No. But you can see ranklist of past contests.

• 2017 : min_25 won with quite a difference. Endagorion, anta, Arterm, W4yneb0t, geniucos, Golovanov399, geniucos, Lewin,etc were in top 25.

• 2016 : tourist, fhlasek, rng_61 finished in top 10.

is it resolved now ?

Yes, we missed typing this detail. Updated statement.

Carmichael Numbers

Problem Statement is updated . Please check again . Apologies for the inconvenience caused

Every cell should be filled

Re-checked. It is correct. Please try again.

It's mentioned in the question : Find the probability that a randomly chosen line in R**2 cross both circles given that the chosen line crosses C1

unordered

If there is only 1 person in group , he is the winner then . Thus there are no matches . Since , it is a knockout tournament , there is a match/game between two individuals and winner of that match goes along and the other is knocked out of the tournament . In case some person is left out i.e he does not get a partner to play with because all other partner in that round are playing with someone else , he automatically goes to the next round . This process continues until one person is left . To know more , you can read up about knockout tournaments .

Yes , they are inclusive

F(g(N)).

I could only work on this contest for 10 hrs :(

I am not sure about correctness of judge answers for Q21 and Q24.

UPD. Q24 is ok.

Can you explain this line for Q21:
Let B be the set of integers made from corresponding elements in bag A such that its elements should be in strictly decreasing order

How is this example working?
A = {2,5,2} then B can be {6,2,1}

No such condition is given.

upto 6 decimal places.

Hint: For recurring tournament, Try to think in opposite direction. i.e Given number of people what will be number of matches.

It is correct.

Also should it be convex or not?

No. Polygon cannot be self-intersecting.

For Q13 think of substituting something in place of (a,b,c)

Hi. On your request, problem statement was updated to simpler and concise explanation yesterday. Check it :)

We can select 2 random indices i,j where i=j or i>j or i<j and swap them.

It is convergent :) It's also very similar to a math competition problem held a few years ago.

Нужно мерять весами с двумя чашками.

I will restate , let me know if this clears it up. You have to choose N containers. You can also choose capacity of each of these N containers. Example : for N=3, you can choose 3 containers with capacities as {1,2,3}. Or you could have chosen {5,5,500}.

You need to measure volume of a given liquid using the N containers you chose. You have to dispose a container after you pour liquid on it. Hence ,there's a range on volume of liquid you can measure for some set of N containers you choose. You optimally choose capacities of containers so as to maximize range of liquid-volumes (which varies from 1 to M) you can measure with your chosen containers.

In this case, following equation holds: a*M — b^N — c = 0, where a, b, c are non-zero constants & M,N are positive integers. Find 100a + 64b + 36c.

No , the slices may not necessarily have center as common point . Numbering can be done starting from any slice (0-indexed or 1-indexed , your wish) , then next number is the slice which is next in the clockwise order and so on . It is just that one guy eats the odd number slices and the other even numbered slices .

How is multiplicative order defined without stating what is the mod ? Do we need to assume things mod n ?

For snooker problem Q23, just solve (2, 3) + (A, B) = 20(u, v) + (x, y), where (x, y) is either (5, 6), (5, 14), (15, 6), (15, 14) and . The direction will be the direction of (A, B) wrt origin. This can be shown by reflecting the diagram. (very well known I guess)

Find all possible (A, B) with A² + B² ≤ d², and calculate the number of directions. I did this by using std::set.

Q28: only the length of cycles are useful. you can sort the length and consider it as state (there at most 5600+ states). code

Just run brute force of what has been written in the problem.

see my comment above :)

Total blue points within a square is equal to the number of unique slopes which is equal to number of unique y / x values. Every point x, y where gcd(x, y) is 1 will gives a unique slope. Euler Totient Function for each x (or y since it is symmetric) will give us number of irreducible fractions of the form y / x and that is the number of unique slopes that particular x or y contributes. Summing totient function from 1 to n / 2 will give us number of blue points in 1 / 8th of the square (why?). Multiplying this value by 8 gives us total blue points and subtracting from (n + 1)² - 1 gives us number of green points for a square of side n.

I'd like to give thanks too, but I want to make some notes about thinks which can be better next time. The list of problems is closed now, so I'll not list everything I could.

• The problem about 100a + 64b + 36c. First of all, it'd be great if you mentioned what you mean by "measuring". As I understood (only after I passed this task), you meant "one can measure x iff it's possible to put some masses we have onto two parts of a balance scale in such a way that one is heavier than another by exactly x". This definition is far from obvious.

• The problem Q15, dracarys. When you write "a rectangular board with dimensions 5x10^18" I imagine a (5·10¹⁸) × (5·10¹⁸) board. You could write "a rectangular board with dimensions 5 and 10^18" or "a rectangular board 5 × 10¹⁸". I don't want to get the correct definition just because you used "rectangular" instead of "square" (actually, I asked the correct statement to my friends, they told me).

• The problem Q16. Every time you talk about probabilities, you must specify the distribution. Otherwise your words make no sense. One can omit this if there is somewhat default distribution (for example, one, reading "chosen randomly from [0, 1)", can accept that the distribution is likely to be uniform), but it's not the case. You can choose a point uniformly from the first circle and then pass a line through it and orthogonal to the radius with this point, and it's one distribution; you can choose angle and distance to the center of the circle uniformly and independently, and this is another distribution. I don't know what you mean by "randomly, given that it passes the first circle". There is NO distribution over lines where the angle is chosen uniformly and the distance to the origin is chosen uniformly. There is no prior distribution for the meant statement.

• Mathtype. You need it. It could help in the 5 × 10¹⁸ problem, it could make the formula "a*m + b^N + c = 0" look "am + bⁿ + c = 0, etc.

This is certainly not everything I wanted to say, but it is about everything. If talk about the essence of problems, they were good, and some of them were great, so thank you for this.

zeliboba. Golovanov399.
Thanks for value feedback. We have taken them positively. Here is screenshot of problem-list raw from portal if you wish to add more to your comment or you can wait for a day, we will put up all problems in our new judge we are making to standardize it for upcoming contests.

And yes, we agree that we should have spent some time on problem statement itself rather than just spending our times to come up with good ideas for essence of problem setting. Apologies for same.

And djdolls is starred. I doubt, if he resides in India.

Yeah, same for me. I'm also Indian and my name is not starred.

Yeah, mine too, (because of my handle, I guess).

Fixed. Pardon for our bad stalking skills :P

I agree with all the above points.

I was stuck for a long long time in the circle problem. And very late I found out about the probability distribution mentioned here.

Similarly, I wished I could solve the harder problems and leave the one problem which I was not getting right. Frustrated enough, I had to leave the contest and when I woke up I realized what was wrong with my approach, but then I had only few minutes (around an hour) time left to solve remaining 7 hard problems.

It was supposed to be a joke -_-

Turns out that we have 2 extra T-shirts left. Congratulations for winning T-shirt :)