ans, abs(a-lid) + abs(lid-b));
                    if(r != -1 && rid != -1) ans = min(ans, abs(a-rid) + abs(rid-b));
                }
            }
            if (ans==1e9) cout << -1 << endl;
            else cout << ans << endl;
        }
    }
}

→ Ответить

godwriter

4 месяца назад, # |

A, B, C, D, F is fucking easy, especially E is just grundy number which is not suitable for an edu round!

→ Ответить

rmz

4 месяца назад, # |

+12

fucking rooms

→ Ответить

hilbertmf

4 месяца назад, # |

I didn't notice the total increase in problem C should not exceed k, I thought we could increse up to k for each element... smh

→ Ответить

sakib.safwan

4 месяца назад, # |

← Rev. 2 →

For problem E, I found out quite early that I had to build grundy numbers for all possible number.

But I completely forgot all the theories. Almost all the time that I took for solving E was trying to figure out grundy again and for reading the question wrong and solving a wrong problem first. :3

→ Ответить

4 месяца назад, # |

← Rev. 5 →

can anybody point out whats wrong with the below idea for D ? (since failing testcase 2 is not visible till end of hacking) Submission: 276681155

Spoiler

→ Ответить

kwangg

4 месяца назад, # ^ |

Sounds right to me.(You mean "closest city to the right of y", not "largest city to the right of y", right?)

→ Ответить

4 месяца назад, # ^ |

yeah i wrote that by mistake. I mean closest missing in the prefix of x and closes missing in the suffix of y. Looks like I effed up the code then. If you can take a look at it please ? (I assure you its written cleanly as much as possible.)

→ Ответить

kwangg

4 месяца назад, # ^ |

Sorry, can't spot a bug immediately. Maybe there's some bug in the implementation detail in finding the closest city to the left or right? Some off-by-one error, maybe?

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4 месяца назад, # ^ |

Yep, no problem, I'll find it somehow. Thanks a lot!

→ Ответить

4 месяца назад, # ^ |

found it: did subsum[l,r] = (psum[r] — psum[l]) instead of (psum[r] — psum[l-1]) :'(

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4 месяца назад, # ^ |

if i understood it properly, that is the solution. the hard part of this shitty problem was the implementation

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Saivishwas

4 месяца назад, # ^ |

idea looks right, you might have made mistake in implementation.

→ Ответить

4 месяца назад, # ^ |

thanks i'll debug against some AC solution I guess then.

→ Ответить

-crystal-

4 месяца назад, # |

← Rev. 3 →

I don't like problem B and E (because I had a hard time with them, obviously). B is a boring classification, and for E you only need to print out the table of SG function values and find the pattern. TBH I didn't feel the joy of solving Cf problems in these two problems.

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T_C_D

4 месяца назад, # |

Can someone help me? 276685070

→ Ответить

https://codeforces.net/contest/2004/submission/276595377

4 месяца назад, # |

← Rev. 2 →

-7

just now I was searching for D solution on youtube and saw problem A,B,C ,E and F were uploaded there during contest

→ Ответить

MonoLith

4 месяца назад, # |

my submission for problem b gave some negative number for first submission- https://codeforces.net/contest/2004/submission/276574322

and same solution got accepted after a few minutes-

if it is some kind of error from system then please recalculate penalty for this problem for me

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-crystal-

4 месяца назад, # ^ |

I don't think they can be called "same solution".

→ Ответить

CSHwang

4 месяца назад, # ^ |

The calc function in the first submission calls R itself in calculating R, which is undefined behavior

→ Ответить

T_C_D

4 месяца назад, # |

My code is producing the[submission:276685070] wrong answer, but I can't figure out why. Could someone help me identify the issue?

→ Ответить

xygzy

4 месяца назад, # |

hmmmm whats wrong with my O(qlogn) solution for Problem D that got TLE on sixth test case? I realized that set::end is O(n) so i precomputed it in the what vector, but still didn't work. all other ops are O(1) or O(logn)

submission

→ Ответить

indresh_hErd

4 месяца назад, # ^ |

Use s.lower_bound instead of lower_bound(s)

Former has O(logn) and latter O(n)

→ Ответить

xygzy

4 месяца назад, # ^ |

yo what the heck! i had no idea. thank you so much. do you know why? same function yet big difference in time complexity.

→ Ответить

indresh_hErd

4 месяца назад, # ^ |

You can see here set time complexity

→ Ответить

sportik

4 месяца назад, # |

← Rev. 2 →

didnt got D in time cause wasted 1 hour on fckn annoying B, for what do shit like that? B in hour, C in 10 mins)

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T_C_D

4 месяца назад, # |

Can someone help me?Your text to link here...

→ Ответить

broErrichto

4 месяца назад, # ^ |

Your code looks correct, but you're getting a Wrong Answer (WA) on test case 5992. Unfortunately, I can't see the specific details of that test case

→ Ответить

4 месяца назад, # ^ |

bro whattt

→ Ответить

s_tapan099

4 месяца назад, # |

← Rev. 2 →

I am getting tle on test 5 can anyone help 276686377 even if the soln is qlog(n)

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late_night_dreams

4 месяца назад, # ^ |

comlexity of your sol is n * q * log(n) see this accepted code using your code 276705976

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s_tapan099

4 месяца назад, # ^ |

thank you so much I got it .

→ Ответить

pessi_penaldo

4 месяца назад, # |

i have no idea why my D is failing, i found the indices of all 6 types of values closest to the left and right and did accordingly am i missing something. solution

→ Ответить

4 месяца назад, # ^ |

You have made exactly the same mistake that I made.


ans=abs(x-pop[x][j][0])+abs(pop[x][j][0]-y);

→ Ответить

https://codeforces.net/contest/2004/submission/276688327

4 месяца назад, # ^ |

← Rev. 2 →

You need to take,


ans = min (ans, abs(x-pop[x][j][0])+abs(pop[x][j][0]-y));

UPDATE :

That is exactly the issue. I copy-pasted your code, and changed one line. Proof :

→ Ответить

pessi_penaldo

4 месяца назад, # ^ |

dude i am crying......, i was burning my blood for this :(

→ Ответить

4 месяца назад, # ^ |

Happens, don't worry. Happened with me also, I got 2 wrong penalty submissions for this.

→ Ответить

AlefHeKaaf

4 месяца назад, # |

+26

E wasn't new...

projecteuler560

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4 месяца назад, # ^ |

This is allowed during Educational rounds. It's one reason why they aren't rated for Div 1.

→ Ответить

shorya1835

4 месяца назад, # ^ |

Source for this please

→ Ответить

https://codeforces.net/blog/entry/21496

4 месяца назад, # ^ |

-8

→ Ответить

shorya1835

4 месяца назад, # ^ |

I don't think anything in the blog implies you can take exactly the same problem from somewhere else.

→ Ответить

4 месяца назад, # ^ |

It's not exactly the same problem.

Project Euler asks you to calculate the number of losing positions with up $$$n$$$ stones in $$$k$$$ piles, while this round asked you to evaluate the winner for just 1 given position.

Looks to me like the CodeForces problem is strictly easier, except of course Project Euler has no fixed time limit.

It's true that the crucial insight behind the two problems is the same, so it's not completely new, but again, that's not unusual for Educational rounds and as far as I'm aware that's intentional.

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shorya1835

4 месяца назад, # ^ |

Fair enough

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g3t_kuhn

4 месяца назад, # |

Can someone explain to me why the code I submitted during the contest was rejected but if I submit the exact code with 0 changes after the contest it gets accepted? In fact I modified the code to get the test cases where it was failing at and that code got accepted when I thought it is obviously wrong.Can someone go to my submissions and check?

→ Ответить

pkhaustov

4 месяца назад, # |

-91

awoo Why for Div2+ rounds you keep giving problems where cheating is extremely simple? No implementation, but only one critical observation which is extremely easy to pass from a cheater to another cheater.

→ Ответить

4 месяца назад, # ^ |

← Rev. 2 →

-18

Since you are giving feedback to Awoo, regarding the contest. I would like to chime in.

I didn't google a shit during the contest. usually I don't. For me, E was really good problem. except for one part, extra BOILER PLATE IMPLEMENTATION for sieve.

I would like to understand what was the motivation to keep pile sizes 10^7 instead of 10^5.

The problem would have been same, if the pile sizes were within 10^5. The core-crux of the solution logic would be same. generating prime till 10^7 and then doing extra implementation things, were more of boilerplate code.

→ Ответить

4 месяца назад, # ^ |

Maybe so you can't completely bruteforce the Grundy numbers and actually have to come up with the "Grundy number of x is the Grundy number of its least prime divisor"? That would make sense in my opinion.

→ Ответить

4 месяца назад, # ^ |

← Rev. 2 →

yeah, makes sense.

→ Ответить

4 месяца назад, # ^ |

← Rev. 2 →

What problems are you talking about? C? For B and D, implementing a solution correctly is IMO much harder than coming up with it. For E, you still need to know about Grundy numbers and not screw up calculating for the small piles. And pretty much every single A is easy to implement, because it's supposed to be really easy, so there's not much organizers can do with that.

→ Ответить

pkhaustov

4 месяца назад, # ^ |

-29

I spoke about problems D and E in particular. Ok, please tell me if these solutions for B and D look hard to implement. Do they?

→ Ответить

4 месяца назад, # ^ |

I didn't realize B is just bruteforce-able like that and was thinking of an O(1) per query solution. I agree doing this is easy. As for D, I don't think your implementation is particularly easy, and looking at the comments, there's a bunch of people with the correct idea that failed to implement it, so I stand by it being difficult to implement.

→ Ответить

shiven

4 месяца назад, # ^ |

B in O(1) per query. Need to block full intersection, additionally extending it by one on each side if needed.

→ Ответить

NexTopper

4 месяца назад, # |

In Problem D, I can't find my mistake, but got wrong answer on test 2. 276676154. and why every one use lower bound to find closest city from x and y? bcz vector was sorted and we can check it O(1)

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4 месяца назад, # |

Not that great of a contest IMO. A and C are fine, but B is just annoying casework, D is basically just implementation, and while I enjoyed E, it's googleable.

→ Ответить

LordLouis04

4 месяца назад, # ^ |

In B u can only check if 'i' is in one segment and i+1 in the other segment for every i in [1,99] then add 1 to the answer, there's no need for casework.

→ Ответить

4 месяца назад, # ^ |

True, didn't realize that doing contest. Doesn't really change my opinion about quality of the problem though, as the fastest solution still suffers from this, and I personally dislike when there's a fast solution that's more complicated than a (significantly) slower one, yet the slower one still passes everything.

→ Ответить

4 месяца назад, # ^ |

I liked problem B. There are different ways to solve it, some of which don't require a lot of case work.

Problem B solution 1

int SolveCase() {
  int l, r, L, R;
  cin >> l >> r >> L >> R;
 
  int answer = 0;
  for (int i = 1; i < 100; ++i) {
    answer +=
        ((l <= i && i <= r) && (L <= i + 1 && i + 1 <= R)) ||
        ((L <= i && i <= R) && (l <= i + 1 && i + 1 <= r));
  }
  if (answer == 0) answer = 1;
  return answer;
}

Problem B solution 2

int SolveCase() {
  int l, r, L, R;
  cin >> l >> r >> L >> R;
  return r < L || R < l ? 1 : std::min(r, R) - std::max(l, L) + (l != L) + (r != R);
}

You complain that problem D is mostly implementation, but what's wrong with that? It's a programming contest, not a math contest.

Going by your username, you probably love math problems, but if all problems are math-based other users will complain about “MathForces”.

And problem E was much more math-heavy, so there was something for everyone. In my view the authors struck a good balance between math and code.

→ Ответить

satyam343

4 месяца назад, # |

+27

So, many $$$n^3$$$ solutions passed in F.

An easy way to fix it would have been to have $$$1 \le a_i \le 2 \cdot 10^4$$$, and decrease the TL to $$$3$$$s maybe. I had to use map because it was not possible to have a global vector of size $$$4 \cdot 10^8$$$. For the suggested constraints, one can simply use a global vector of size $$$8 \cdot 10^7$$$, and find $$$f(a[l,r])$$$ in $$$O(1)$$$ by traversing over $$$(l,r)$$$ in the increasing order of $$$r-l$$$.

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aryanc403

4 месяца назад, # ^ |

$$$O(n^2)$$$ using your idea 276838722

→ Ответить

cheaters_169

4 месяца назад, # |

← Rev. 5 →

+24

some leaked codes that yall best be on lookout for

Problem A

int n;
cin>>n;
vi a(n,0);
read(a);
if(n>=3){
	cout<<"NO"<<endl;
}
else{
	if(abs(a[0]-a[1])==1){cout<<"NO"<<endl;}
	else{
		cout<<"YES"<<endl;
	}
}

Problem A

#include<stdio.h>
#include<stdlib.h>
int main()
{
    int test;
    scanf("%d", &test);
    while(test--)
    {
        int n;
        scanf("%d", &n);
        int *a = (int *)malloc(n * sizeof(int));
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
        }
        if(n >= 3)
        {
            printf("NO\n");
        }
        else
        {
            if(abs(a[0] - a[1]) == 1)
            {
                printf("NO\n");
            }
            else
            {
                printf("YES\n");
            }
        }
        free(a);
    }
    return 0;
}

Problem B

int a,b,c,d;

cin>>a>>b>>c>>d;
if(c>b or a>d){
	cout<<1<<endl;
	continue;
}

if (c < a){
	swap(a,c);
	swap(b,d);
}
// overlap-1
// left right dekh lena
int ans=0;
maps<int,int> m;
for(int i=a;i<=b;i++){
	m[i]++;
}
for(int i=c;i<=d;i++){
	m[i]++;
}
int l=-1;
int r=-1;
for(int i=1;i<=100;i++){
	if(m[i]==2){ans++;
		if(l==-1){l=i;}
	r=i;
	}
}
ans-—;
if(min({a,b,c,d})<l){ans++;}
if(max({a,b,c,d})>r){ans++;}
cout<<ans<<end;

Problem C

#include <bits/stdc++.h>
using namespace std;
int long long void solve()
{
	int n, k;
	cin >> n >> k;
	vector<int> a(n);
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	sort(a.rbegin(), a.rend());
	for (int i = 0; i < n; i++)
	{
		if (k == 0)
		{
			break;
		}
		if (i == n - l)
		{
			break;
		}
		int diff = a[i] - a[i + 1];
		int r = min(diff, k);
		a[i + 1] += r;
		k -= r;
		i++;
	}

	int oddindexsum = 0;
	int evenindexsum = 0;
	for (int i = 0; i < n; i++)
	{
		if (i % 2 == 0)
		{
			evenindexsum += a[i];
		}
		else
		{
			oddindexsum += a[i];
		}
	}
	cout << abs(oddindexsum - evenindexsum) << endl;
}

int32_t main(){
	int t;
	cin >> t;

	while(t--){
		solve();
	}

	return 0;
}

Problem D

#include <bits/stdc++.h>

using u32 = unsigned;
using i64 = long long;
using u64 = unsigned long long;

const std::string color = "BGRY";

constexpr int inf = 1E9;

int id(int x, int y) {
    assert(x != y);
    if (x > y) {
        std::swap(x, y);
    }
    return y * (y - 1) / 2 + x;
}

void solve() {
    int n, q;
    std::cin >> n >> q;
    
    std::vector<std::array<int, 2>> c(n);
    
    for (int i = 0; i < n; i++) {
        std::string s;
        std::cin >> s;
        
        for (int j = 0; j < 2; j++) {
            c[i][j] = color.find(s[j]);
        }
        assert(c[i][0] < c[i][1]);
    }
    std::vector<std::array<int, 6>> pre(n), nxt(n);
    for (int i = 0; i < n; i++) {
        if (i == 0) {
            pre[i].fill(-1);
        } else {
            pre[i] = pre[i - 1];
        }
        pre[i][id(c[i][0], c[i][1])] = i;
    }
    for (int i = n - 1; i >= 0; i--) {
        if (i == n - 1) {
            nxt[i].fill(n);
        } else {
            nxt[i] = nxt[i + 1];
        }
        nxt[i][id(c[i][0], c[i][1])] = i;
    }
    
    while (q--) {
        int x, y;
        std::cin >> x >> y;
        x--;
        y--;
        
        if (x > y) {
            std::swap(x, y);
        }
        
        int ans = inf;
        
        for (auto a : c[x]) {
            for (auto b : c[y]) {
                if (a == b) {
                    ans = y - x;
                } else {
                    int w = id(a, b);
                    for (auto z : {nxt[x][w], pre[y][w]}) {
                        if (0 <= z && z < n) {
                            ans = std::min(ans, std::abs(x - z) + std::abs(y - z));
                        }
                    }
                }
            }
        }
        if (ans == inf) {
            ans = -1;
        }
        std::cout << ans << "\n";
    }
}

int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    
    int t;
    std::cin >> t;
    
    while (t--) {
        solve();
    }
    
    return 0;
}

Problem D

#include <bits/stdc++.h>

using namespace std;

const int N = 2e5 + 10;
const int inf = 1e9;

int n, q, val[N], prv[N][16], nxt[N][16];

void solve() {
    cin >> n >> q;
	auto get_idx = [](char c) {
		if (c == 'B') return 0;
		if (c == 'G') return 1;
		if (c == 'R') return 2;
		return 3;
	};
	for (int i = 1; i <= n; i++) {
		string s; cin >> s;
		val[i] = (1 << get_idx(s[0])) | (1 << get_idx(s[1]));
	}
	for (int i = 0; i < 16; i++) prv[0][i] = nxt[n + 1][i] = 0;
	for (int i = 1; i <= n; i++) {
		for (int j = 0; j < 16; j++) prv[i][j] = prv[i - 1][j];
		prv[i][val[i]] = i;
	}
	for (int i = n; i >= 1; i--) {
		for (int j = 0; j < 16; j++) nxt[i][j] = nxt[i + 1][j];
		nxt[i][val[i]] = i;
	}
	while (q--) {
		int x, y; cin >> x >> y;
		if (x > y) swap(x, y);
		if (x == y) {
			cout << "0\n";
			continue;
		}
		if (val[x] & val[y]) {
			cout << y - x << '\n';
		} else {
			int res = inf;
			for (int i = 0; i < 4; i++) if (val[x] >> i & 1) {
				for (int j = 0; j < 4; j++) if (val[y] >> j & 1) {
					int mask = (1 << i) | (1 << j);
					if (prv[x][mask]) res = min(res, abs(x - prv[x][mask]) + abs(y - prv[x][mask]));
					if (nxt[x][mask]) res = min(res, abs(nxt[x][mask] - x) + abs(nxt[x][mask] - y));
				}
			}
			if (res == inf) res = -1;
			cout << res << '\n';
		}
	}
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t = 1;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

Problem E

#include <bits/stdc++.h>
using namespace std;
const int N = 1e7+10;
int grd[N];
int main () {
 ios_base::sync_with_stdio(0);
 cin.tie(0);
 int cnt=0;
 memset(grd, -1, sizeof grd);
 grd[0] = 0;
 grd[1] = 1;
 for(long long i = 2;i<N;i++) {
  if(grd[i] == -1) {
   cnt++;
   if(i == 2)cnt=0;
   if(i == 3)cnt=2;
   for(long long j = i;j<N;j+=i) {
    if(grd[j] == -1)
     grd[j] = cnt;
   }
  }
  if(grd[i] == -1) {
   cnt++;
  }
 }
 int t=1;
 cin >> t;
 while(t--) {
  int n;
  cin >> n;
  int x=0;
  while(n--) {
   int a;
   cin >> a;
   x^=grd[a];
  }
  cout << (x==0?"Bob":"Alice") << "\n";
 }
}

→ Ответить

sahaun

4 месяца назад, # ^ |

+17

All of them has the useless check in their code for problem E lol.

if(grd[i] == -1) {
  cnt++;
}

→ Ответить

NexTopper

4 месяца назад, # |

← Rev. 3 →

In Problem D, I got wronganswer on test 2. I can't find my mistake.can any one help me? 276676154, and why everyone use lower bound to find the closest point from x and y, cz the vector are sorted and we can find it O(1).

update : problem solved

→ Ответить

arutsudar

4 месяца назад, # |

Can someone help me understand what I missed in Problem D ?

Submission: 276664135

Idea:

If there is a Portal with same color available on source and target, then, it is target - source.

If not, then :

Let's say for example, source is BG and target is RY, then, I check for the minimum value of below logic with these 4 portal pairs — BR, BY, GR, GY. (i.e., trying to see if at-least 1 color matches). If no suitable answer, then return -1.

If any of these is present inbetween source and target, then, it is target - source.

Else if, it is present after target index, then, (foundRight - source) + (foundRight - target).

Else if, it is present before source index, then, (target - foundLeft) + (source - foundLeft).

Else, INT_MAX

→ Ответить

Bytecode1234

4 месяца назад, # ^ |

But four more cases are possible- RB, RG, YB, YG.

→ Ответить

SuperMo

4 месяца назад, # ^ |

← Rev. 2 →

→ Ответить

aureliony

4 месяца назад, # ^ |

The problem statement clearly stated only 6 types. Unlucky.

→ Ответить

arutsudar

4 месяца назад, # ^ |

Right. But, I took that into account in my code. I just missed to mention it in my comment above.

Thank you for your reply.

So, the mistake I did was that, I missed to take the minimum of these two scenarios :

Math.min(

(foundRight - source) + (foundRight - target),

(target - foundLeft) + (source - foundLeft)

)

→ Ответить

4 месяца назад, # ^ |

← Rev. 2 →

Else if, it is present after target index, then, (foundRight — source) + (foundRight — target).
Else if, it is present before source index, then, (target — foundLeft) + (source — foundLeft).

You need to take the minimum of these two, not just the first one that's possible.

(Let's say you want to go from BG at 4, to RY at 6, and there are BR cities at 3 and 9, then clearly it's better to go via the city on the left at 3. But if BR cities are at 1 and 7 instead, then it's better to go via the city on the right at 7.)

edit: Also remember to handle the case where target < source. You can simply swap the indices in that case, since distance is symmetric, but if you forget, some of your math might not work out correctly unless you put some abs() calls in there.

→ Ответить

arutsudar

4 месяца назад, # ^ |

That's exactly what was missed. Thank you very much for your reply.

Can't believe I missed this. (When initially reading the problem statement I thought of this scenario to take minimum of the two. But, later somehow forgot after starting to code, and got deviated from it).

(I did take into account the target<source scenario. I just didn't mention it in my comment above.)

→ Ответить

4 месяца назад, # |

← Rev. 2 →

+17

I tried so hard to see the pattern from this sequence, which is the grundy value in the case we picked $$$y$$$ such that $$$gcd(x, y) \neq 1$$$. After contest I read the statement one more time and... F!!!!!!!!!!

→ Ответить

Akil_Newaz_Chy

4 месяца назад, # |

Although I am still gray while solving 1500++ problems but I think I have very bad luck. It is 1st time I solved 4 problems but could not submit the 4th one in time and the 2nd problem also took so much time for some silly edge case. I never cheated in any contest and I am happy for my hard work. I do not know if people cheated in this contest or not but I am happy for my rank as I did my best. But I am also mad for my bad luck. I do not know why every time I messed up for some silly edge cases. It happens in every contest specially in codechef ( yesterday also I messed up the 2nd problem but did 4th one ) as the give more observation base problems.

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Md._Tanbirul_Islam

4 месяца назад, # |

← Rev. 4 →

For problem c:

276686828

// this is code

 a[j+1]=a[j+1]+min((a[j]-a[j+1]),k);
           k=k-min((a[j]-a[j+1]),k);

shows wa..

But- 276688213

// this is code

 ll h=a[j]-a[j+1];
            
        a[j+1]+=min(k,h);
            
        k=k-min(k,h);

Shows ac with same logic...

N.B.: All variables and array in long long data type

But why?? Is there any difference for min() function??....

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s3yoonpark

4 месяца назад, # ^ |

a[j+1] changes in the first code ...

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Md._Tanbirul_Islam

4 месяца назад, # ^ |

Thanks.. I have got this....

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s3yoonpark

4 месяца назад, # ^ |

yea xD i made the same mistake if you check my submissions.

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Md._Tanbirul_Islam

4 месяца назад, # ^ |

I checked:)...

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sstrong

4 месяца назад, # |

+14

Why would you refer to the operation in G as "process", when the most reasonable meaning of this word is a sequence of operations, not exactly one.

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Karatinos

4 месяца назад, # |

-107

Petition for action on copied task E by recognizing it as unrated or in other possible ways.

The authors directly violated one of the obligations of the authors of the round: copied the task from an existing one, replacing some words that were not essential in solving the problem with others.

If you vote for this post, it is similar to digitally signing this petition. Let MikeMirzayanov and the authors see us and take action.

Thank you.

На русском:

Петиция о принятии мер насчет скопированной задачи E путём признания его нерейтинговым или другими возможными способами.

Авторы напрямую нарушили одно из обязательств авторов раунда: скопировали задачу с уже существующей, заменив некоторые несущественные в рамках решения задачи слова на другие.

Если вы голосуете за этот пост, это аналогично цифровому подписанию данной петиции. Пусть MikeMirzayanov и авторы нас увидят и примут меры.

Спасибо.

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4 месяца назад, # ^ |

The problem statement is simple enough for the problem setter to come up with it independently and just failing to find it online (which would mostly be a failure of the coordinator). Still a big issue, but one of negligence, not of malice.

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Karatinos

4 месяца назад, # ^ |

It is possible that the authors did not do this on purpose. But it is their responsibility to prevent such negligence. And a lot of users have found this task on the Internet, which further aggravates the situation

like indian_rounds_LULW said 2 hours ago in comments:

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Jayki

4 месяца назад, # |

I can't see why my submission for C is failing:

Sorted the array in ascending order.
Tried to equal pairs of elements
Then alculated the score as the difference between each pair

I would really appreciate if someone could help

276684985

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AlefHeKaaf

4 месяца назад, # ^ |

In the "increase to make i = i+1 for all i" part, in "k >= a[i + 1] — a[i]" state you should swap 2 lines in your if, otherwise k won't change at all.

I think it's the only problem.

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Jayki

4 месяца назад, # ^ |

totally missed that :(, thank you!

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4 месяца назад, # ^ |

Compare and feel... 276693772 (I copied your solution I changed a little bit).

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Jayki

4 месяца назад, # ^ |

thanks!

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4 месяца назад, # |

My best chance to reach orange but stuck at F for an hour :(

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4 месяца назад, # ^ |

Erm, actually, yellow is different from orange.

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4 месяца назад, # ^ |

you're right. I get them reversed.

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TheUchiha

4 месяца назад, # |

Its difficult to reach expert without doing graph. I skipped trying D question because I thought it was from graphs, I tried E then.

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sahaun

4 месяца назад, # ^ |

It's just binary search.

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4 месяца назад, # ^ |

← Rev. 2 →

Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)

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4 месяца назад, # |

Well, my D sol got hacked, but I have no idea how I could improve the time of it. It's QlogN and it barely passed during the contest, so I'm not sure if it's just not the intended solution, or if my code is unoptimized (thought I tried to speed it up during the contest and didn't find anything) : 276659141

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ALESTA

4 месяца назад, # ^ |

dont use unordered map ever

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4 месяца назад, # ^ |

It has only 6 entries though

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ALESTA

4 месяца назад, # ^ |

oh didnt see

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4 месяца назад, # ^ |

use (auto &m: cnt) instead of (auto m: cnt).

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4 месяца назад, # ^ |

Sped it up 10x, thanks! Completely forgot about auto making copies, but I'm happy that at least I got the correct solution

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RahulC7

4 месяца назад, # |

Where does this solution: https://codeforces.net/contest/2004/submission/276689812 fail? I've tried so many different test cases...

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Shiv_Narayan

4 месяца назад, # |

From Problem D why 276671958 gives a TLE but when I replace mp with a vector of vector it passes?

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4 месяца назад, # ^ |

Use for (auto &it : mp) instead of for (auto it : mp). You are making copies of it everytime.

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4 месяца назад, # |

← Rev. 2 →

need help in problem D, I keep getting TLE on test 12 (Python)... 276697598

My idea is binary search to get best choice for x on left and right (binary search 2 times) then minimize it.

UPD: Resolved TLE, it's list.insert(0, list) which I think O(1) but it's O(n)

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cxlian

4 месяца назад, # |

I think this should be a pretty bad round. The first three questions are too simple, while the fourth question combines both simplicity and complexity. As for question E, it truly educated me：）

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L_______

4 месяца назад, # |

← Rev. 4 →

-20

276687644

In probelm D, why does this give TLE on test 5, but when I changed unordered map to vector, it got accepted?

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4 месяца назад, # ^ |

Just replace for(auto it : mp) with for(auto &it : mp).

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L_______

4 месяца назад, # ^ |

Yeah, just noticed it now

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L_______

4 месяца назад, # ^ |

forgot putting '&' in for(auto it : mp)

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4 месяца назад, # |

This one please

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SonuGupta001

4 месяца назад, # |

I was so stupid that I spent 1 hour figuring correct cases for problem B and even then it failed 6 times.

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kul_ved18

4 месяца назад, # |

I don't know why I am getting wrong answer on C on this submission 276671135 and right answer on 276689047 Can someone please help me with this

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here https://codeforces.net/contest/2004/submission/276858015

4 месяца назад, # ^ |

a[i]+=min(k,a[i+1]-a[i]); changes a[i] and affects the following k-=min(k,a[i+1]-a[i]);.

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kul_ved18

4 месяца назад, # ^ |

Oh such a silly mistake , thanks for pointing it out

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NKN

4 месяца назад, # ^ |

Made the same mistake.. took me 40 mins to figure out :(

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kul_ved18

4 месяца назад, # ^ |

Oh man, I was so frustrated coz of it that I left the contest midway.

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ShashiPreetam

4 месяца назад, # |

Hiya, could someone let me know why this solution for D is giving TLE? Is it the cuz of the lower bound calls on the sets? 276676038

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ljb

4 месяца назад, # ^ |

I think that's because you use set, which will increase the complexity when inserting, and the order itself during traversal is already sorted, so if you replace set with vector, it can be accepted.

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ShashiPreetam

4 месяца назад, # ^ |

Yeah I completely forgot about the insertion complexity, I was way too focused on the lower bound part. There really was no need to use a set, such a silly mistake. But yeah thanks a bunch for the help!

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ljb

4 месяца назад, # ^ |

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HassanAdel0

4 месяца назад, # ^ |

← Rev. 2 →

lower_bound in the using set is slow you have to do it like that st.lower_bound(x); like that it won't TLE

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SeaYellow

4 месяца назад, # |

My 311 ms B was hacked to tle. The pretests are awful.

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indian_rounds_LULW

4 месяца назад, # ^ |

Get a load of this guy, so many notorious coincidences.

First B solution got hacked, second one was FST'ed:

Codes for D are nearly identical:

...so on and so forth

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SeaYellow

4 месяца назад, # ^ |

I swear to god I never shared my code. I am thinking about someone hacking either my computer or codeforces server right now. Horrifying. I'm about to share this to my university group chat.

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SeaYellow

4 месяца назад, # ^ |

← Rev. 2 →

By the way I have recorded the whole coding process. It was all alone.

==============

Figured out, I shared my account once and now I have to change my password.

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prince705236364273

4 месяца назад, # |

why my 1 solution for question giving TLE and other not solution 1:itration on the map and applying binary search 276711680 soltion 2:iteration on vector colors = {"BR", "BG", "BY", "GR", "RY", "GY"}; 276711617

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Himiitp

4 месяца назад, # |

Which test case am I missing? For question 2004B - Game with Doors. Here is my code. 276640556

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HusseinFarhat

4 месяца назад, # |

My solution for D is in O(n + q) (so optimal time) using 6 prefix and 6 suffix arrays that are precomputed, that tell the earliest/latest (depending on if its the prefix or suffix) occurence of some 2 letters combination in O(n), now if a[x] and a[y] share anything then it's just y-x, if not, iterate through all other letter combinations, then look them up in the precomputed arrays and get the result from that

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AliFarhat

4 месяца назад, # ^ |

Bet

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henrenzhenderen

4 месяца назад, # ^ |

cool! its same as jiangly's

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link : https://mirror.codeforces.com/contest/2004/submission/276598378

4 месяца назад, # ^ |

Mine is the same too took a lot of time to implement

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cockatooo

4 месяца назад, # |

Love problem F. Very simple implementation for a hard problem. Shows how elegant CP can be.

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subrat0018

4 месяца назад, # ^ |

Can you share your approach?

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AlephNullams2008

4 месяца назад, # |

The number of people having their D solution hacked is CRAZY (including me)

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wukaichen888

4 месяца назад, # |

← Rev. 2 →

+19

$$$O(\frac{1}{6}N^3)$$$ solution for problem F.

Is it right?

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akkafakka

4 месяца назад, # ^ |

I'm curious why you mentioned a constant 1/6 explicitly.. is there any reason?

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Cubyte

4 месяца назад, # ^ |

In my opinion,it's not the intended solution but obviously the problem setter wants to let $$$O(n^3)$$$ solutions pass.

Or there won't be awful constraints like $$$n\leq 2000$$$ and $$$5\rm s$$$ Time Limit which makes some $$$O(n^3)$$$ solutions with small constant like yours unhackable.

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AshrafSustS19

4 месяца назад, # ^ |

← Rev. 2 →

+16

My solution was O(n^3), it didn't pass. Slightly less than that actually

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Yanami-Anna

4 месяца назад, # |

At first, I thought what we do in problem E, is pick y stones in the pile which has x stones, guaranteed the gcd(x,y) is not equal to 1. As I do this problem what I thought for almost half an hour, I realize that the statement says:

It is not allowed to take from the pile a number of stones y such that the greatest common divisor of x and y is not equal to 1.

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red_one_fine_day

4 месяца назад, # |

What was the idea for F?

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october_rain

4 месяца назад, # |

How do I report suspicious submissions? I found several users that either have identical code to this comment or added useless snippets in their code (e.g. doing ans ^= 23333; twice) to avoid detection.

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0x5f

4 месяца назад, # |

Why is the judging queue so long?

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jatrophalouvre

4 месяца назад, # |

← Rev. 2 →

If the number of colors for D is not 4 but on the scale of $$$2\times10^5$$$, what will be the solution to the new problem? Can the new problem be solved with the given time limits?

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hermes999

4 месяца назад, # |

In problem B, I implemented O(100^3) and give me TLE. I think my solution should be pass https://codeforces.net/contest/2004/submission/276617200

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0x5f

4 месяца назад, # ^ |

But there are at most $$$t=10^4$$$ testcases while no extra constraints on the sum of $$$l,r$$$ were declared. So your exact time complexity is $$$O(tl^3)$$$ in which $$$l$$$ is the length of the interval, and total operations is about $$$10^{10}$$$.

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hermes999

4 месяца назад, # ^ |

Never happend me before I will pay attention to the constraints. Thank you

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harshbksp98

4 месяца назад, # |

I made an Undirected graph for d and applied dijkstra for each query and ended up getting tle, D initially looked like a graph problem

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sushi_666

4 месяца назад, # ^ |

Could it be done using hashmap??

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4 месяца назад, # ^ |

its a lot simpler than that

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Baby_Bird

4 месяца назад, # ^ |

Using Dijkstra to find the minimum cost would be even more computationally heavy than just traversing all the cities during each query.

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hussain_11

4 месяца назад, # |

Where is editorial

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shaasshh_14

4 месяца назад, # |

i am trying to submit but getting in queue from so long... anyone knows why? the contest was over 12 hrs back

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devansh_khandelwal

4 месяца назад, # |

All the people saying B was implementation heavy and included a lot of edge cases, just didn't get the solution right. You just need to traverse i from 1 to 99 and increment the answer if the room i lied in the first range and i+1 lied in the second range and vice-versa. There was just one edge case when there is no intersection between the ranges, the answer would be 1 in that case. You can check this submission for reference 276586168

I don't think B was a bad problem at all and the authors did a great job!

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4 месяца назад, # ^ |

You don't even need to traverse it It's just simple O(1) calculation submission

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devansh_khandelwal

4 месяца назад, # ^ |

Yup. That's another way to solve it. Don't know why everybody is hating this problem.

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4 месяца назад, # ^ |

I have no idea

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4 месяца назад, # ^ |

fr, in fact i missed an edge case on C but solved B fairly easily. i used hashing in it, here's my solution:276611369

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__tomato__

4 месяца назад, # |

Here is my submission 276639740 for C but I got WA on test 4. Who can explain why??

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msvardhan19

4 месяца назад, # ^ |

int overflow. use long long.

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__tomato__

4 месяца назад, # ^ |

← Rev. 2 →

but I've defined int long long already UPDATE: I declared arr[2e5] :) feeling so dumped now

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TheRoadLessTaken

4 месяца назад, # |

Can someone please explain me this? My contest submission for Problem D was hacked https://codeforces.net/contest/2004/submission/276679036 But when I submit the same solution now, it gets accepted https://codeforces.net/contest/2004/submission/276743351

Also, I had 2 solutions (using map & unordered_map) which got accepted during the contest. What happens if one of them fails the systests? Is it the final submission that is tested or if any of them works then we get an AC? I was getting a TLE in D so I was trying different things and was not sure what was the best way to do this.

Thank you

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1.618034

4 месяца назад, # ^ |

if you have 2 subs in contest one hacked and one still AC. u will take points for the accepted sub by its submission time. (This for edu rounds only)

For D, why it works because it's probable that system tests have not been updated by hack tests yet. system testing didn't happen.

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will_james_m

4 месяца назад, # |

Oops.

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TCWW

4 месяца назад, # |

Why is my game still unrated now?

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bestLM10

4 месяца назад, # ^ |

me too,probably wait for more time.

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its_priyans

4 месяца назад, # |

Why are the ratings not yet updated?

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4 месяца назад, # ^ |

System testing is not completed yet.

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polsage

4 месяца назад, # |

Can anybody tell me what is wrong in my code for C problem.

Solution Link: https://codeforces.net/contest/2004/submission/276717142

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polsage

4 месяца назад, # |

Can anybody help me find the error in my code for problem C

Solution Link: https://codeforces.net/contest/2004/submission/276717142

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4 месяца назад, # ^ |

Same mistake as this.

Note that alice -= Math.min(alice, k); affects k -= Math.min(alice, k);

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polsage

4 месяца назад, # ^ |

oh great, I am dumb

thanks a lot

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Lavaloon_xsh

4 месяца назад, # |

F with $$$\mathcal{O}(n^3)$$$ complexity passed, but my solution in $$$\mathcal{O}(n^2 \log n)$$$ got TLE because I used "push_back" $$$2\times 10^6$$$ times. However, the problems are very nice, thanks for the problem providers!

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_ALFarouK_

4 месяца назад, # |

I think this one is one of the easiest divs 2. even D was solvable. Big Thanks for the big efforts.

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ckxian

4 месяца назад, # |

Why is this contest rated so low ?

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hellotherehello

4 месяца назад, # |

sabbirsajids45 solved problems in Ruby, Java, C, C++, and Kotlin. He is a really versatile coder.

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4 месяца назад, # |

TL47 on F, BLYAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAT

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4 месяца назад, # ^ |

Ok, nvm, 4.968s / 5s xD ultra lucky.

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awoo

4 месяца назад, # |

Auto comment: topic has been updated by awoo (previous revision, new revision, compare).

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nyanspyglu

4 месяца назад, # |

It's very interesting to write from two accounts https://codeforces.net/submissions/keshavspam2800 https://codeforces.net/submissions/keshavgarg8800

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4 месяца назад, # ^ |

This guy is the next LGM. Solving every question in <10 minutes. LOL keshavgarg8800

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keshavgarg8800

4 месяца назад, # ^ |

-15

coming from a person who solved 10 questions in his entire life and he is at expert level

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Moksh930

4 месяца назад, # ^ |

← Rev. 2 →

please have a life! Cheating in contests won't take you anywhere. your submissions have literally [................................................] these big variable names. clearly you have cheated!

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stalker.vip.pro.01

4 месяца назад, # |

I don't know exactly that maspy account was hacked or that is cheating but the submission 276630607 and 276683562 of Sparsh_Mittal are the same during this contest.

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caustique

4 месяца назад, # ^ |

+17

This submission 276683562 was done 14 minutes after the end of the contest when solutions were already available to everyone.

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am_advik

4 месяца назад, # |

Hi everyone, why is the contest showing unrated for me even tho my rating is less than 2100. Any idea?

→ Ответить

aqua4

4 месяца назад, # ^ |

ratings weren't updated yet

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chiragsingh876

4 месяца назад, # |

I implemented an Nlog(N) solution for problem D (276652556), but since I coded it in Java, which is slower than C++, I got a Time Limit Exceeded error. This happened after the contest, putting me behind 2.5k more participants. Later, I submitted the same logic in C++ (276815660) with the help of ChatGPT, and it got accepted. This isn't the first time I've experienced this issue—same logic gives TLE in Java but passes in C++. I hope MikeMirzayanov can look into this and resolve the issue. I was close to becoming an expert after this contest, but due to the language (not the logic), I lost that chance.

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4 месяца назад, # ^ |

if youre aware of this, why not just code in c++.. mike cant make java run faster

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chiragsingh876

4 месяца назад, # ^ |

Mike can do something about this. On Codeforces, certain languages like Python often get a time limit multiplier (2x or even 3x) compared to C++. This allows users to code in other languages too.

As for why I don’t just switch to C++—I feel more comfortable coding in Java at the moment. Since Java is an available option for submission, it should be viable. If uniformity across languages can’t be ensured, then perhaps it’s better to remove Java as an option altogether.

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FuyuMai12

4 месяца назад, # ^ |

On Codeforces, certain languages like Python often get a time limit multiplier (2x or even 3x) compared to C++.

This is false for official rounds.

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4 месяца назад, # ^ |

that is like saying you can't win an F1 race because you used a 2nd hand honda civic, there's a reason why almost everyone here codes in cpp but i'm sure you could've done something to optimize it in java itself, try looking it up online

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chiragsingh876

4 месяца назад, # ^ |

I get your point, but the analogy isn't quite right. Codeforces supports multiple languages, so it's fair to expect some parity. I've already optimized my Java code (276816388), and it still runs close to the TLE (just 79ms away, might give tle if we run it again). On the other hand, unoptimized logic in Python (276837719) is giving runtime close to C++ (276815660) because of better time scaling, even though Python is slower than Java. So, even a 3rd hand Honda Civic can win if the rules are well-structured.

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motorsponge

4 месяца назад, # ^ |

I notice that you are using the standard Java Scanner class. This is generally not recommended. Here is your first solution with my modified version of Kattio, 276843722.

I used to be a Java enjoyer for competitive programming. And I still enjoy using Java for other tasks. But it only took me a week to start using C++ comfortably (at least in CP). It is not an undue burden to expect the contestants to be mindful of what language they use.

Two reasons I wouldn't want some sort of time scaling:

The CF judging servers are already stressed out. Adding a time multiplier for certain, usually more burdensome languages, would add on to this stress.
Some problems are intended with certain time limits in mind. With a time multiplier, this can sometimes, unfairly, be bypassed. For example, in the USACO, certain O(n^2) solutions that wouldn't normally pass in C++, would pass only in Java because of the time limit multiplier. This goes against the intentions of the problem writers.

People have gone to fairly high places on this website with alternative languages, such as Java. Worry about improving your problem solving skills, because at our level, the language choice really doesn't matter that much.

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jigyashugupta

4 месяца назад, # ^ |

← Rev. 2 →

Dudee, check out my Java code for D 276647600

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4 месяца назад, # |

when will ratings update?

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AryanSeth

4 месяца назад, # |

When are our ratings going to be updated after this round? It's been over 24 hours now, or is this a normal thing?

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4 месяца назад, # ^ |

I don't think it's a normal thing, ratings are generally updated in <2 hours after system testing.

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ishuboi07

4 месяца назад, # ^ |

Something's really fishy these days

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ishuboi07

4 месяца назад, # |

Why are the ratings still not updated in the platform?? It's over 24 hrs!!

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4 месяца назад, # ^ |

They will probably update rating for this round after recalculating the rating changes of the last div1+2 round. Lot of cheaters were skipped but rating hasn't changed yet

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4 месяца назад, # ^ |

oh ok, that makes sense

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4 месяца назад, # |

still unrated....

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htoshiro

4 месяца назад, # ^ |

wait

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itsprathamj5

4 месяца назад, # |

Why my rating has not upgraded or degraded after this round, i solved 3 questions in this round and am a newbie..

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daukah

4 месяца назад, # |

I swear to god if i wake up and ratings still not updated

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4 месяца назад, # ^ |

fr, i'm literally coming here to check every hour lmao

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AryanSeth

4 месяца назад, # ^ |

Extremely real, I have no idea what's happening this round

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cockatooo

4 месяца назад, # ^ |

congratz on the positive delta :)

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Habiba_09

4 месяца назад, # |

Ratings when?

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GryllsP

4 месяца назад, # |

Not satisfied with my performance in this contest. I will try to improve in the next contest.

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LokLambert

4 месяца назад, # |

Well My solution got a plag warning. But seriously I haven't copied from any source but I Googled to read a few sources to optimise my solution as it failed to pass all the test cases I implemented in my IDE. Any suggestions on how to make our code more unique and plagiarism-free are highly appreciated.

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dpd

4 месяца назад, # ^ |

Stop cheating.

I never thought about creating unique solution and never got skipped verdict.

And also explain these lame excuse to those who have never coded in their life.

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4 месяца назад, # ^ |

Your previous 2 contests were skipped. You are clearly cheating lol.

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godofwarnings

4 месяца назад, # |

Just googled what is nim numbers in hoping to find something related to it. https://codeforces.net/blog/entry/66040 This CF article is literally what I did in my solution to E using something similar to Sieve of Eratosthenes. Kind of modified the code used here: https://www.geeksforgeeks.org/combinatorial-game-theory-set-4-sprague-grundy-theorem/

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AnkushRaj

4 месяца назад, # |

-9

Dear Codeforces Team,

I have received a notification that my solution for problem 2004E (submission ID: 276630140) significantly coincides with the solutions of other participants. I would like to clarify that my approach was independently developed using well-known concepts from combinatorial game theory, specifically the Sprague-Grundy theorem, which is a standard method for solving problems of this nature.

My solution is based on calculating Grundy numbers to determine the outcome of the game. This technique is well-documented and is commonly used in problems involving impartial games, such as Nim. Given that these concepts are fundamental and widely taught, it is likely that multiple participants arrived at similar solutions independently.

For reference, here are some sources that explain the theory behind my approach:

Sprague-Grundy Theorem on CP-Algorithms Game Theory: Sprague Grundy Theorem on GeeksforGeeks Nim Game and Grundy Numbers — TopCoder I assure you that I did not engage in any form of code sharing or plagiarism. My work was done independently, and I am committed to upholding the integrity of the competition. If needed, I am open to providing further details or discussing the specific steps of my solution to demonstrate its originality.

Thank you for your understanding

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Azm1t

4 месяца назад, # ^ |

Your Code for E

#include <bits/stdc++.h> 
using namespace std; 

const int MAX_N = 1e7 + 10; 
int grundy[MAX_N]; 

int main() { 
    ios_base::sync_with_stdio(0); 
    cin.tie(0); 
    
    int primeCounter = 0; 
    memset(grundy, -1, sizeof grundy); 
    grundy[0] = 0; 
    grundy[1] = 1; 
    
    for(long long i = 2; i < MAX_N; i++) { 
        if(grundy[i] == -1) { 
            primeCounter++; 
            if(i == 2) primeCounter = 0; 
            if(i == 3) primeCounter = 2; 
            for(long long j = i; j < MAX_N; j += i) { 
                if(grundy[j] == -1) 
                    grundy[j] = primeCounter; 
            } 
        } 
        if(grundy[i] == -1) { 
            primeCounter++; 
        } 
    } 
    
    int testCases = 1; 
    cin >> testCases; 
    while(testCases--) { 
        int numPiles; 
        cin >> numPiles; 
        int xorSum = 0; 
        while(numPiles--) { 
            int stonesInPile; 
            cin >> stonesInPile; 
            xorSum ^= grundy[stonesInPile]; 
        } 
        cout << (xorSum == 0 ? "Bob" : "Alice") << "\n"; 
    } 
}

Coincidentally, your solution and the leaked solution both have this useless check

check

if(grundy[i] == -1) { 
    primeCounter++; 
}

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4 месяца назад, # ^ |

LOL AnkushRaj Someone leaked solution so that guys like you would be able to trace easily

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alfia

4 месяца назад, # |

My code coincide with my other account because I didn't know multiple accounts are not allowed on cf hence kept the other one kalfiaaa for practicing

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alfia

4 месяца назад, # ^ |

But will not go against rules of cf for sure

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sushi_666

4 месяца назад, # |

-11

Attention!

Your solution 276582979 for the problem 2004B significantly coincides with solutions sushi_666/276582979, Farhan.WaheedSS/276587437, AryanSeth/276637757, parshiv_k/276649425. 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.net/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. Hello I got a message about my code being plaged in B. I dont know any of the other three remotely, and by looking at the submission time I hope its evident that my code isnt something like a copied code. I also didnt use IDEONE so thats out of question. This was one of the best contests I ever had so please remove the plag check if you belive Im innocent which I can guarantee you that I am. I went to pupil after a long struggle and dont wanna lose the rating for something I didnt even do. Please awoo help me with this one.Thank YOu

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https://codeforces.net/contest/2004/submission/276709586

4 месяца назад, # ^ |

The submissions are the EXACT SAME, except for random newlines that don't do anything.

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ParthK.

4 месяца назад, # |

Can someone Help me in D problem, why am i getting a tle at test 6, here is my solution link

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_Kee

4 месяца назад, # ^ |

Replace for (auto it : msi) with for (auto& it : msi).

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ParthK.

4 месяца назад, # ^ |

← Rev. 2 →

Thanks a lot bro, it worked, but why there was a tle? and what difference does the & play? also when we need to use auto & and when is it not necessary? can you provide some clarification please..i was just using auto it for all my answers till now, is it working because auto it was creating copy of itself in O(n) and hence giving tle, and auto& it is passing it by reference and hence don't give tle?

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_Kee

4 месяца назад, # ^ |

The auto version copies the entire vector for each iteration, in the same way as pass-by-value function arguments.

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ESL2M

4 месяца назад, # |

based

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morphinecode

4 месяца назад, # |

← Rev. 2 →

Your solution 276648513 for the problem 2004D significantly coincides with solutions davit_tsibadze/276637520, morphinecode/276648513.

I want to clarify that this is merely a coincidence, I don't even know the other person, we don't even belong to the same country, I wasn't using any third party code editor and this isn't even a case of leak as he is the only person whose code shares similarities.

I want to point out that there are significant syntactical differences between both solutions, it is just the binary search part that matches and since the binary search was repeated 6 times so it caused a lot of same syntax.

I wish Mikemizayanov, awoo, and the authors adedalic, BledDest and Neon to take a look at it and solve this issue.

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vedu1004

4 месяца назад, # |

I am writing in response to the email regarding my solution (ID: 276675706) for problem 2004E, which has been flagged for significantly coinciding with other users' solutions. I would like to provide some context regarding how my solution was developed.

For problem E, for E, my solution vedu1004/276675706 is not even close with yash3003/276594150 , as both are in different languages and code structure is also different.

For problem D, I asked ChatGPT to generate a helper function. Given that the approach is relatively straightforward, it is possible that another user who used a similar tool might have ended up with code that appears similar to mine.

I understand the importance of adhering to the rules, particularly the clause that states solutions and test generators must use source code completely written by the participant, except for code published or generated using tools available before the round. My intention was not to violate any rules, and I hope this explanation clarifies the situation.

I wish @Mikemizayanov, @awoo, and the authors @adedalic, BledDest and Neon to take a look at it and solve this issue.

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vedu1004

4 месяца назад, # |

-14

I am writing in response to the email regarding my solution (ID: 276675706) for problem 2004E,and 2004D which has been flagged for significantly coinciding with 1 other user solutions. I would like to provide some context regarding how my solution was developed.

For problem E, both my code vedu1004/276669741 and other user code yash3003/276669660 are in different language and the code structure is also very different.

I want to clarify that I did not share my code with anyone, nor did I receive code from others. The code I submitted was based on my own work and public resources that were available before the contest. I wish @Mikemizayanov, @awoo, and the authors @adedalic, BledDest and Neon to take a look at it and solve this issue.

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