Codeforces Round #826 (Div. 3)

2 years ago, # ^ |

If you are over 25 and own a computer, you must try this game

→ Reply

yamen_mahmoud

2 years ago, # ^ |

give me the link to downlaod it abto

→ Reply

2 years ago, # ^ |

You are not over 25

→ Reply

yamen_mahmoud

2 years ago, # ^ |

you are 15

→ Reply

2 years ago, # ^ |

14* and I don't play this game. I am eagerly waiting for my 25th birthday

→ Reply

yamen_mahmoud

2 years ago, # ^ |

if i was on whatsapp i will give you a like

→ Reply

2 years ago, # ^ |

Thanks

→ Reply

suyasho786

2 years ago, # ^ |

sir can you give the hint for the third question of contest how to think for the problem and the basic apporoach should come in our mind it will be greatful if you give me some guidance. Hope for the reply sir thank you

→ Reply

Mueez_Aizdi

2 years ago, # |

← Rev. 2 →

BTW I like the cat in the profile picture of Vladosiya.(◠‿◠)

→ Reply

mikeac

2 years ago, # ^ |

Me too(^_^)

→ Reply

onebit1024

2 years ago, # ^ |

looks like a dog to me lol :)

→ Reply

Mueez_Aizdi

2 years ago, # ^ |

Don't make fun of that cute little kitty.

→ Reply

onebit1024

2 years ago, # ^ |

It was a compliment :))

→ Reply

DeletedUser2343

2 years ago, # |

ismail-but-noob I hope we finally reach pupil

→ Reply

ismailfateen

2 years ago, # ^ |

→ Reply

Iliya-Aghazadeh

2 years ago, # |

Please put strong test cases, I hate to get hacked in div3/4 contests.

→ Reply

2 years ago, # ^ |

Then make stronger solutions?

→ Reply

Rohancoder

2 years ago, # |

Best of Luck everyone!Hoping for a positive delta

→ Reply

ericskh2

2 years ago, # |

Glad to see Div. 3 again after a month.

→ Reply

ostrich_

2 years ago, # |

Thanks Vladosiya to arrange a lot of Div3 contest for beginner. Hopefully this contest will be more interesting);

→ Reply

mikeac

2 years ago, # |

According to the rating was rolled back, so I participate in this round will be rated or not?

→ Reply

Tkm_algo

2 years ago, # |

As a pupil, I'm hoping for the Specialist in this round

→ Reply

dwight_schrute15

2 years ago, # |

hopefully,I'll become green in this round :)

→ Reply

Hello-Tourist

2 years ago, # |

Best of luck everyone

→ Reply

chlalstjr

2 years ago, # |

Me after solving E

Waiting for contest end

→ Reply

SilverWing05

2 years ago, # ^ |

Is E really that easy for 2700+ (I am sure it would go above 3000 by the end of the contest) people to solve?
I can't think of anything at this point :')

→ Reply

MrOtter

2 years ago, # ^ |

maybe memoization will help you

→ Reply

2 years ago, # ^ |

-16

Lol 😂 😂 its simple dp problem why r u so dumb

→ Reply

rishisgsits_22

2 years ago, # ^ |

for tourist all problems are easy but for most of people it isn't, i am also too dumb to not able to solve E

→ Reply

soggybiscuits

2 years ago, # ^ |

Do I need to know dp to be able to solve E ?

→ Reply

tosivanmak

2 years ago, # ^ |

It would be much better to solve E with dp. I don't think greedy could solve this problem.

→ Reply

AlexMari

2 years ago, # |

+22

Great contest!Congrats to authors!

→ Reply

10g0

2 years ago, # |

← Rev. 2 →

Problem E can be solved with DFS.
my solution: 175655607

→ Reply

Jhin

2 years ago, # |

Am I the only stupid one who was trying to solve C without considering the elements should be next together and wasted half of them time on that

→ Reply

saideira

2 years ago, # |

How to solve the problem C?

→ Reply

Mohamed_Hesham

2 years ago, # ^ |

Brute Force O(n*n*logn)

→ Reply

2 years ago, # ^ |

-8

Note that the sum of each segment needs to be same. So given the total sum as $$$S$$$, we must partition the array into $$$P$$$ segments with sum $$$\frac{S}{P}$$$. Now we can bruteforce by iterating $$$P$$$ from $$$1$$$ to $$$N$$$. (remember to skip when $$$P$$$ cannot divide $$$S$$$)

→ Reply

sjain8078

2 years ago, # ^ |

← Rev. 2 →

i did same but still failing, could you please let me know where i was wrong update:- https://codeforces.net/contest/1741/submission/175632745

int main() { int t=1,i,j,k,n,m,q; cin>>t; while(t--) { cin>>n; int mx=0,ans=0,count=0,ans1=INT_MAX; long long totalsum=0; vector v(n); for(i=0;i<n;i++){ cin>>v[i]; totalsum+=v[i]; mx=max(mx,v[i]); } j=n; long long ch,sum=0; while(j>0){ ans=-1; sum=0; if(totalsum%j==0){ ch=totalsum/j; if(mx<=ch){ for(i=0;i<n;i++){ count++; sum+=v[i]; if(sum==ch){ ans=max(ans,count); sum=count=0; } else if(sum>ch){ ans=-1; break; } } if(sum!=0){ ans=-1; } } } if(ans!=-1) ans1=min(ans,ans1); j--; } cout<<ans1<<endl; }

return 0;

}

→ Reply

2 years ago, # ^ |

-8

Check my submission — 175591036.

→ Reply

hnust_liuyunzhi1

2 years ago, # ^ |

Dichotomous answer and O(n^2)check. So it is O(n*n*log(n))

→ Reply

MrLolthe1st

2 years ago, # ^ |

175601238 — $$$O(n\cdot log(n))$$$

→ Reply

koqo

2 years ago, # |

Perfect Div.3 round. Congrats.

→ Reply

Phantom_Performer

2 years ago, # |

Great problems! enjoyed D and F the most!

→ Reply

bangan

2 years ago, # |

how to solve E ?

→ Reply

epsilon_573

2 years ago, # ^ |

You can use DP.
Video Solution explaining the DP Solution.

Hint:

→ Reply

2 years ago, # |

My God ,solved E in the last minute

→ Reply

__underscored__

2 years ago, # ^ |

how to solve E?

→ Reply

https://codeforces.net/blog/entry/107823?#comment-961906

2 years ago, # ^ |

→ Reply

2 years ago, # ^ |

Same for me here, but with F. Got my submission accepted after the contest had already finished. It was tense.

→ Reply

The_Alchemist

2 years ago, # ^ |

How to solve F can you please explain. Thanks

→ Reply

2 years ago, # ^ |

Not sure I'm gonna be capable of providing a better explanation or solution than the editorial, but here is what I did:

1) I sorted all the segments by their left coordinates (their starts)

2) On the sorted sequence, I grouped all the consecutive segments of the same color

So using this as an example, after my grouping we would have the following array: {{1,3}, {2}, {5}, {4}}

3) I looped through every group of segments and for each segment calculated the smallest distance from it to another segment of a different color, for this I checked two things:

First: The distance of the end of the segment to the start of the first segment on the next group (the segment of a different color with the closest start)

Second: The distance of the start of the segment to the end of the segment of a different color of a previous group (previous group = starts before the current one) with the furthest end

To do the second part you have to always keep track of the furthest end so far, which is easy, but for each segment you want to compare it with another segment of a different color, so you have to keep track of the segment with furthest end so far and the segment with the furthest end that doesn't have the same color as that one, doing that you are good and can always apply the method above to calculate the answer for any segment.

→ Reply

18o3

2 years ago, # |

Missed my first Div3 Ak by 30seconds T_T

→ Reply

hnust_liuyunzhi1

2 years ago, # |

How to write question E? I have no idea

→ Reply

2 years ago, # ^ |

← Rev. 2 →

look carefully , every element of array may/maynot have the potential of being the right/left endpoint indicating the length of a segment so from each element we can have at most 2 probable segments, so we can get highest 2*n segments ,then the problem comes down to figure out whether you can build an array of n elements using some of these segments which are non-intersecting ; you can do that through dfs which I did

→ Reply

dd_07

2 years ago, # ^ |

← Rev. 2 →

For every element just check if it can cover the area to it's left and right. If it covers the area to it's left then update the value of that position (dp[i]) with dp[i-left-1] and if it covers the area to it's right then update the value of dp[i+arr[i]] with dp[i-1]. And check for dp[n] at the end. Here is my submission 175628640

→ Reply

callmepandey

2 years ago, # |

Loved problem G, but failed to implement during contest.

→ Reply

The_Alchemist

2 years ago, # ^ |

Hi can you please explain your approach of problem F. Thanks

→ Reply

harsh_it2003

2 years ago, # ^ |

If you want, just have a look at my code. It is very readable. (I just solved it. https://codeforces.net/contest/1741/submission/176072445

→ Reply

2 years ago, # |

Difficulty-wise, which Div3 problem is equivalent to Div2 D?

→ Reply

Wandoka

2 years ago, # ^ |

probably the last one

→ Reply

2 years ago, # ^ |

Yea, I have 75% chance of solving Div2C, never solve Div2D. I never solve Div3F for the past two contests.

→ Reply

2 years ago, # ^ |

F or G

→ Reply

2 years ago, # ^ |

Yea, I have 75% chance of solving Div2C, never solve Div2D. I never solve Div3F for the past two contests.

→ Reply

aKILLL

2 years ago, # |

Was thinking dp for E but each state of my dp has a loop so thought time complexity will be n * n. Saw some solution has passed which have done the same.

→ Reply

2 years ago, # ^ |

you have miscalculated the complexity, it's not n * n

→ Reply

luobogao

2 years ago, # ^ |

← Rev. 2 →

Two nested loops is a naive approach and yeah too slow at O(n^2). But if you think about how updates to your dp table are happening there, you can massage inner loop's updates into the outer loop and have just a single loop for O(n) runtime.

Alternatively think of it as a graph problem, two edges at each index going back/forward, determine if start is connected to the end.

Really cute problem btw with cute simple solution.

→ Reply

superden85

2 years ago, # |

+11

problems were cool! thanks to the authors!

→ Reply

toba

2 years ago, # |

Pretty Nice Contest! Btw, what is the test that makes solutions for E fail?

→ Reply

lazy_loop

2 years ago, # ^ |

just try setting n to 2e5 and all values to 1.

→ Reply

toba

2 years ago, # ^ |

Yeah, I tried that. I think the only solutions that are hacked that way are the ones with no memorization.

→ Reply

lazy_loop

2 years ago, # ^ |

yeah many of them are n*n

→ Reply

29logN

2 years ago, # |

← Rev. 3 →

Got WA in problem E using memorization during contest.
But just after the contest got AC with the same code only after removing the memorization part
why did it not exceed time limit??
submission
upd: hacked

→ Reply

jeal0uspengu1n

2 years ago, # |

Felt like was giving ABC, so yeah nice contest ^_^

→ Reply

17vineet

2 years ago, # |

The cheaters are gonna cheat. There were many videos on YT and many codes where shared among various groups on discord and telegram. This ruins the original rankings and rating of fair participants.

→ Reply

ayush_n

2 years ago, # ^ |

That's Why No of Accepted Submission Increases for Problem C-D

→ Reply

2 years ago, # |

can anybody hack my E ?

→ Reply

YMXL

2 years ago, # ^ |

your code editing is a little bit ugly

→ Reply

2 years ago, # ^ |

my hand writing was ugly in school too , same goes for my code writing

→ Reply

2 years ago, # |

I use dfs to solve E.But it looks like can be hacked.Can anyone show me the way to hack these solutions?

→ Reply

2 years ago, # ^ |

why do you think it can be hacked?

→ Reply

2 years ago, # ^ |

I have saw some these solutions had been hacked

→ Reply

2 years ago, # ^ |

ok mine is hacked ,forgot to do memoization lol

→ Reply

2 years ago, # ^ |

I did memoization.But may be it can be hacked too.

→ Reply

https://codeforces.net/contest/1741/submission/175667416

2 years ago, # ^ |

ok let's find out , can anyone hack this solution of E :

→ Reply

2 years ago, # ^ |

It seems that c++14 is much faster than c++20.May be I will use c++14 the next time.

→ Reply

https://codeforces.net/blog/entry/107879

2 years ago, # ^ |

I wrote a dp solution using python and so far so good, don't really see how it could be hacked...

→ Reply

sergeev.PRO

2 years ago, # |

← Rev. 2 →

cool F. Really disappointed about not solving it in time(

→ Reply

Tempboy696996

2 years ago, # |

-47

Problem Statement of problem C was bad. For literally around 2 hours and 15 minutes I tried to figure out whats wrong with me. To the author of this problem, you could have written subarray instead of segments. For around 2 hours and 15 minutes I just thought I thought about subsequence. I know it was not completely authors mistake. I should have been more careful. But still you should have avoided the confusing words. Rating goes rrrrrrb :(

→ Reply

jimm89

2 years ago, # ^ |

+23

The statement is perfectly clear, and illustrated with examples. I have also failed to read questions properly in the past and paid the price, but that is not the fault of the author.

→ Reply

svorogaze

2 years ago, # |

-38

F is too easy for its position

→ Reply

toba

2 years ago, # ^ |

Do not agree with that honestly.

→ Reply

NokSyte

2 years ago, # ^ |

Kinda cringe tbh

→ Reply

Beevo

2 years ago, # ^ |

+21

Yes, clearly.

→ Reply

JustJie

2 years ago, # ^ |

Maybe you're just too good? Lol. There's only 351 solves atm

→ Reply

Sarinha

2 years ago, # |

-22

Is the contest going to be unrated after this?

→ Reply

Zakaas

2 years ago, # |

Kudos to authors for clear problem statements.

→ Reply

Adarsh8409

2 years ago, # |

+24

Problem E was a perfect 1D dynamic programming question and the best problem for div3 E.

Overall Great questions. Congrats to authors.

→ Reply

ARMINIUS

2 years ago, # |

another great contest by these legends

→ Reply

Neerajjoshi2003

2 years ago, # |

Can someone can give a small hint for D I have enough time but still couldn't make it

→ Reply

Adarsh8409

2 years ago, # ^ |

← Rev. 2 →

you can split the array in 2 parts every time. and you have to check that the elements in first part is always less than second part for every subtree. if not, increase the count and repeat the same process till the leaf of the tree.

Use recursion for solving the problem.

You can take reference from my code-> https://codeforces.net/contest/1741/submission/175603313

→ Reply

2 years ago, # ^ |

Just complementing, recursion is a good path, but also not the only one

→ Reply

SilverWing05

2 years ago, # ^ |

In one way, if you know merge sort, you basically know the solution for this problem.
Would recommend you to learn the idea of merge sort from GFG or someplace else and try it on your own.
Just for reference 175631361

→ Reply

Math_is_Good

2 years ago, # ^ |

Think of divide and conquer technique.

→ Reply

A_G

2 years ago, # |

Did anyone else use min cost max flow on G? I think it works in $$$O(km\log m)$$$.

→ Reply

AlexLorintz

2 years ago, # ^ |

That's insane, do you think you can describe your approach?(I only managed to solved it with bitmasks)

→ Reply

A_G

2 years ago, # ^ |

← Rev. 3 →

For a vertex $$$v$$$ in the original graph, let there be an entrance node $$$2v-1$$$ and exit node $$$2v$$$ in the flow graph. The source is vertex $$$1$$$ (the entrance node for vertex $$$1$$$). Let $$$2n+1$$$ be the sink. The edges are:

$$$2v-1\to 2v$$$ with capacity $$$\infty$$$, cost $$$0$$$ for $$$1\leq v \leq n$$$.
$$$2u\to 2v-1$$$ with capacity $$$\infty$$$, cost $$$0$$$ for an edge $$$u, v$$$ in the original graph if $$$d(v) = d(u)+1$$$ (where $$$d$$$ is distance from $$$1$$$).
$$$2v-1 \to 2v$$$ with capacity $$$1$$$, cost $$$-x_v$$$ where $$$x$$$ is the number of friends without cars at vertex $$$v$$$, for $$$1\leq v\leq n$$$.
$$$2v\to 2n+1$$$ with capacity $$$y_v$$$, cost $$$0$$$ where $$$y_v$$$ is the number of friends with cars at vertex $$$v$$$, for $$$1\leq v\leq n$$$.

The answer $$$k$$$ plus the minimum cost flow of this graph. (Or maybe I'm wrong and you can hack me.) Stop at $$$k$$$ units of flow instead of a maximum flow because the minimum cost will be reached within $$$k$$$ units. Otherwise, the time complexity is $$$O(nm\log m)$$$, I think.

→ Reply

AlexLorintz

2 years ago, # ^ |

← Rev. 2 →

Sounds nice, if I got it right it is computing the "maximum cost maximum flow" to find out the maximum number of friends that can be driven and then subtract this from K to get the answer. The transitions are also pretty logical, I don't know if I can find any counter-example. Anyway, I would have never thought of modelling it with a flow chart, gg:)

→ Reply

Phoenix_8

2 years ago, # |

Hi, Can someone help with F? I saw some solutions with segtree, but they were difficult to understand.

→ Reply

people_plus_plus

2 years ago, # ^ |

← Rev. 3 →

Try think this way: there is no colors in this problem, task is the same but without segment colors.

→ Reply

onebit1024

2 years ago, # ^ |

let's assume that the segments are sorted in increasing $$$l$$$. You can split the problem into two parts -

i) find the greatest $$$r[j]$$$ ($$$j$$$ < $$$i$$$) among all $$$l[j]$$$ which are less than $$$l[i]$$$ and with different color than $$$color[i]$$$

ii) find the least $$$l[j]$$$ ($$$j$$$ > $$$i$$$) among all $$$l[j]$$$ greater than $$$l[i]$$$ and with different color than $$$color[i]$$$

Let's see how we can solve the first part. We can store an array $$$mx$$$ where $$$mx[k]$$$ indicates the greatest $$$r$$$ value for a given color $$$k$$$. Store this $$$mx$$$ array in a segment tree so that we can retrieve the maximum value of $$$mx$$$ with updates. Now, for this case, you may wonder why you would actually require a segment tree and not just maintain a set or something similar. The thing is, if we are at the $$$ith$$$ segment, the maximum $$$r$$$ value of any segemnt $$$j$$$ ($$$j$$$ < $$$i$$$) might be of color similar to that of the $$$ith$$$ segment. In which case, we have to find the max $$$r$$$ value of a color which is not that of the $$$ith$$$ segment color. Say the color of the $$$ith$$$ segment is $$$k$$$. We can do this by updating $$$mx[k]$$$ in our segment tree to some very small number and then finding maximum in segment tree and later resetting $$$mx[k]$$$ to what it was initially. Ie make sure that $$$mx[k]$$$ is ignored in our max segment tree.

Now, as we move through all $$$i$$$ in increasing order of $$$l$$$, we can easily figure out the maximum $$$r$$$ value of a different color using our segtree (explained above). Once we find this maximum $$$r$$$, we we can update the answer for this segment to $$$max(0,segment[i].l-max_r)$$$

The second part of our problem can be solved in an almost similar way, except we would require a minimum segtree to find the minimum $$$l$$$ value of a segment $$$j$$$ ($$$j$$$ > $$$i$$$) of different color. Alternatively,we could use the same segment tree (max segment tree) except store values in the segment tree as (-(actual value)), now, the abs of the max value will be the least $$$l$$$ value. Once we find this minimum $$$l$$$, we can update the answer for this segment to $$$max(0,min_l-segment[i].r)$$$ We'd also have to make some minor changes like redefining $$$mx[k]$$$ to minimum $$$l$$$ value for a given color $$$k$$$

submission

→ Reply

Phoenix_8

2 years ago, # ^ |

Thank you so much!

→ Reply

NokSyte

2 years ago, # |

+15

Awesome round!!! Keep it up creators! ^D

→ Reply

2 years ago, # |

Great round

→ Reply

SouLPegasiuuuS

2 years ago, # |

Someone please hack my submission for E. 175641835

→ Reply

shelbytom.23

2 years ago, # |

Can someone explain DFS approach to solve problem E.

→ Reply

people_plus_plus

2 years ago, # ^ |

← Rev. 2 →

I think DP solution much easier to understand

→ Reply

people_plus_plus

2 years ago, # ^ |

But DP idea almost like DFS solution. So let me tell you the DP idea first and then DFS

1) DP solution. Let $$$dp_i$$$ be the True, if the array $$$b[1..i]$$$ can be build by array $$$a$$$ and False otherwise. Our base is $$$dp_0 = True$$$. Then, if $$$dp_{i-1}$$$ is True that means we can build array $$$b[1..i-1]$$$, that we have to do with $$$b_i$$$? Let's say it's the count of numbers in next subarray, we set the $$$dp_i+b_i = 1$$$. And if $$$b_i$$$ was the counter in previous subarray we need to check if we can build array $$$b[1..i-b_i]$$$. The answer is in $$$dp_n$$$ now.

2) DFS solution. Let's now say from index $$$i$$$ we can go to indexes $$$i+b_i$$$ and $$$i-b_i$$$. Because $$$b_i$$$ is count of numbers in next or previous segments. Let's call this edges and $$$i$$$ as nodes. We have to be able to move from node $$$1$$$ to node $$$n$$$. The answer is $$$dfs(1)$$$ or $$$dfs(n)$$$ now.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

-8

We can use the visited array to our advantage in the DFS solution, we can simply start DFS from 0 and then check $$$vis[n]$$$ (basically right after the last cell). If the string is valid $$$vis[n]$$$ will be true, otherwise it will be false.

→ Reply

riddle_7

2 years ago, # |

Great problems but weak test cases for E killed me.

→ Reply

omeganot

2 years ago, # |

Nice contest, thanks! I think d and g are pretty nice.

I was originally kind of annoyed by F but i looked at some other people's submissions after contest and they're pretty cool, so that was mostly my fault for being an idiot.

→ Reply

itz_Parv

2 years ago, # |

← Rev. 3 →

How to approach D anyone?

→ Reply

koqo

2 years ago, # ^ |

Its a MergeSort problem, you just have to keep track of the left and right part for each segment using a 2d array and then on each iteration see if you can merge i and i+1.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

-8

Also you can track the first element of the left and right subsegment and swap them if left is greater than right. You don't have to swap the whole array, and this way you get an $$$O(n)$$$ time complexity due to $$$\Sigma_{k=0}^{n-1}{2^k} = 2^n-1$$$ and $$$n$$$ was in the form of $$$2^k$$$. If the absolute difference of the two elements are equal to the difference of the indices for each step, the tree is valid. otherwise the answer is $$$-1$$$

→ Reply

nayemZR

2 years ago, # |

Why E no was with so weak pretest? Forgot to use memorization.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

You can survive only by using dp.Memorization is also not enough if you use dfs.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

-8

I think your dfs got TLE hacked because the time complexity was $$$O(V^2)$$$ due to scanning for segments on every index. You can enumerate segments beforehand, this will lead to a maximum of $$$2V$$$ edges, and worst time complexity is $$$O(V)$$$ if you do this

→ Reply

2 years ago, # ^ |

Why for E, my dfs + memo got hacked as TLE? https://codeforces.net/contest/1741/submission/175640727 method is called getCanSent the memo map has at most 2n entries.

→ Reply

2 years ago, # ^ |

← Rev. 3 →

-11

Well, I can't exactly know where the hack happened. But your code will (and did) work very slow, due to multiple reasons. Reasons may include hash tables being simply slow, and the recursion leading to a deep copy.

→ Reply

Citypop

2 years ago, # ^ |

Your transition can potentially take O(N) time. And you have N states, so it's actually a O(N^2) solution.

→ Reply

Tom66

2 years ago, # |

← Rev. 2 →

I find G easier than F and I've hacked my friend DitaMirika and another participant ha___il for their carelessness.

detail: Their time complexity can be as large as $$$O(2^n)$$$

→ Reply

2 years ago, # |

Why for E, my dfs + memo got hacked as TLE? https://codeforces.net/contest/1741/submission/175640727

method is called getCanSent

the memo map has at most 2n entries.

→ Reply

SHA_BI

2 years ago, # |

F is too hard for me :(

→ Reply

2 years ago, # |

why this 175643343 n^2 solution is passing for problem E?

→ Reply

SouLPegasiuuuS

2 years ago, # ^ |

I tried to submit it again and it is giving TLE now but in standings it is an accepted solution even after system testing :(

→ Reply

2 years ago, # ^ |

SysTests aren't done yet as far as I know. Been looking for hacks and Systests didn't go off since hacking phase ended, it probably will start soon.

→ Reply

SouLPegasiuuuS

2 years ago, # ^ |

Oh yes, my bad. That solution will get caught in system tests then.

→ Reply

2 years ago, # ^ |

← Rev. 2 →

when will system test start?

→ Reply

2 years ago, # ^ |

Most likely soon, maybe an hour or two later at most. As far as I have seen, it doesn't take much longer usually.

→ Reply

2 years ago, # ^ |

Usually its written there that "system test pending". but its not the case for now.

→ Reply

2 years ago, # ^ |

I have noticed it in previous rounds with hacking phases as well. In Div2/1/Global rounds it gets 'Pending' status but not in the Div3/4/edu as far as I have noticed. It might change a minute or two before the SysTests start, but it's marked as completed mostly.

→ Reply

2 years ago, # ^ |

Hey i can only message 1 time in 10 minutes. but it looks like this is not the case for you why?? and i think some things should be common in all the contest not like this happens in this that happens in that

→ Reply

2 years ago, # ^ |

← Rev. 2 →

I guess because you got -73 contribution, so CF is trying to revoke your 'privileges'.

Well, it def should be, but I guess there are some technical difficulties, like this one isn't pending in SysTests yet, but analyzing hacks to add in SysTests. Once it's analyzed, it will be SysTested. I definitely am not the best person to ask this to, but this might be the case.

→ Reply

2 years ago, # ^ |

← Rev. 3 →

If my interpretation of this experience is correct, it's definitely not a direct correspondence to contribution. I believe it is some "rate limit" happening, sometimes it goes to 4 messages every 60 minutes. Getting it down to 4/60 had nothing to do with contribution for me though.

→ Reply

Taekki

2 years ago, # |

May I ask whether I would be rated for this round or not? By the rules stated in the announcement, I am not a trusted participant yet(since I took < 5 rated rounds). But according to this line "Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you.", I assume I should be rated for this round. Thanks in advance to your answers

→ Reply

2 years ago, # ^ |

As far as I know, it should be rated for you.

→ Reply

Taekki

2 years ago, # ^ |

But it appears in the unrated contests in my profile page haha

→ Reply

2 years ago, # ^ |

It's because the rating changes isn't out yet, so it's marked as unrated for everyone yet. It probably will take a few more hours to update.

→ Reply

an1ket_62

2 years ago, # |

Does hacks improve delta in div3/div4/educational round ? if yes, how ?

→ Reply

2 years ago, # ^ |

-10

As far as I know, in Edu Div2 rounds you get +100 points for successful hack and -50 for unsuccessful. So yes, in Edu Div2 you improve your score.

In Div3/4 I don't think that is the case, since it's not based on scores but amount of solved problems.

→ Reply

2 years ago, # ^ |

EDIT:

Well, I guess I messed up with Div2 edu and standard Div2 round, but downvoting instead of correcting and answering the question isn't really helping a lot.

→ Reply

mahendra_2890

2 years ago, # |

How do people target solutions for hacking? Is it random?(just look at people's solutions and try to find vulnerabilities) or it there some other way?

→ Reply

2 years ago, # ^ |

I think they find a general case which might hack some of the approaches, then check for that specific problem if it's implemented via the approach you know how to hack. Not certainly sure tho.

→ Reply

kasiru_69

2 years ago, # |

Yesterdays contest was good Though I am going to get a negative But it was fun :) :) :) :)

→ Reply

_AIH_

2 years ago, # |

As a Competitor, I competed this round.

→ Reply

kavi_khalique

2 years ago, # |

My rating is 935 and i participated in Codeforces Round #826 (Div. 3) but its showing unrated for me and i haven't got any ratings. Can anyone please help me out because in notice its also written that " Regardless of whether you are a trusted participant of the third division or not, if your rating is less than 1600, then the round will be rated for you."

→ Reply