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agrim07's blog

Invitation to CodeGuerra — OpenCode by IIIT Allahabad
By agrim07, 9 months ago, In English

Greetings CodeForces!

GeekHaven — the Technical Club of IIIT Allahabad is excited to invite you to CodeGuerra, a flagship event within OpenCode — a month-long open-source event hosted by students at IIIT Allahabad.

The contest will take place on December 22, 2023, at 20:00 IST.

Many thanks to all the people who made this round possible:

You will be given 7 problems and 2 hours and 15 minutes to solve them. The difficulty of the round will closely resemble Div. 3, 4 rounds.

We hope you will enjoy the contest!

UPD: Registration is open now

UPD: Editorial is out

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9 months ago, # |
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22

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9 months ago, # |
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very much excited, #IIITA

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9 months ago, # |
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Excited for this one!! agrim07 thanks for inviting.

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9 months ago, # |
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Lets go !!

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9 months ago, # |
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Looking forward!!

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9 months ago, # |
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Congratulations! I Hope there will be many contests ahead organised by this team☺️

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9 months ago, # |
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Looking forward to giving this contest!

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9 months ago, # |
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Is this rated or just?

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9 months ago, # |
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Excited !!

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9 months ago, # |
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Excited !!

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9 months ago, # |
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As a tester i think all of the problems are very nice and you will definitely learn something from all of them .

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9 months ago, # |
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very excited for this :D

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9 months ago, # |
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:tamil:

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9 months ago, # |
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pvtr aiyo

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9 months ago, # |
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Excited and looking forward to give the contest

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9 months ago, # |
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Excited

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9 months ago, # |
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any registration link?

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    9 months ago, # ^ |
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    You will be able to register 6 hours before the start of the contest using this link

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9 months ago, # |
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Thanks for inviting us. Hopefull to get a good rank

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9 months ago, # |
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so there are no prizes whatsoever ?

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9 months ago, # |
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Excited

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9 months ago, # |
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Excited!!!

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9 months ago, # |
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excited for this contest

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9 months ago, # |
  Vote: I like it +4 Vote: I do not like it

enjoyed solving problems A to E , thanks for the contest

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9 months ago, # |
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what was the approach for problem C ??

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    9 months ago, # ^ |
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    Spoiler
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    9 months ago, # ^ |
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    for f(i) = f(i — 1) + f(i — 2) we can see that, whenever i is odd, the parity of f(i) is odd, just try out some examples you will figure it out.

    Now for g(i) = count(f(i), 1) — count(f(i), 0), we can see that whenever i % 3 = 2, the parity of g(i) is even, else parity of g(i) is odd, try out some examples.

    It is also easy to prove the above pattern.

    void solve() {
    int n; cin >> n;
    if(n % 3 == 2) {
        if(n & 1) cout << "NO\n";
        else cout << "YES\n";
    } else {
        if(n & 1) cout << "YES\n";
        else cout << "NO\n";
    }
}

    Hope it helps.

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    9 months ago, # ^ |
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    Numbers that are congruent to {1, 2, 3} mod 6

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9 months ago, # |
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What would be the approach for the problem E i tried binary search + two pointers but not get it can someone tell

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9 months ago, # |
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I was getting wrong answer on test2 for D.

My approach: Obtain a new string by adding the substrings from index (0 to l-1) and (r to last-1). Now check if str is a substring of this new string.

Here is my code , can you help where i am wrong...

My Code
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    9 months ago, # ^ |
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    Please read the question one more time. We need to search if there is subsequence of string s(0..l-1) + s(r + 1, ... n — 1), equals to str, not substring of s(0..l-1) + s(r + 1, ... n — 1).

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