Блог пользователя agrim07

Invitation to CodeGuerra — OpenCode by IIIT Allahabad

Автор agrim07, 3 года назад, По-английски

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:

The problems were authored and prepared by pvtr, shekharh433, Fremder, NirbhayPaliwal and me.
18o3, Ishaan_23, ALESTA, LaVa_ and coding_monk3 for testing and providing detailed feedback.

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

announcement, contest

agrim07
3 года назад
38

Комментарии (31)

Показать архивные | Написать комментарий?

shashank_lol

3 года назад, скрыть # |

← Rev. 2 →

→ Ответить

patilshashank

3 года назад, скрыть # |

very much excited, #IIITA

→ Ответить

HanumanBhakt

3 года назад, скрыть # |

Excited for this one!! agrim07 thanks for inviting.

→ Ответить

breakthe_rule

3 года назад, скрыть # |

Lets go !!

→ Ответить

rishika_g31

3 года назад, скрыть # |

← Rev. 2 →

Looking forward!!

→ Ответить

Mr.Roamer

3 года назад, скрыть # |

Congratulations! I Hope there will be many contests ahead organised by this team☺️

→ Ответить

devansh_ladha

3 года назад, скрыть # |

← Rev. 6 →

Looking forward to giving this contest!

→ Ответить

Pirated_Codes

3 года назад, скрыть # |

← Rev. 2 →

Is this rated or just?

→ Ответить

Fremder

3 года назад, скрыть # ^ |

No it's not rated

→ Ответить

0mBudsman

3 года назад, скрыть # |

Excited !!

→ Ответить

rainbowtable

3 года назад, скрыть # |

Excited !!

→ Ответить

ALESTA

3 года назад, скрыть # |

As a tester i think all of the problems are very nice and you will definitely learn something from all of them .

→ Ответить

aayush_chhabra

3 года назад, скрыть # ^ |

ALESTA sir orz

→ Ответить

slurpieboy

3 года назад, скрыть # |

pvtr aiyo

→ Ответить

audi_007

3 года назад, скрыть # |

any registration link?

→ Ответить

agrim07

3 года назад, скрыть # ^ |

← Rev. 2 →

You will be able to register 6 hours before the start of the contest using this link

→ Ответить

-Anuj_Kumar-

3 года назад, скрыть # |

Thanks for inviting us. Hopefull to get a good rank

→ Ответить

anjkk01

3 года назад, скрыть # |

so there are no prizes whatsoever ?

→ Ответить

rn_das_2004

3 года назад, скрыть # |

excited for this contest

→ Ответить

rainbowtable

3 года назад, скрыть # |

enjoyed solving problems A to E , thanks for the contest

→ Ответить

x123x123

3 года назад, скрыть # |

what was the approach for problem C ??

→ Ответить

rainbowtable

3 года назад, скрыть # ^ |

Spoiler

if(((n-2)%3==0&&n%2==0)||((n-2)%3!=0&&n%2!=0)){
      out.println("YES");
}
else{
      out.println("NO");
}

→ Ответить

anubhav100rao

3 года назад, скрыть # ^ |

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.

→ Ответить

Anupam_Ghosh

3 года назад, скрыть # ^ |

Numbers that are congruent to {1, 2, 3} mod 6

→ Ответить

grey_is_my_life_idk

3 года назад, скрыть # |

What would be the approach for the problem E i tried binary search + two pointers but not get it can someone tell

→ Ответить

VinCenzo___

3 года назад, скрыть # ^ |

Think of prefix or suffix array.

→ Ответить

grey_is_my_life_idk

3 года назад, скрыть # ^ |

← Rev. 2 →

I told E not D

→ Ответить

VinCenzo___

3 года назад, скрыть # ^ |

E was similar to this question https://www.codechef.com/problems/CHANDWEIGHT

→ Ответить

prakharmishra

3 года назад, скрыть # |

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

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

int main()
{
    int n,m,q,l,r,i;
    string s,str,news;
    cin>>n>>m;
    cin>>s>>str;
    cin>>q;
    for(i=0;i<q;i++)
    {
        cin>>l>>r;
        news=s.substr(0,l-1)+s.substr(r,n-r);
        bool temp= news.find(str)!=string::npos;
        if (temp)
        {
            cout<<"YES"<<endl;
        }
        else
        {
            cout<<"NO"<<endl;
        }
    }
    return 0;
}

→ Ответить

anubhav100rao

3 года назад, скрыть # ^ |

← Rev. 2 →

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).

→ Ответить