fixed

As far as I know, No it means that the scores of problems won't be changed during the contest ,this means if problem C is for 1500 points before the contest it can't be changed to 1250 or 1750 points during the contest even if number of submissions for that problem increase or are relatively less.

Next time we are going send T-shirts again =) This round is created mostly for participants of Petrozavodsk Training Camp.

Send me your tracking number.

div 1

Maybe your browser can't display it, but the word "frequency" is crossed out.

I think you mean very tough.

In div 2 the standing was very unusual, from the 27th to the 1500th all of them solve A and B only !!!

very easy A,B and very hard C,D,E. I think the difference in difficulty should be normally increasing not exponentially.

If nothing bad will happen it will be.

It's usual rating contest, except that it's prepared by us

I believe this round should have the same rules as other rounds. See the last Aimfund Round.

yes

you must be registered before start of the contest. usually register is open 24 hours before the contest begin.

You're welcome :)

Thanks for this

First observation: each 'b' symbol will be connected with every other symbol in the string.

So, find all symbols connected to all others and add them to first set ('b').

First symbol that is not 'b' will be 'a'. Add this symbol and all symbols connected to him to a second set ('a').

All other symbols not in first two sets will be in third set ('c').

Now validate sets 'a' and 'c' so each symbol in the set connected to each other (set 'b' is valid by its definition). If not — answer 'No', if yes — print the answer.

Complexity is O(N^2).

Same here. Fucked up on indecies (+/- 1)

Factor a[0]-1,a[0],a[0]+1, a[N-1]-1,a[N-1],a[N-1]+1. Then for each prime from that factors, do:

• in first pass count best solutions for left prefixes
• in the second (reversed) pass get best solutions for suffixes and join with the best solutions for prefixes.

Solution for prefix is basically a run of modifications such that gcd(run) = p and then run of deletions. Deletions from prefix solution join with deletions from suffix solution.

PS: there may be a mistake in idea, I didn't get AC.

PS: it's actually easier to make one pass of DP for each prime

My idea was that either the left element or the right one will be present in the final array, so try for every prime factor of those numbers (plus 1 and minus 1 too) the minimum cost to build an array with all numbers multiple of that prime.

I did it with prefix/suffix DP and some operations of addition in between. I don't know if my solution is correct though.

Note that you only need to consider prime divisors of a₁, (a₁ - 1), (a₁ + 1), a_n, (a_n - 1), (a_n + 1) as possible GCDs. Trying all the possibilities should be easy enough.

Stupid mistake, when factoring by trial division forgot to check if the rest of the number is > 1 and then it should be added to prime list too.

9th test is the first when a large prime appears.

I am not sure why it should work, but here is my approach, which passed pretests. First of all guess each friend once (order doesn't matter now). After that I always greedily guess the friend which gives me the biggest profit as a matter of probability that you win in that turn. I simply do that with a heap and while I have enough time.

I had the same problem: my A solution got hacked 9 min before contest's end, but I got the notification only when it was over :( So I spent 9 last minutes on D rather than finding bugs in A. Seems like manual page refresh is a must :(

Why So Many Ones?

3 1 1 1

is enough

Well I think ABC tasks was very good for div2: difficulty is raised from A to B and from B to C significantly (there were like 3000+, 2000+ and 1000+ solved participants).

But D and E was too hard (only 15 and 1 solved). Imo, D should be actual E and there needs to be a simplier task between C and D to be perfect problemset for Div2 :)

Please, say that acca is the right answer :)

Both are supposed to be accepted right?

I think I also had it bad. Forgot to print "Yes" in the special case ( n = 2 and m = 0 ) and got WA78. But I fixed it after the contest and it got accepted. I kind of hate myself right now.

I even asked a question, are 'a' and 'z' considered neighbours.

Well share the submission link/username so that i know where to look incase i don't get something from editorial.

Please see fofao_funk; maybe you can be friends.

There are two possible solutions: the first is using Chinese Theorem (calculating the answers modulo some numbers about 10⁶, then obtaining the answer in long arithmetics), or using the next trick: take and represent each coefficient as a_i = b_i·M + c_i, where c_i < M and b_i < M. Then the multiplication of polynomials with coefficients a_i is reduced to 3 or 4 multiplications of the polynomials with coefficients b_i and c_i. There coefficients are less than M, so the usual FFT in doubles is OK for them. For instanse, check Endagorion's solution.

nice graph!

Either use cin or cast to double when working with long double. It's known problem on Codeforces.

I think you should use statement like: cout << fixed << setprecision(10) << ans << endl;

Haha glad i wasn't the only one. The only solution i was able to come up with is n*sqrt(max(a_i)): go over all the possible divisors up to sqrt(max(a_i)) and for each number a_i, calculate if it's best to delete this number (we can do deletions infinitely so we don't need to care about contiguous segments) or to change it to be divisible by the potential divisor. At 30 billion such operations this is too slow for the given time limit though. We can improve this by taking only into account prime divisors, which brings it down to at most 3 billion operations, which is also too slow, but getting close at least.