applepi216's blog

Codeforces Round 986 (Div. 2)
By applepi216, 6 weeks ago, In English

Hello Codeforces!

I am glad to invite you to Codeforces Round 986 (Div. 2), which will start on Nov/10/2024 18:35 (Moscow time). Note the slightly unusual start time.

The contest will run for 2 hours and have 6 tasks. The contest will only be rated for those with a rating not higher than 2099, but we welcome higher rated users to participate out of competition.

Holding the contest would have been impossible without:

and of course a special thank you to all of you for participating!

Score Distribution: $$$500-1000-1500-1750-2000-2500$$$.'

UPD: The Editorial is now published.

UPD2: Congratulations to the winners!

Div 1:

  1. jiangly
  2. A_G
  3. BurnedChicken
  4. MAKMED1337
  5. maspy

Div 2:

  1. SUPERLWR-beta
  2. mychecksdead
  3. sweet_dream
  4. contee
  5. Sunflower233

First solves:

  • Vote: I like it
  • +131
  • Vote: I do not like it

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6 weeks ago, # |
  Vote: I like it +33 Vote: I do not like it

As a tester, this is indeed a Codeforces round.

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6 weeks ago, # |
  Vote: I like it +45 Vote: I do not like it

As a tester, I almost planned to register until getting tagged.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

As a tester, i couldn't solve all the problems.

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    WOW! ok, it'll be a bit hard◉_◉

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      You don't need to solve all the problems to get a really big rating boost.

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6 weeks ago, # |
  Vote: I like it -11 Vote: I do not like it

As a participant, I wish we can learn something from the contest

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6 weeks ago, # |
  Vote: I like it +29 Vote: I do not like it

as a participant, I will start from Problem F, or die trying.

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6 weeks ago, # |
  Vote: I like it +31 Vote: I do not like it

As a tester, I want to know why you're not doing Universal Cup.

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6 weeks ago, # |
  Vote: I like it +12 Vote: I do not like it

I hope the people who are genuinely solving tasks(by themselves) get a positive delta.

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6 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

As a participant Hope to solve 3 problem

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6 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

I hope i can reach pupil , its nearly impossible ig but ill try my best .

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6 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

As a participant, I wish I can solve at least one problem

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6 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

I hope I will reach M in this round.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Hoping for some positive delta.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

the time start is so good 22:35 utc+7, also it time to go pupil! :-D

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

hope to solve 4 problems :)

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

As a participant, I believe I will learn something new.

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6 weeks ago, # |
  Vote: I like it -9 Vote: I do not like it

I wish every CF round started at 9, so that I could have time to eat my disgusting mess food in peace.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I'll try to make my day with this Round. Good luck to everyone!

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

As a participant, I wish I won't get wa on test 2

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6 weeks ago, # |
  Vote: I like it +6 Vote: I do not like it

As a Codeforces user, I don't understand all these ass comments everywhere, what is the story behind them? why are they considered cool?

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

hope to solve 4 problems -_-

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Who else came here and the contest haven't started?

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

I wont breath until someone rep this

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6 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

As a tester, goodluck.

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6 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

As a participant, this is my first unrated Div.2 :D

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Can't register after contest starts? I registered and still couldn't submit

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6 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

jiangly OP!

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6 weeks ago, # |
  Vote: I like it -14 Vote: I do not like it

it's dev1 in disguise

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6 weeks ago, # |
  Vote: I like it -7 Vote: I do not like it

Rays of hatred for changing D

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6 weeks ago, # |
  Vote: I like it -11 Vote: I do not like it

As a participant, i am ded †.

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6 weeks ago, # |
← Rev. 2   Vote: I like it -8 Vote: I do not like it

i'm cooked

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

This is the one of the CF rounds of all time

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6 weeks ago, # |
  Vote: I like it +19 Vote: I do not like it

as a participant i can say the solution was out on telegram. Please take strict action on cheaters a request. I am fedup of this cheating can't you block their account permanently and ask everyone on the platform to link their account with their phone numbers so that if one cheats he will need to create a new account with a phone number and a person can have limited phone number not like gmail account so he or she might not do the same with other phone number account. The problems where good indeed

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6 weeks ago, # |
  Vote: I like it +9 Vote: I do not like it

great b and c. Makes you really think.

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6 weeks ago, # |
  Vote: I like it +25 Vote: I do not like it

C is much easier than B

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    6 weeks ago, # ^ |
      Vote: I like it +11 Vote: I do not like it

    D is much easier than B

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    how to solve C?

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      two pointers

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        6 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        something like sliding window?

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          6 weeks ago, # ^ |
          ← Rev. 4   Vote: I like it 0 Vote: I do not like it

          What i know that thery are the same, anyway yes u can use sliding window,
          But before that pre-calculate prefix & suffix to know max good groups u can made in ranges [1, l] or [r, n] in O(1).
          Finally the range [l, r] is optional answer, if (pre[l — 1] + suff[r + 1]) is greater than or equal to m.

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6 weeks ago, # |
  Vote: I like it -6 Vote: I do not like it

Div 1.33

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

how to solve F?

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Why q 2 j 4 is invalid in problem D for the second test case? where: Q: 2 3 1 4 K: 1 2 3 4 J: 1 4 2 3

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    6 weeks ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    q 2 is invalid because q_1 < q_2

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      6 weeks ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      my bad. mistaken as position. thanks!

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      sorry, I don't get it, which rule does that violate? q values 1 more than 2 according to their list, so it should be okay to swap?

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Realized my mistakes at the end of the contest...

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6 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

B burned me to charcoal, C reduced me to ashes and buried me in the mantle

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    6 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    B took all my time, then after solving B, I realised that C was way simpler

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Still have no idea how many rounds of simulation to run for pA... was guessing 100 * 10 would be enough?

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    10 should obviously be enough, i think. Since max destination coord can be (10, 10), and if we ever reach there, we would have to make some progress at each iteration of sequence. The bare minimum progress if (1, 1) (since each iteration is identical and we cant move (0, 1) one time and (1, 0) the other). Since max x/y is 10 and each time we get closer to it by atleast 1, 10 rounds would be enough.

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      6 weeks ago, # ^ |
        Vote: I like it +4 Vote: I do not like it

      10 is not enough, surprisingly...

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        6 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        yup, right. luckily i iterated it 20 times during contest, lol. nice problem.

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      6 weeks ago, # ^ |
      ← Rev. 2   Vote: I like it +7 Vote: I do not like it

      $$$10$$$ is not enough. Consider a path where Alice goes $$$3$$$ down, $$$1$$$ right, $$$1$$$ left, and $$$4$$$ up. The point $$$(1, 10)$$$ is eventually reached,but it is only reached in the $$$14$$$-th repetition. I'm pretty sure $$$14$$$ is enough, but I can't prove that.

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        6 weeks ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        aaaahhh i see, that's cool. Luckily i iterated it 20 times instead of 10 during contest for some good measure, or i would have been cooked even worse than I already am :(

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6 weeks ago, # |
  Vote: I like it +7 Vote: I do not like it

C<B

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6 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

pls dont make these shits again.

thank you.

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6 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

I liked B and C, but the problem statement for D made me lose braincells

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6 weeks ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

In the second test case for Div2D, can you please help me understand ? I am able to reach from type-1 to type-4 card in this manner.

1) Give the type-1 card to queen, and get type-3 card from queen

2) Give type-3 card to king , and get type-2 card from king

3) Give type-2 card to Jack, and get type-4 card from Jack

What is the problem ?

I couldn't solve B, or C or D. Implemented all 3 questions, but kept on getting WA on pretests. Quite imbalanced contest IMO.

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    6 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    Alice can't exchange card a for card b if b < a. Haha it's funny how we all wrote at the same time

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    The problem is you are disrespecting Alice's trading preferences. Alice would never trade card 3 for card 2 as she prefers 3 over 2.

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    We can only go up in types, we cannot go from type-3 from type-2, it is Alice preference.

    Also spent like 20 minutes on it

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Alice won't exchange type-3 card to type-2 card because Alice values type-3 card more than type-2 card

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I didn't consider Alice's preferences. I simply ignored it in problem statement. :( .

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    6 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Step 2 is wrong , Alice won't trade for a lower card as given in the question.

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    6 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    You can't trade with the queen in the second test case,

    both Alice and Queen wants to trade to a higher number.

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    6 weeks ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    Alice (meaning you) isn't willing to trade a higher card for a smaller card, so your trade of giving $$$3$$$ to receive $$$2$$$ can't happen.

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You can't give to king type-3 and get type-2, because 2 is smaller than 3. Alice can change cards only in increasing order according to her preferences

    "Alice's preferences are straightforward: she values card a more than card b if a>b, and she will also only trade according to these preferences."

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6 weeks ago, # |
  Vote: I like it +18 Vote: I do not like it

B was harder than C.

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6 weeks ago, # |
  Vote: I like it -29 Vote: I do not like it

Problem C is GARBAGE.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

nice C and D, retarded A and B.

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6 weeks ago, # |
← Rev. 2   Vote: I like it -33 Vote: I do not like it

shittiest div2 ever applepi216 please do better next time

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Solved A in 4 minutes just to stare at 5 wrong submissions on B....

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

for B, if b == 0 and c == 0 => if(n <= 2) ans = n — 1 else ans = -1;

if(c >= n) ans = n;

otherwise ans = n — (n — 1 — c)/b — 1;

what is wrong with the approach?

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6 weeks ago, # |
  Vote: I like it +2 Vote: I do not like it

Terrible weekend. Got humbled twice I don't belong here.

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6 weeks ago, # |
  Vote: I like it +15 Vote: I do not like it
B topic
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6 weeks ago, # |
  Vote: I like it +3 Vote: I do not like it

It was so close for AC :(

The feeling when you skip B and can solve D & E is indescribable, feels good.

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6 weeks ago, # |
  Vote: I like it +11 Vote: I do not like it

I felt like A < C = D < E < B by difficulty

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    6 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    B is so easy, if you carefully analyzed each case, it's more annoying than difficult in my opinion.

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      can you share your approach

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        6 weeks ago, # ^ |
        ← Rev. 3   Vote: I like it 0 Vote: I do not like it
        hint

        I analyzed cases in the following order:
        1. handle case when n = 1
        now 1 < n
        2. handle cases when b = 0, a [c, c, ...,c]
        now 1 < b
        3. from given test cases, I realized it may be better to also handle case when b = 1 and c = 0, basically a = [0, 1, .., n-1], so we need not to do anything
        now b*(i-1) + c is always greater than i-1
        4. using the hint, it's possible to see that for each big number (> n-1), when we apply the operation, if there is still big numbers, then max is one of such integers and it's replaced by a good number (in [0,n-1]), which means that for each big number, we will need a single distinct operation to delete it, after removing all big numbers there will be no evil.
        So, now the task is to count how many big numbers in array a.

        hint
        solution
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6 weeks ago, # |
← Rev. 2   Vote: I like it +3 Vote: I do not like it

Wow, I think I solved D in the last 2 minutes of contest, hoping it passes system testing!

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6 weeks ago, # |
  Vote: I like it +46 Vote: I do not like it

I spent maybe 10 minutes rereading the problem statement for D until I figured out that the Queen/King/Jack of Hearts are supposed to be the players and are not cards of the game. Other than that, great contest :)

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

what the hell was be

can someone help me by telling what case i missed

https://codeforces.net/contest/2028/submission/290946207

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    6 weeks ago, # ^ |
    ← Rev. 2   Vote: I like it 0 Vote: I do not like it

    I made a small change in your code and it got accepted. Here is the accepted solution. You missed the case when n == 2 in the if condition where a==0 && b==0.

    Spoiler
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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Implementation skill issues wasted so much time Left contest in frustration Will work on myself

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6 weeks ago, # |
  Vote: I like it +6 Vote: I do not like it

I submitted my solution for problem C in the last 30 seconds without even testing it!

I am really happy!

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I'm so close to submit but It's 2 minute late...

    Now I'm waiting for the practice mode

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

B ruined everything

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6 weeks ago, # |
← Rev. 2   Vote: I like it +5 Vote: I do not like it

Is problem D a graph problem ? In the entire contest, I try to build the graph and reduced the edges so that the graph size is O(n) instead of O(n^2). I try the mono stack or segment_tree but failed very badly.

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I did solve it using bfs but there might be other non-graph methods

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It can be interpretted as a graph problem. To reduce the number edges you can keep track of which nodes are already visited and only consider edges ending in unvisited ones (using a set or something).

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I was trying the same idea using mono stack for around 40 minutes, but understood that prefix max and position of maxn on the prefix is enough.

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      I did use mono stacks to reduce edges, but I don't know if it is a $$$\Theta(n)$$$ solution: 290933314.

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6 weeks ago, # |
  Vote: I like it +5 Vote: I do not like it

first time solving a problem which has less than 500 submissions. (problem E)

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

C is actually good problem, sadly I lose all my time for A/B because I'm bad at edge cases.

Just sad.

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6 weeks ago, # |
← Rev. 3   Vote: I like it +1 Vote: I do not like it

Wasted almost an hour on C by somehow concluding that Alice's segment can be non-contiguous and now I'm actually curious how to solve the modified problem
Ex: For $$$m=2,v=2$$$ and $$$a=[100,2,100,1,1,100]$$$ answer would be 300 where we give segment $$$[1,1]$$$ and $$$[3,4]$$$ to the creatures and rest to Alice.
$$$O(n.m)$$$ is obvious but can we do better ?

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    6 weeks ago, # ^ |
    ← Rev. 12   Vote: I like it 0 Vote: I do not like it

    I am actually curious what the $$$O(n.m)$$$ solutions is!?
    All I can think of is complete search on the set of pieces allocated for Alice, and for each check whether we can make monsters happy using the rest. It's $$$O(2^n.n)$$$

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      6 weeks ago, # ^ |
        Vote: I like it +1 Vote: I do not like it

      Let $$$dp[i][j]$$$ denote the minimum sum of the j pieces to satisfy j creatures among the first i elements.
      $$$dp[i][j] = min(dp[i-1][j],dp[k-1][j-1] + sum[k,i])$$$ where k is the maximum index less than i such that $$$sum[k,i] >= v$$$
      Answer would be $$$sum[1,n]-dp[n][m]$$$
      We would need to do binary search to find k so the TC would actually be $$$O(n.logn+n.m)$$$

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    6 weeks ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    I wasn't able to spend much time on C in the contest (got cooked by B like many others), but the approach I had was quite straightforward. I did end up upsolving it after contest ended, with an O(n + m) solution, which is this case is just O(n).

    Key Observation: Alice's section can only be at the beginning, end or somewhere in the middle (since she can only take a single contiguous section) — part of the question, but easy to miss.

    Approach: 1. Create modified prefix arrays of potential sections in both forwards and backwards directions (adding to array whenever the sum crosses v — creating sections greedily in both directions). 2. If length of above arrays isn't over m, then a solution is not possible (as we couldn't create m sections even using greedy). 3. Now it's just a matter of maximizing Alice's section (m+1th) while considering m sections from a combination of forwards and backwards arrays — we can iterate i in 0..m and consider max of (total - forwards[i] - backwards[m-i]) as the final answer.

    AC submission without using 2 pointers or binary search: 290960412

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Sorry I should have been clearer. I meant how to solve the modified problem that I misunderstood not the original problem.

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6 weeks ago, # |
  Vote: I like it +4 Vote: I do not like it

Why this solution to problem B is giving TLE on pretest 4?

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    6 weeks ago, # ^ |
      Vote: I like it +17 Vote: I do not like it

    Scanner in Java is incredibly slow.

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      6 weeks ago, # ^ |
        Vote: I like it +4 Vote: I do not like it

      Thanks, just submitted the same logic with bufferedreader and it got accepted.

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6 weeks ago, # |
  Vote: I like it +15 Vote: I do not like it

D is extremely hard for me to understand the statement.

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6 weeks ago, # |
  Vote: I like it 0 Vote: I do not like it

Thanks for the contest! Feels good getting pB in 1 try!

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6 weeks ago, # |
  Vote: I like it +25 Vote: I do not like it

I have a simple question Was not there other names in the WORLD except King, Queen ans Jack ?

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    6 weeks ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    The names imo are justified in literary context, as they are directly linked with Alice's Adventures in Wonderland.

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      6 weeks ago, # ^ |
        Vote: I like it +13 Vote: I do not like it

      I like how the descriptions naturally reflect this concept, although the q k j order on D got me a few times because I'm just so used to the k q j or j q k order due to all the playing card games I've played so far.

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6 weeks ago, # |
  Vote: I like it +1 Vote: I do not like it

What can I learn from problem A? If constraints are really small, just run the simulation enough number of times which is very large than the constraints?

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6 weeks ago, # |
← Rev. 4   Vote: I like it 0 Vote: I do not like it

can someone explain why in D, in the the second test case.

Alice exchanges to card 3 with queen and then to card 4 from jack. Why isnt this a valid workaround

In other words,

why isnt the soln

3 q

4 j

correct?

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    6 weeks ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    She cannot exchange card 4 with jack because p[4] = 3, and p[3] = 2 therefore p[3] > p[4] doesn't hold.

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      6 weeks ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      yea, my bad I was debugging it wrongly

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6 weeks ago, # |
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I think that problem B needs too many steps as the second problem in Div2,but after minutes of thinking,I solve it.I use very easy to use and convenient translator today which I didn't know in former.But my parents asked me not to use the translator on some competitions that are not important or do not increase the rating value.

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6 weeks ago, # |
  Vote: I like it +35 Vote: I do not like it

To be fair, I don't believe the submissions from podyapolskiy.yaroslav are works done by a single person without using AI tools. If you look at their submission history, the person switched to a new random question every 3-5 minutes, and was able to produce a fully written solution that passes the sample but failed at some later pretests (except A). Even worse (and that's why we could discover this) -- they don't even bother hiding it with spacing out the submissions by a few more minutes, or at least trying them in order.

The final judgement obviously needs to be made by the admins, but it's fairer to award the first solve to the genuine first solver if the admins found this case guilty.

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6 weeks ago, # |
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can anyone tell why my approach fails 290961099 ? can we solve it by binary search ?

problem c.

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6 weeks ago, # |
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Nice Contest!

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6 weeks ago, # |
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What was the point of even having D, it was just simple greedy and implementation

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6 weeks ago, # |
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B had a bit hard concept. C and D were excellent, I enjoyed implementing it. They really teach us to visualize the problems and to handle indices carefully. All in all, a great round.

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6 weeks ago, # |
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As a participant and a beginner, I can't even solve B.

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6 weeks ago, # |
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first time getting a first solve :D

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6 weeks ago, # |
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We can see that the first person to solve problem A, podyapolskiy.yaroslav, used AI on all of the questions.

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6 weeks ago, # |
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Oh may bad , I got WA on A , because of limit of repeating the process , if I got it correct , then I'll top 3k now

Meme

Just not in python

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6 weeks ago, # |
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orz

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6 weeks ago, # |
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"Amazing problem! It really tested my understanding of edge cases and forced me to think creatively. Loved how every observation brought me a step closer to the solution. Looking forward to more challenges like this!"

"Finally solved it after hours of debugging! The problem was tricky but super rewarding once I figured out the approach. Thanks, Codeforces!"

"Great problem! Loved the unique constraints – they made me rethink my entire approach. Big shoutout to everyone who shared tips on handling tricky edge cases!"

"One of the best problems I’ve solved recently. It perfectly balanced complexity with elegance. It felt satisfying to apply different techniques to reach the solution. Kudos to the author!"

"Spent half my day on this but learned so much! From wrong attempts to the right approach, the journey was worth it. Excited for more problems like this!"

Each of these comments is around 60 words or less, focusing on appreciation, learning experience, and excitement for future challenges.

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6 weeks ago, # |
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POV : B is close to This

I've just noticed while upsolving AC problems

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6 weeks ago, # |
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what is wrong with the rating system man???

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6 weeks ago, # |
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I will try to solve more than 3 problems

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6 weeks ago, # |
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it’s a great idea to highlight the name of the first solver for each problem!

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6 weeks ago, # |
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MikeVazovsky is the strongest competetive programmer ever

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