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The top 50 contestants at least 18 years old will be invited to participate in the onsite finals.

I got it! Thank you very much.

I compute the Minkowski sum with D&C.209634929

Sorry to bother you, but can you explain how the code in the solution computes the Minkowski sum? I only know how to calculate the Minkowski sum of two convex hulls:)

here are codes 205875904 205853237

Hi MikeMirzayanov, I recieved a mail.

Attention!

Your solution 205875904 for the problem 1827B1 significantly coincides with solutions paula/205853237, propane/205875904. Such a coincidence is a clear rules violation. Note that unintentional leakage is also a violation. For example, do not use ideone.com with the default settings (public access to your code). If you have conclusive evidence that a coincidence has occurred due to the use of a common source published before the competition, write a comment to post about the round with all the details. More information can be found at http://mirror.codeforces.com/blog/entry/8790. Such violation of the rules may be the reason for blocking your account or other penalties. In case of repeated violations, your account may be blocked.

I don't have substantial evidence, and I haven't used an online IDE or multiple accounts to participate in the competition. I haven't even heard of this person from Croatia, so it's just a coincidence. Although the logic of these two programs is almost identical because the logic of the program is simple, there are still noticeable differences in the coding style.

my question is a little ambiguous. i want to know What rating can be considered as "high" enough to prepare a good contest.

The quality of problems depend on many things indeed. But what i want to know is that, to prepare a good contest(school contest etc.), what rating should i have at least? I think there may be some basic requirements for a good author.

What rating can be considered as "high"? black-red or red or orange?

The first step to find a not-1 position costs $$$\lfloor \frac{n}{2} \rfloor$$$ querys. The second step cost about $$$n(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots)$$$ querys. Why the total cost can be limited to $$$2n$$$?