By csacademy, 5 months ago, ,

Hello, Codeforces!

We are going to host a new contest at csacademy.com. Round #69 will take place on Wednesday, February/14/2018 15:05 (UTC). This round will be a Div. 2, affecting users with a rating below 1800.

Recently, Facebook has reintroduced the possibility of recurring events. If you choose "Interested" here, you will be notified before each round we organise from now on.

#### 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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• +53
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 » 5 months ago, # |   -13 Nice position to be in ;p
 » 5 months ago, # |   -20
•  » » 5 months ago, # ^ |   0 Again memes about Valentine's Day.
•  » » » 5 months ago, # ^ |   +70 Thanks, I've been wondering what it could be about.
 » 5 months ago, # |   0 How to solve E?
•  » » 5 months ago, # ^ |   0 I solved it by coding a dp solution and then trying to guess a recurrence relation from the results for smaller values of n. After finding the recurrence relation, then it was a simple matrix exponentiation. I could not solve the problem during the contest as the whole process took me a bit more than 2 hours. Can anyone who solved it during the contest, tell a bit more about how to approach these types of problems? The editorial did not help much here.