Блог пользователя rk_98

Автор rk_98, история, 4 года назад, По-английски

"Somewhere something incredible is waiting to be known"

What good is an idea if it remains an idea in itself? Try. Experiment. Fail and then Try again.

Tecnoesis is back with a bigger, better and improved version of NIT Silchar’s annual Inter College Coding Contest under NITS HACKS 4.0 which will entice you into improving your coding and real-life problem-solving skills.

The coding contest will be held in two phases, prelims and finals (both online). The top 25 teams from the prelims will compete in the finals.

PRIZES WORTH 25K

Rules

  1. A team of 2-3 members is required from the same college to take part in the contest.
  2. Each participant must have a CodeForces account.
  3. A team must be created in Codeforces comprising of the members.
  4. Finally, you should register in the contest using the formed team.

Registration

Click on this link to register yourselves. You won't be considered as an official participant if you don't register here. You also have to register in Codeforces for taking part in the contest.

https://docs.google.com/forms/d/e/1FAIpQLSf5QG5V2s1IiVPj-ZWUJ8CzlMsSBO6zxvbvtEC5NAzuXKrJjQ/viewform

Contest Time

The contest will be on 25th April at 20:05(IST). The contest duration will be 2-2.5 hours.

Contest Link

Click on the link below to take part in the contest. Registration for the contest will start 6 hours before the contest. http://bit.ly/nits_hacks_prelims

I'd like to thank my friends amit_dwivedi, ani37, sprkgoyal, Abhi60, dark_ninja, Padmanabha, gibbous, R_K_C and a guy who doesn't want his name mentioned (funny guy, ik XD) for problem setting and testing.

See you on the leaderboard.

UPD: The form link has been updated

UPD2: There will be 8 problems and you will be given 2 hours and 15 mins to solve them.

UPD3: Those who can't access the contest using the above link can use the below link

https://codeforces.me/contestInvitation/5b1abd498d0a5044b7c5701103d178d7c6c8c2bf

UPD4: The list of participants who will be eligible for the finals will be announced shortly. The date of the finals hasn't been finalized yet (will be in this week though). Thank you all for participating, will post the editorial soon.

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4 года назад, # |
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The Google Form link doesn't seem to work.

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4 года назад, # |
  Проголосовать: нравится +60 Проголосовать: не нравится

Clashes with Codejam Round 1b

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4 года назад, # |
  Проголосовать: нравится +6 Проголосовать: не нравится

You should definitely consider postponing it as it clashes with CodeJam-1B.

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    4 года назад, # ^ |
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    Codejam starts at 21:30 IST. You can take part till then. Contest got postponed once already due to Codechef Lunchtime. We'd have shifted it but we're a bit tight on schedule. And last minute changes would be troublesome.

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4 года назад, # |
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Clashes with IPL xD

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4 года назад, # |
  Проголосовать: нравится +3 Проголосовать: не нравится

Please shift the competition one hour in advance otherwise it will be difficult to participate. Please.

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4 года назад, # |
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I am unable to register my team there is blank space

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4 года назад, # |
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rk_98, how are teams considered to advance to the Final Round ?

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    4 года назад, # ^ |
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    The team members need to be from the same College. We will send an email to the team leader of the teams who has filled the above google form. We will get the details of the team members. Then we will consider the top 25 official teams from the leaderboard where 5 seats will be reserved for NIT Silchar.

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4 года назад, # |
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We can not enter the contest link. I talked with a few participants. They can not enter too. Please look into it.

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4 года назад, # |
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Auto comment: topic has been updated by rk_98 (previous revision, new revision, compare).

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4 года назад, # |
  Проголосовать: нравится -8 Проголосовать: не нравится

Can anyone suggest me ideas on how to approach C

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    4 года назад, # ^ |
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    Same question!

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      4 года назад, # ^ |
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      First preprocess the given array to find the frequency of each element in range[0,100] for a every subarray.If frequency of atleast 1 element is greater than 1 then answer is 0 else find the xor pair product of the elements of that subarray.Since max distinct no are only 101 you can find it by brute force.You can find my sol here https://pastebin.com/ARgCqLV2

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    4 года назад, # ^ |
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    The problem requires just a simple observation that a[i]<=100, which means choosing a query of size(R-L+1) > 100 will definitely result in repetition of some elements (Pigeonhole Principle). If any of the two elements are equal, then their pairwise XOR will be 0 and hence, the whole product will become 0. Now, for the remaining queries whose size <= 100, just calculate XOR of all the possible pairs and calculate their product.

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4 года назад, # |
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U should have considered increasing the time limit for python in B Question My fellow teammates got TLE 3 times :( without his FastIO template. Anyways very nice contest , though didn't qualify but really appreciated the Questions. Would like to see another contest from Ur side soon.

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4 года назад, # |
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Can u please provide a small editorial for the problems. It would be very helpful for upsolving .