Codeforces Round #983 (Div. 2)

3 недели назад, # |

+33

As an author, I'm honored to bring you the score distribution for this round. Good luck.

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programmer5777

3 недели назад, # |

That announce was made 8days ago mean 10day before contest. Seems like they are prepared...(Excited)

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Robi_Robi-yt

3 недели назад, # ^ |

Yessss

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Rabeeeee

3 недели назад, # |

you forgot to thank Mike Mirzayanov.

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hydroshiba

3 недели назад, # |

OMG Vietnamese round

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papa-ka-para

3 недели назад, # |

While saving Cats and Dogs, We kept celebrating with Chickens Pigs and Goats. Hypocrite Humans.

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hundunlilun1

3 недели назад, # |

Good KFC.

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shash4321

3 недели назад, # |

+18

omg kuroni round

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DylanSmith

3 недели назад, # ^ |

omg kuroni round

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Khka

3 недели назад, # |

Damn.. did you guyz really forgot to thank our lord and saviour MikeMirzayanov...

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thanhchauns2

3 недели назад, # ^ |

Damn, I'll alert him.

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LovelyMicas

3 недели назад, # |

← Rev. 2 →

-8

looking forward the result.

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Navi4ok

3 недели назад, # ^ |

bro you ain't a tester, who are you trying to fool.

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-is-this-BIT-

3 недели назад, # |

-8

As a contestant, how can I become a tester $$$??$$$

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Navi4ok

3 недели назад, # ^ |

same question here.

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-berozgar-hu-bhai-

3 недели назад, # ^ |

usually testers are authors' friends

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-berozgar-hu-bhai-

3 недели назад, # ^ |

you can't

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AlephNullams2008

3 недели назад, # |

Mary skelter fan spotted omg

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3 недели назад, # ^ |

Another brethren? Here?

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AlephNullams2008

3 недели назад, # ^ |

how bro feels saying 'brethren' and 'admirable'

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goldrevboy

3 недели назад, # |

Special thanks to KFC was what really hit me <3

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-333

3 недели назад, # |

Is KFC preparing a round?

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Aritro_Sarkar

3 недели назад, # |

Guys, my exam is starting in a month (I gave a practical exam today).

I have all of my syllabus to read.

Should I do this contest?

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-is-this-BIT-

3 недели назад, # ^ |

Cf contest >>>>>> School

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TCL

3 недели назад, # ^ |

My last exam finished today and I guess you studed more then me

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whyalwaysmezzz

3 недели назад, # |

hope full solve

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moderax_draxx

3 недели назад, # |

I think the team loves chicken nuggets more than their code :D

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weirdflexbutok

3 недели назад, # |

Pray that i solve atleast 1 problem(s) this time :')

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Titan493ASST

3 недели назад, # ^ |

Yo, this guy's my senior at the college and he freakin' wants to be unrated!

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Hamza_taigo

3 недели назад, # |

time to celebrate and rank up

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mohamedbouafif

3 недели назад, # |

+13

If we write down all the red letters we get the word : Kangokutou Which is a term that is often used in Japanese literature and media to refer to isolated islands designated for imprisoning criminals, similar to the concept of Alcatraz Island in the United States. It may also appear in fictional works to evoke themes of isolation, confinement, and punishment SOS?

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ItsAJHarsh

3 недели назад, # |

OMG This looks like a Halloween edition !!!!! :)

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Hyperbolic

3 недели назад, # |

+16

As a tester, I'm curious about AkiLotus's profile character

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programmer5777

3 недели назад, # ^ |

← Rev. 2 →

As a anime fan, I'm also curious..

If this character is from an anime please tell me the name of the anime.

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3 недели назад, # ^ |

I hope I can officially announce her soon in a future round. ;)

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nhphuc

3 недели назад, # |

as a 20yrs fan, kuroni orz

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SilverCipher

3 недели назад, # |

As a compiler, I cant complain.

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tpy_plays_lavaloons

3 недели назад, # |

wish this contest will be interesting

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extremelyLonely

3 недели назад, # |

-86

I am writing to request that MikeMirzayanov postpone the contest, as it is Diwali in many regions of India today.

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3 недели назад, # ^ |

unfortunately the world doesn't revolve around India

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papa-ka-para

3 недели назад, # ^ |

+20

I am from India. And I am embarrassed seeing these kind of immature comments from people from India.

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extremelyLonely

3 недели назад, # ^ |

-12

Haa kattue tujhe kya fark padega

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papa-ka-para

3 недели назад, # ^ |

← Rev. 2 →

I am Hindu. And I am embarrassed seeing these kind of immature comments from people from our Religion.

Btw, Belated Happy Diwali to you toooo !!!

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extremelyLonely

3 недели назад, # ^ |

Happy Diwali, but what was so embarrassing when someone ask to postpone the contest?, this year during eid one guy asked to postpone the contest, and it was indeed postponed, but when I asked, I get hate from my own people.

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3 недели назад, # ^ |

who cares

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SaveliyIsaev

3 недели назад, # |

← Rev. 2 →

return specialist?

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husc24.Serge_

3 недели назад, # |

good luck

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SuperJimmy

3 недели назад, # |

Good luck everybody!

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Nafis424

3 недели назад, # |

Goodluck for everyone

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3 недели назад, # |

excited

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thietgia3452002

3 недели назад, # |

← Rev. 2 →

(D) is Mucho Texto

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3 недели назад, # |

B had so many edge cases but atleast i solved it

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3 недели назад, # ^ |

Really ? 289203244

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3 недели назад, # ^ |

wow sick, for me it's a little bit more 289255586

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3 недели назад, # ^ |

quite the same TBH

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3 недели назад, # ^ |

i meant i didnt think about odd even got 3 WA there lol i also solved similarly though

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3 недели назад, # ^ |

I see

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3 недели назад, # ^ |

So close yet so far, i actually couldn't debug it :/ 28928125

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3 недели назад, # ^ |

You kind of over-complicated it.

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3 недели назад, # ^ |

Reason why i am still a newbie :'/

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3 недели назад, # ^ |

well you should practice more problems ,and that too consistently, if you actually care about ratings.

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3 недели назад, # ^ |

Yes, at least i don't wanna end this as a newbie

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3 недели назад, # ^ |

don't debug for simple code/problem, hack yourself and you will grow

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3 недели назад, # ^ |

hack myself what-

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3 недели назад, # ^ |

take a 1 min break then try to find a counter example to break your own code.

I do this every time I got WA in live contest.

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3 недели назад, # ^ |

Ohhh- Well actually I came up with the logic in like around last 10 minutes & sadly couldn't find a counter example even after trying a number of tests :/ will have to sleep over this one ig

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adepteXiao

3 недели назад, # |

+19

Samples in C are useless

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frissky9

3 недели назад, # ^ |

It's a trend ig, it's always been like this.

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11 дней назад, # ^ |

I don't get this. Care to explain yourself?

I do believe the samples have covered most of the general idea of the problem, like what is being done in each operation and how the final criterion is visualized.

If anything, I think I haven't yet explained too deep into how an operation sequence is invalid, but that should be the work of the Notes section, not the samples.

And furthermore, I believe samples' purposes only stop at illustrating what the problem asks, not giving any insight of how to solve it (though the 4th sample was there as sort of a sanity check for you guys when coding).

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adepteXiao

4 дня назад, # ^ |

Sorry for the late reply. I didn't like that in all the examples you can just sort the array and replace some prefix of the numbers with the largest number in the array. Because of this, the tests seemed monotonous to me.

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4 дня назад, # ^ |

Pretty sorry for the experience, but I still don't believe that is something possibly faultable. Again, the testset only needs to illustrate the exact specs of the problem's components (operation, validity, etc.), not hinting or distinguishing right and wrong approaches.

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Paulieaditya

3 недели назад, # |

time to touch grass

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SXZZ5

3 недели назад, # |

Ofcourse it is on us to read statement carefully, but it would have been so nice if the fact that we are given an identity permutation was clearly stated in Problem B. No way a problem with such a trivial solution has taken so much time for so many people to AC.

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Paulieaditya

3 недели назад, # ^ |

how did you solve it?

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3 недели назад, # ^ |

← Rev. 2 →

my solution I think its simple ~~~~~

int n,k;
    cin>>n>>k;
    if(n==1){
        if(k==1){
            cout<<1<<endl;
            cout<<1<<endl;
        }
        else cout<<-1<<endl;
        return;
    }
    if(k==n){
        cout<<-1<<endl;
        return;
    }
    if(k==1){
        cout<<-1<<endl;
        return;
    }
    if(k%2){
        cout<<5<<endl;
        if(k==n-1){
            cout<<-1<<endl;
            return;
        }
        cout<<1<<' '<<2<<' '<<k<<' '<<k+1<<' '<< n<<endl; 
        return;
    }
    cout<<3<<"\n";
    cout<<1<<' ';
    cout<<k<<' ';
    cout<<k+1<<endl;

~~~~~

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SXZZ5

3 недели назад, # ^ |

← Rev. 2 →

since given array is identity permutation we can simply split the array into three parts $$$[1,..,k-1], [k], [k+1,...,n]$$$. Since everything in first subarray is smaller than $$$k$$$, its median is also smaler than $$$k$$$. Similarly median of the third subarray is greater than $$$k$$$. Now, for median of medians part we have only three numbers with one each on either side of $$$k$$$.
Only edge case is that size of subarrays should be odd.
$$$[k]$$$ is already odd.
If $$$[1,...,k-1]$$$ is of even length then $$$[k+2,...,n]$$$ must also have even length (otherwise $$$even +1 + odd = even \neq parity(n)$$$. So just split them further into $$$[1]$$$ and $$$[2,...,k-1]$$$. Similarly split $$$[k+1,...,n]$$$ into $$$[k+1]$$$ and $$$[k+2,...,n]$$$ if need be.

So you either split into $$$3$$$ subarrays or $$$5$$$ subarrays always. In both cases we have equal number of medians less than $$$k$$$ and greater than $$$k$$$, so overall median is still $$$k$$$.

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peti1234

3 недели назад, # |

+76

D is solvable with $$$n-3$$$ queries. We have to find the array $$$p$$$. After the first positive value $$$p$$$ is strictly increasing.

Iterating from $$$n-2$$$ to $$$1$$$ we can find $$$p_{n-1}$$$.

$$$p_{n-1}=i$$$ if the $$$(n-1, i)$$$ path doesn't cross $$$0$$$ for the first time.

After that we can do the same for $$$p_{n-2}$$$ iterating from $$$p_{n-1}-1$$$. And so on. Using the observation that node $$$1$$$ has exactly $$$2$$$ adjacent nodes we can save one more query.

Here is my solution: 289236088

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e3e

3 недели назад, # ^ |

did you know that function res() does not return?

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zfnu

3 недели назад, # |

Is E really just a knowledge check on how to solve a system of equations for the specific matrix? Or is there an easy solution?

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p__ce1052

3 недели назад, # ^ |

just get the relation between v[i] and v[i+2] for all i and make them all non-negative by adding minimum value. then check whether it's valid

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3 недели назад, # |

WAforces, I'm feeling -3 intel for making 4WAs in B and 1 WA in C

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WTF man why is my solution for C wrong?

3 недели назад, # ^ |

real. it's so stressful when you keep getting WAs lol

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equeen

3 недели назад, # |

ABC was SpeedForces for me, but D was a great problem... too bad I couldn't solve it lol

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Astroflexx

3 недели назад, # |

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3 недели назад, # ^ |

hope this helps 289278230

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Astroflexx

3 недели назад, # ^ |

What is your solution?

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3 недели назад, # ^ |

← Rev. 3 →

Step_1: sort array a

Step_2: binary search a[i] + a[i+1] in a. Called the result j for example (it infers the array a[i] a[i+1]... a[j] fits the problem requirement, every triplet will fit x+y>z etc.)

result = n - max(step_2)

UPD: seriously my code is simpler, just read it. It's 4-5 liner code only...

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Astroflexx

3 недели назад, # ^ |

Yes I got it thanks

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3 недели назад, # ^ |

int n; cin >> n;
  vi v = inp(n);
  sort(all(v));
  dbg(v);
  int i = n - 1;
  while (i > 0 && v[i - 1] + v[i] > v[n - 1]) i--;
  int j = 2;
  while (j < n && v[0] + v[1] > v[j]) j++;
  cout << min(i, n - j) << endl;

could u tell why this approach fails? cant check the failing testcase

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3 недели назад, # ^ |

← Rev. 2 →

try this:

2
6
1 2 2 3 3 4
7
3 3 4 5 6 7 7

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3 недели назад, # ^ |

2 and 2 is the output

which i believe is correct?

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3 недели назад, # ^ |

← Rev. 2 →

correct, nvm

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3 недели назад, # ^ |

i figured out that my code fails if there is a x + y <= z pattern in middle of array where i or j dont reach

1 1 1 3 4 5 9 10 11 22 30

i stops at 11 and j stops at 3 so res is 8 but correct res is 7 considering the correct seq to be 5 9 10 11 and changing the rest of the values

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MKasirloo

3 недели назад, # |

Problem D gives correct answer on my compiler, but wrong answer on CF

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_Gawd_

3 недели назад, # |

+24

After doing D

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coordinatebash

3 недели назад, # |

Was there an easy solution for C? I tried to memoize with both start and end index of sorted lengths and got TLE

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Noborita

3 недели назад, # ^ |

Yes, Try to check for every position in a sorted array, If this was the minimum, how much are valid ?, it can be done in O(N*Log2(N))

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sahnigwl

3 недели назад, # ^ |

Damn for C I boiled down to that all numbers revolve around median of the sorted array, so any number i.e. greater than median + 2 and any number smaller than median — 1 would be modified. Something like this, was not able to get it right for the sample test cases and time got over. Might be I was overthinking.

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3 недели назад, # ^ |

My approach:

It is sufficient to ensure that a1 + a2 > mx, where a1 and a2 are the two smallest elements of the array, and mx is the largest.

We can change the maximum element of the array. So, sort the array and iterate from the end. Let's stay mx1 = largest element, mx2 = second largest and so on...

For each mxi, we find the number of operations required such that a1 + a2 > mxi. To do that, we find the first pair ai and ai+1 such that ai + ai+1 > mx, and number of operations is simply i - 1, as we must change all elements smaller than ai.

The answer is min(op1, op2, ...), where opi = minimum number of operations for mxi

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ayush118

3 недели назад, # ^ |

Implementation

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sahnigwl

3 недели назад, # ^ |

← Rev. 2 →

But I saw n was in range 2*(1e5), I thought a N^2 algorithm would result in TLE, Im saying O(N^2) since you're checking for each

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3 недели назад, # ^ |

← Rev. 2 →

It can be done in Nlog(N). Check the implementation above. You can store v[i] + v[i+1] in an array, then to find the first pair with ai + ai+1 > mx, just find the upperbound of mx in this array.

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sahnigwl

3 недели назад, # ^ |

gotcha!

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TAIYANGFENG

3 недели назад, # |

← Rev. 2 →

+39

I think maybe I can't go to sleep tonight.

I practice very hard for two months, and I finally make it.

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frissky9

3 недели назад, # ^ |

congrats

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3 недели назад, # ^ |

becoming master?

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TAIYANGFENG

3 недели назад, # ^ |

It seems I get rank 5 and become Master.

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3 недели назад, # ^ |

then congrats MASTER :)

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Titan493ASST

3 недели назад, # ^ |

Inspiring!

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sarvagya2545

3 недели назад, # ^ |

Congrats!

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krism

3 недели назад, # ^ |

Congrats MASTER!

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3 недели назад, # ^ |

-8

damn man big W

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alpha1215

3 недели назад, # |

in B i thought there's actually an array given in the input :(

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3 недели назад, # |

How to solve E? Also Is it possible to construct the answer(operations) array from the final element (if we know the element which will be equal after the operations are done)?

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_Gawd_

3 недели назад, # ^ |

+10

It should be something around difference array.

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SkyWave2024

3 недели назад, # ^ |

← Rev. 2 →

+13

Hint 1

Hint 2

Hint 3

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3 недели назад, # ^ |

What these operations can do is change any 2 elements by either of [+1, +1], [-1, -1], [+1. -1], [-1, +1] these operations. From these it will always be possible to make all the elements equal since n is odd. Is this correct? Can you explain Hint 3, I didn't get it.

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SkyWave2024

3 недели назад, # ^ |

← Rev. 2 →

+15

Sorry for the late reply (I went to sleep after posting the commnt :()

Actually you only need the operation [+1,+1] or [-1,-1]. Let's define $$$inc(i,x) $$$ means that let $$$i,i+1$$$ both increase $$$x$$$ and $$$dec(i,x)$$$ for the same.

Consider element $$$i $$$ and $$$i +1$$$,

If $$$a_i < a_{i +1}$$$, do $$$dec (i,a_i)$$$ so that $$$a_i = 0$$$.
Otherwise, do $$$inc(i+1, a_i)$$$ and now $$$a_i < a_{i +1}$$$.

Keep doing it until there are only two non-zero number in the array and decrease them By there min so only one of them left.

Now we have $$$n-1$$$ zeros and using $$$inc(i,\text{left_number})$$$ to make them equal.

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3 недели назад, # ^ |

Oh that's really cool. Thanks.

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3 недели назад, # ^ |

← Rev. 2 →

The answer to your second question is yes. It is easy to construct the array of operations given the final state, since we just need to solve the linear system

$$$ \begin{bmatrix} 2 & 1 & 0 & \cdots & 0 & 1\\ 1 & 2 & 1 & 0 & \cdots & 0 \\ 0 & 1 & \ddots & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & 1 & 0 \\ 0 & \cdots & 0 & 1 & 2 & 1 \\ 1 & 0 & \cdots & 0 & 1 & 2 \\ \end{bmatrix} x=Mx=s-a $$$

where $$$x_i$$$ is the number of times you applied the operation with index $$$i$$$, $$$a$$$ is the input and $$$s$$$ is a vector with the final state (in the problem all the entries of $$$s$$$ are the same however this method applies for every $$$s$$$). It is trivial to see that the $$$k$$$-th row of $$$M^{-1}$$$ is a cyclic shift by k of the upper row of $$$M^{-1}$$$, so you can solve this system easily with convolution/fft.

However, for the problem itself there are way simpler solutions.

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c.young_sb

3 недели назад, # ^ |

Can you explain a little detailed how to use FFT to solve such equations?

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3 недели назад, # ^ |

← Rev. 2 →

+12

Sure, for vector $$$v$$$, denote by $$$w_{v,i}$$$ the vector $$$v$$$ cyclically shifted by integer $$$i$$$.

We know that

$$$ M^{-1}= \begin{bmatrix} w_{v,0} \\ w_{v,1} \\ \vdots \\ w_{v,n-1} \end{bmatrix}. $$$

To solve the system, just multiply $$$M^{-1}$$$ by $$$u=s-a$$$. Of course, you can't do this naively because it would be $$$\mathcal{O}(N^{2})$$$. However, notice that on the vector

$$$ M^{-1}(s-a)= \begin{bmatrix} w_{v,0} \\ w_{v,1} \\ \vdots \\ w_{v,n-1} \end{bmatrix} u= \begin{bmatrix} w_{v,0} \cdot u \\ w_{v,1} \cdot u \\ \vdots \\ w_{v,n-1} \cdot u \end{bmatrix} (\star) $$$

every single entry is the dot product of a certain cyclic shift of $$$v$$$ and $$$u$$$. If you consider a vector $$$r=\begin{bmatrix} v,v\end{bmatrix}$$$ then $$$(\star)$$$ is just some of the entries of the convolution of $$$r$$$ and $$$u$$$.

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zackscott286

3 недели назад, # ^ |

← Rev. 2 →

Could we use this to solve? It got me WA

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3 недели назад, # ^ |

The link you gave is for tridiagonal matrices, $$$M$$$ is not tridiagonal because of the "corner entries" $$$M_{0,n-1}$$$ and $$$M_{0,n-1}$$$.

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zackscott286

3 недели назад, # ^ |

Take a look at the Variants section, there is a solution for the matrix which is the same as in this problem.

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zfnu

3 недели назад, # ^ |

← Rev. 2 →

Yes, you can solve with it, but you'll need to store the intermediate results as fractions instead of doing double division and then do a little bit of reconstruction on top of that to get the integer answer.

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3 недели назад, # ^ |

Then yes, you can solve it with that. I used fft because I am more used to it, but I would expect that there is a linear solution like the one described there.

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zackscott286

3 недели назад, # ^ |

It got me WA on test 6 however, though I copy pasted the whole function from wiki. It's hard to deal with precision here. Maybe FFT is more reliable in this situation.

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3 недели назад, # ^ |

Thanks for sharing. Will go through these concepts to get better at it.

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DuongForeverAlone

3 недели назад, # |

← Rev. 2 →

+11

ACs on problem D rose drastically at the end._.

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devvikram22

3 недели назад, # ^ |

Can u tell me how to improve speed solving greedy problems? That's only stopping me from Expert

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DuongForeverAlone

3 недели назад, # ^ |

Write out some cases and see the pattern. That's quite a good strategy actually.

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Rohak

3 недели назад, # |

damn i am not skilled at greedy problems at all...

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Andimeo

3 недели назад, # ^ |

Greedy is hard. Especially the proof part.

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Sibycoin

3 недели назад, # |

Is there anyone who also used a[N] instead of a[2N] on Problem A to declare the array? Just wasted almost 0.5h on it and the unstable mindset finally results in the thorough collapse:(

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risingStark

3 недели назад, # ^ |

How many you solved finally?

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Sibycoin

3 недели назад, # ^ |

only ABC, and the "pretests accepted" solution of C is even submitted on the very last time of the round (only 10+ minutes left). I think at least I could solve 3~4 problems much quicker without this annoying bug:( But it's totally my own fault I have to admit, and nothing to accuse about the authors and the statements.

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A5HU705H

3 недели назад, # ^ |

misread the question thought a[i] = a[j] only allowed if i <= j. Thought about the problem like this for half an hour. What a waste :(

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3 недели назад, # ^ |

← Rev. 2 →

I make other mistake of overthinking the problem. Also cost 0.5h

After all the fog clears, it's just res_min = cnt1&1, res_max = min(cnt0, cnt1)

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Sibycoin

3 недели назад, # ^ |

lol I've discovered the fun fact that it's exactly how CF Div.2 AB features. The solution is as easy as you could hardly believe, but if you fell into the trap of overthinking then it would just be rather complex or even impossible to solve

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3 недели назад, # ^ |

Might someday we will brave enough to rainboy the contest (doing problem in E-D-C-B-A order)

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3 недели назад, # ^ |

Very true. It's very hard to remain calm and "think simple" during contests. I always overcomplicate under pressure

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wuhudsm

3 недели назад, # |

+31

My E solution

Consider the difference array.

Let $$$d[i] = a[i] - a[(i-1) \mod n]$$$ (0-indexed).

We can see that the sum of $$$d[i]$$$ does not change. Thus, if the sum of $$$d[i]$$$ is not zero, there is no solution. Otherwise, we can always construct a solution.

Our target is to make $$$d[i] = 0$$$ for all $$$0 \leq i < n$$$.

An operation for $$$i$$$ (denoted as $$$\text{op}(i)$$$) is equivalent to: $$$d[i-1]++, d[i]++, d[i+1]--, d[i+2]--$$$ (mod $$$n$$$).

Merging operations:

$$$ \text{op}(i) + \text{op}(i+2) \implies d[i-1]++, d[i]++, d[i+3]--, d[i+4]-- $$$

Since $$$n$$$ is ODD, $$$\text{op}(i) + \text{op}(i+2) + \ldots + \text{op}(i-1)$$$ results in $$$d[i-1]++, d[i+1]--$$$.

Next, make $$$d[i-1]++, d[i+1]--$$$, then perform $$$d[i+1]++, d[i+3]--$$$, etc. to achieve $$$d[i]--, d[i-1]++$$$.

The implementation is painful. Can't finishit in the contest ;)

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methanol

3 недели назад, # ^ |

+20

Sum of $$$d_i$$$ is always $$$0$$$, so there always exists a solution for an input.

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alpha1215

2 недели назад, # ^ |

it took me much time to arrive at this solution but the implementation is very simple 289836310

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risingStark

3 недели назад, # |

What is the problem with me??? Every contest its one thing or the other.

Sometimes I am able to find out the solution of a problem but unable to implement it quickly. I figured out some solution for D. But I was unable to implement it properly in 40 mins. I was juggling between different approaches. How do I calm my mind and implement the solution first properly? Is it just the lack of practice and nervousness that gets to me or is there something wrong with my time management?

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3 недели назад, # ^ |

It's lack of confidence if you lose your focus on your own algorithm without a clear WA/TLE verdict.

Practice daily to gain confidence. Confidence isn't grow by words mate.

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SkibidiPilav

3 недели назад, # ^ |

I had the same problem as you, and there is no trick. Just compete in more contests. Today i had solution for D in like 5 mins, but had no idea how to implement it, but i didnt panic and figured it out quite quickly after writing down some simple code. It all comes with experience, you cant speed it up.

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risingStark

3 недели назад, # ^ |

I am unable to put my mind to focus..Its just that it keeps wandering here and there.

I am really demotivated right now because it is happening repeatedly with me now. I find some solution during contest, it doesn't work. Just after the contest (when my brain knows the contest is over, there's no time limit) I submit it and get an AC within 10 mins. But in the last 30 mins of the contest ending, I am just unable to even debug my code properly.

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ONEHUNDREDPUSHUPS

3 недели назад, # |

почему гениальные идеи приходят когда я сру

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SigmaGoku

3 недели назад, # |

In problem D can any other node except root have degree more than 2?

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yvmr_33

3 недели назад, # ^ |

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SigmaGoku

3 недели назад, # ^ |

Why?

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3 недели назад, # ^ |

cuz everything else is path thus they need to be a straight line, thus a degree of at max 2 for every node except 0.

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yvmr_33

3 недели назад, # ^ |

1. Each node that was initially adjacent to the node 0 will be the end of some path. 2. A path is a tree whose vertices can be listed in the order v1,v2,…,vk such that the edges are (vi,vi+1) (1≤i<k ).

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msvardhan19

3 недели назад, # ^ |

If we remove root node 0 and all adjacent edges, this tree will turn into a forest consisting of only paths.

think this through — can you figure out now?

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SigmaGoku

3 недели назад, # ^ |

Thx got it now.

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LordLouis04

3 недели назад, # |

missed problem D cause i swaped 2 variables in my code bruh

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treasuryonly

3 недели назад, # |

Thank you guys and Code Mely for the amazing contest!

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dewanganprashant47

3 недели назад, # |

Please anyone can explain me 3rd question

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alexwice

3 недели назад, # |

E has a cute trick. Write W[i] = V[i] + V[i+1]. https://codeforces.net/contest/2032/submission/289289897

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Noobish_Monk

3 недели назад, # ^ |

So now we have an operation $$$[+1, +3, +3, +1]$$$, how is it easier?

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alexwice

3 недели назад, # ^ |

A[i+1] + W[i] + W[i+1] = T differences, W[i+2] = W[i] + A[i+1] — A[i+2]. Say W[0] = w. Use above eqn to fill out W. Each W[i] = aw+b will have a=1. Sum all rows to find the real W[0] ("extra = ..." in the code)

Now from W solve for V, its easier to solve the n equations W[i] = V[i] + V[i+1]

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no_oneplb

3 недели назад, # |

+20

The problems were mostly good ,the issue with D is it just gives away the solution because there are so many limitations(conditions) so you just implement it, and the difference between D and E is too huge.

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The_Logic

3 недели назад, # |

Special thanks to everyone for They breathe oxygen. and seriously for you.

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9mar3

3 недели назад, # |

it 's unrated.

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mamalunga

3 недели назад, # |

← Rev. 2 →

Why is the result for this test case for problem C 1 and not 2?

1 1 1 2 Even if we replace the middle 1 with 2, then we get 1 2 1 2, and the triple 1 2 1 is degenerate.

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yvmr_33

3 недели назад, # ^ |

← Rev. 3 →

we can change 2 to 1. i.e [1,1,1,2] -> [1,1,1,1]

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msvardhan19

3 недели назад, # ^ |

← Rev. 2 →

replacing 2 with 1 is optimal — 1 1 1 1

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Dr.Zain

3 недели назад, # |

what is wrong with this solution for C? :

always check the two smallest numbers and if thier sum <= biggest number then replace the samllest with the biggest

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LMeyling

3 недели назад, # ^ |

you can also replace the biggest with the smallest

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SkibidiPilav

3 недели назад, # ^ |

Consider 1 7 1 1 1 1 1 1 10. Your code replaces all thr ones with 10, but you can also replace the 10 with 1

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coordinatebash

3 недели назад, # ^ |

consider the case

1, 1, 1, 1, 10.

Is it better to replace the first 3 ones with a 10? or the last 10 with a one?

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3 недели назад, # |

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

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zrnstnsr

3 недели назад, # |

← Rev. 2 →

For E, one can add $$$1$$$ to $$$a_i$$$ and $$$a_{i+1}$$$ equivalently by applying $$$\frac{n+1}{2}$$$ operations on $$$i+1,i+3,\dots,i-2,i$$$, and what remains is simple.

P.S. cf logout my account automatically before the contest start, so I thought I hadn't registered. MISS THE CHANCE TO BECOME MASTER AGAIN!!

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heavenly_

3 недели назад, # ^ |

It was possible to late register

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Alenygam

3 недели назад, # |

How to stop being stupid?

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Le_Gusto

3 недели назад, # |

finally got expert :)

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qm02192803

3 недели назад, # ^ |

wow, amazing :0 I hope I will reach green tomorrow.

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qm02192803

3 недели назад, # |

Can anyone teach me problem E?

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HajdeIk

3 недели назад, # |

Can someone explain this. At problem C there is this part in test case 3 test line 13 and its like this : 4 1 1 2 3 The correct answer by the judge is 2 but if we remove 2 than that would mean no triplet. Can someone see to this or am I tripping. (the error : wrong answer 13th numbers differ — expected: '2', found: '1')

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qm02192803

3 недели назад, # ^ |

you need to replace, not remove. So after replace 1 1 to 3 3, we will get 4 3 3 2 3 which is the most optimal operation that satisfy the condition.

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HajdeIk

3 недели назад, # ^ |

Makes sense. I forgot that when I was making the solution on the paper I thought of it for one moment as removing since it would make it simpler and got carried out on it.

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cry

3 недели назад, # |

top1codeforces is clearly using GPT. Why is he still getting first solves credits on A?

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3 недели назад, # ^ |

Interesting. I actually didn't diagnose every code (relying on the submission order to credit), and I also didn't know how to distinguish LLM-generated code from regular ones. Would love some enlightenment, and if feeling solid enough, I could change it accordingly.

(Might take a day or two however, it's midnight in my place right now)

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Pirate_King

3 недели назад, # ^ |

He got banned after solving A-F in today's div3 in 6 mins 💀

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3 недели назад, # ^ |

Updated :skull:

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chuka231

3 недели назад, # |

Mistook the operation ai += aj in C and got WA forever...

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Frogmaster

3 недели назад, # |

I really enjoyed problem D, the first interactive problem I have ever solved during a contest

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3 недели назад, # |

int n; cin >> n;
  vi v = inp(n);
  sort(all(v));
  dbg(v);
  int i = n - 1;
  while (i > 0 && v[i - 1] + v[i] > v[n - 1]) i--;
  int j = 2;
  while (j < n && v[0] + v[1] > v[j]) j++;
  cout << min(i, n - j) << endl;

why exactly does this approach fail for C ? cant really see the testcase

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christmas964

3 недели назад, # |

Thank you very much for the contest.

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sunkuangzheng

3 недели назад, # |

My hack on problem F got Unexpected verdict, AkiLotus can you fix it please?

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3 недели назад, # ^ |

Apparently a tester solution also failed your hack. I'm trying to work on it, thank you for pointing out.

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3 недели назад, # ^ |

The fix is done. I don't know if the hack could be redone though (I never saw such happen before), but I believe you could hack that solution again.

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sunkuangzheng

3 недели назад, # ^ |

Thank you so much! I've rehacked it and it works :)

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Ghulam_Junaid

2 недели назад, # |

Hi! I have received a message from the system that my submission 289199140 for 2032B - Medians coincides with 289209093. What should I do about this?

This was a relatively easy problem, and I won't be surprised if 10 other people have written the exact same code as mine. Will I be considered a cheater for this by CF now? How can I erase this suspiciousness on me?

This is the first time, I have received such a message. I hope this was the right place to post this comment.

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