As an experiment the Educational Codeforces Round 33 will be rated for Div. 2. ×

By csacademy, 3 months ago, ,

Hello, Codeforces!

We are going to host a new contest at csacademy.com. Round #42 will take place on Wednesday, August/16/2017 15:00 (UTC). This round will be a Div. 2, affecting users with a rating below 1675.

#### Contest format:

• You will have to solve 5 tasks in 2 hours.
• There will be full feedback throughout the entire contest.
• Tasks will not have partial scoring, so you need to pass all test cases for a solution to count (ACM-ICPC-style).
• Tasks will have dynamic scores. According to the number of users that solve a problem the score will vary between 100 and 1000.
• Besides the score, each user will also get a penalty that is going to be used as a tie breaker.

• Computed using the following formula: the minute of the last accepted solution + the penalty for each solved task. The penalty for a solved task is equal to log2 (no_of_submissions) * 5.
• Solutions that don't compile or don't pass the example test cases are ignored.
• Once you solve a task you can still resubmit. All the following solutions will be ignored for both the score and the penalty.

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• +73
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 » 3 months ago, # |   0 I can't understand why ? `The penalty for a solved task is equal to log2 (no_of_submissions) * 5.`
•  » » 3 months ago, # ^ |   0 That way, easier tasks will have higher penalty.
•  » » » 3 months ago, # ^ |   +1 so...why ?for example why csacademy can't use no_of_submissions * 5 or no_of_submissions ?why log2 ?
•  » » » » 3 months ago, # ^ |   0 I think it is just a formula they came up to balance the scoring, also don't know why it would be different from what you suggested.
•  » » » » 3 months ago, # ^ |   +38 One reason is the following: imagine that you try to solve a problem and you miss some corner case. You're running out of time, get frustrated, start submitting a lot, hoping that you're going to pass. In the end you get AC, but now you are sooo low in the standings. You lose all your motivation, and even though you still have time to solve another problem, you quit. But with log2, the penalty isn't so bad, so you still have a chance.Another reason is that 2 WAs on the same problem is better than 2 WAs on different problems.
•  » » » » » 3 months ago, # ^ |   0 oh ok ! I got it :)
•  » » » » » 3 months ago, # ^ | ← Rev. 2 →   0 I think one should feel sad for someone with 2 WA's on different problems rather than single problem.And also one should be careful enough about not doing multiple wrong submissions on single problem because one of the main goals of the contests for many would be to prepare for ICPC,where this format carries a wrong impression and recklessness.I prefer something like after (say 10 submissions) . U take penalty as 10*5 and before that take it as 5*x .
 » 3 months ago, # |   +14 Just a reminder, the round starts it 6 hours.
 » 3 months ago, # |   0 Is there a way to solve the last problem when we are only provided some random matrix and are asked to find the submatrix with largest xor sum?
•  » » 3 months ago, # ^ |   0 I know O(n^3 log X) solution. Fix l, r, then calculate array, where c[i] = xor of all a[j][i] for l <= j <= r. Then solve 1D case in O(n log X) using trie.