Hello everyone!

The second round of MemSQL Start[c]UP 3.0 will take place this Saturday, September 30, at 10:05am US/Pacific.

There will be an onsite round at MemSQL HQ and parallel online rounds that will be open for div1 and div2 participants. The online finalists (placed in top 500 in Round 1) will be given 100 MemSQL Start[c]UP T-shirts.

The onsite contestants will be participating from MemSQL HQ using their personal laptops. Breakfast will be served starting at 8:30 am, contest starts at 10:05 am and lunch will be served after the contest. The three winners of the onsite round will be awarded Amazon gift cards for $1000, $500 and $250. Please come early to set everything up.

Here is the list of people who agreed to participate: aandrew, al13n, Ali.Sh, architkarandikar, Belonogov, BIT-silence, chenmark, farmersrice, lewin, LiChenKoh, _M_, NgocHai, ntfoim, Paraboli4eskyHyperboloid, SaveVMK, scott_wu, sdya, SnapDragon, winger, Wo_oW, xiaowuc1, yum, yzyz.

If you haven't accepted the invitation yet or you think that we missed you, please message MikeMirzayanov or cerealguy.

We will update this post with more contest details later.

Good luck!

UPD: For onsite finalists: you are already registered for the corresponding round. You don't need to register anywhere else. For everybody else: choose the corresponding division round. All rounds will be rated for everybody, and we'll make special standings for the official participants of Round 2 (who placed in top 500 in Round 1).

UPD2: All online finalists will be registered in Div. 1 round automatically.

UPD3: Huge thanks to round testers: ashmelev, Errichto, cyand1317, gritukan!

UPD4: You will be able to see results of offsite finalists using this link.

Congratulations to winners!

Onsite finalists:

- scott_wu — $1000 Amazon gift card!
- Belonogov — $500 Amazon gift card!
- xiaowuc1 — $250 Amazon gift card!
- LiChenKoh
- sdya

Other finalists and div. 1:

Div. 2:

Round 1's announcement said "Only people who finished in the top 500 in Round 1 can participate." about Round 2. Now it says the online rounds are open for div1 and div2 participants.

So anyone can participate, or only people who finished in top 500 in Round 1?

These two statements don't seem exclusive to me. It doesn't say the round is open to

alldiv1 and div2 participants.Well, it doesn't say it explicitly, but that's what I concluded. Maybe first announcement's statement only applies to onsite round.

based on the last round, it is rated for all contestant, but the official contestant are the top 500. I think only the top 500 is eligible for t-shirt.

Who cares, they don't send them anyway

what about offline round?

if you dont know telavshi tovs and you can come with us for seeing tovli without any top 500 whatg do you think? p.s subscribe my YT channel "qurdi baxbaxa" and iqurde with me

I hope it is not open for everyone, it collides with Codechef.

Actually, it doesn't. Codechef Lunch time ends right before this round starts.

Who cares about Codechef when there is a Codeforces round :D !!!

Is this rated??

Yes. It is rated.

rated

Dude you made history.

Dude, it's an accedent :P

Maybe he got upvotes instead of downvotes because he changed "it" to "this" ?

You made another history :P

This is the first time i've seen "Is this rated"/"Is it rated" get upvotes

"Take Risks In your life"-Swami vivekanada

What? why is there a separation between div 1 and div 2? so how is the top 100 selected? Or does the t-shirt not really matter as Fcdkbear said

Sorry if it was unclear, only those who placed in top 500 in Round 1 can win a t-shirt. They will be registered in Div. 1 automatically.

Edit: I am registered in div1 indeed. Disregard content below.

Thanks

This is not true. I can't register in div 1 and I'm registered in div 2.

I also think that the idea of separating the round is stupid, without separating top 500 in round 1 from the rest. It means you should have 4 contests or 3 if you join 25 onsite with 475 online.

Otherwise I guess you will provide different problems for div 2 and how are you going to compare these participants in the official standings?

Of course you can't register now because you

willbe registered automatically. Within a few hours.It will be my first div 1 contest while being in div 2 :P Maybe it was worth to do all these stupid mistakes during last contests, to make this historical event happen ;)

You should also disable registration in div2 round. Otherwise it would be the first time I will participate in both div1 and div2 rounds at the same time. I wonder how will my rating be calculated in that case :)

We will unregister everybody from top-500 from div2 closer to the beginning of the round.

Will there be a place for everyone to plug in laptops? My laptop's battery is broken.

I called up MemSQL and they said there is. Consider this resolved. :)

We will have enough room for everyone to plug in their laptops.

Don't forget your chargers, this round will be longer than regular CF rounds!

Number of problems ?

I am registered in both divisions right now. So, which division will be considered for me? :/

Div 1, read KAN's comment above.

Yes, I read that comment, but in that comment he didn't make it clear what would happen if we're registered in both divisions so I was confused. He cleared it later though. Thanks anyway :)

Why are there div2 participants in div1 registration list?

because they placed top 500 in the first round of memsql

moejy0viiiiiv and tourist both register,Who will win?:)

Petr

Both of them are legends, but tourist, i mean come on...

23 : 05 , 22 : 05 are best time for contest Codeforces should take all contest is that time.

good timing for you is a bad timing for others. http://codeforces.com/blog/entry/54744?#comment-387598

Ho! sorry I thought about only myself.I live in Bangladesh.But I sleep in day and work at night.So its quite easy for me to attend contest At night.But At time < 21 i have different prayer time.So its better to me to attend contest at time >= 21.

Why registration is completed for div.1?

Edit: Ok looks like I'm automatically registered. It's pretty unobvious.

Scoring?

I registered in Div. 2 round but it seems that there is a bug in Codeforces that considered me as Div. 1 round participant.

Now I can't submit my solution to Div. 2 round.

You are in top-500 of Round 1, so you should compete in Round 2, which is a div1 contest.

18 minutes in and not a single Div.2 contestant solved more than A and B, this round is pretty bad tbh

Why second example in problem A div 2 return YES ??.

The fourth note for Div2A is so funny lmao.

Div2 C's test 7 is an immovable test case xD

really ? I passed the problem without any problem . i had a first try compile too which is an strange thing for me .

How to solve Div2-C ?

I solved it greedily

*Pretests i mean

Greedy.First construct an best solution without caring if you will require an extra pizza.After that try removing slices which minimise loss for 2 cases(one less pizza for type1 and type2).

Yeah, i did the same

Is E greedy?

How to solve Div2 E?

Can we solve it with maximum matching?

Will you please elaborate?

I tried a greedy approach to match each small query with nearest and largest but got a runtime error.

Segment trees can never make your life easy. :(

I didn't solve the problem, but I've started thinking in the direction of graphs.

When we choose to take some number

a_{i}, we need to find a pair for this numbera_{j}, such thatj>ianda_{j}>a_{i}. The weight of the edge (a_{i},a_{j}) isw_{ij}=a_{j}-a_{i}. When we choose an edge (a_{i},a_{j}) that excludes all the edges which come intoa_{j}.Okay, I am being able to relate my idea with this idea.

Can you tell what will be the best choice for an edge then such that it can be taken into count?

According to my idea, I sorted all elements of array

aand then start taking them one by one.In each iteration, my idea was to find the maximum value in the array from

pos+ 1 tonwhereposis the actual position of that element in array a.Your idea looks better =)

But still getting WA. :(

Here's what I did for Div 2 E (without segment trees): Have two sets(multisets) of unsold and sold items respectively. Every day, when you encounter a new price, you may choose to do one of the following: 1. sell the smallest element in unsold. 2. insert the smallest element in sold to the unsold set, and replace it with the new price in sold set for better profit, or 3. insert this new price in the unsold set(when both cases bring negative profits.) This has overall time complexity O(nlogn). submission

I liked the problemset. Too bad it was too hard for me.

The whole point of my participation in the competitions.

Can div1D/div2E be solved with treap + binary search? I came up with this, but there were too little time to code it.

Can you provide some resouce for learning treap datastructure? TIA.

You can check this https://e-maxx-eng.appspot.com/data_structures/treap.html

How to solve C, D, E?

C: My solution is to fix answer by binary search!

I tried binary search but all I got was a "WA on pretest 7". Can u please elaborate your approach? :)

Fixed the final answer by binary search, populated the DP table and verified my assumption at the end.

Code

Yeah, I also used this approach. I'm so mindblowned until now, :")

The DP approach is kinda straightforward, where the state is

`DP(at_which_level, total_time_up_to_this_level)`

. The answer is on`DP(0,0)`

, but the problem is it can be cyclic to call`DP(0,0)`

again at some points.Binary search the answer is to fix a double value of

`X`

, which we think it is the answer i.e.`DP(0,0) is X`

. So whenever the the DP call`DP(0,0)`

on the recurrence, replace it with`X`

.Calling the "real value" of

`DP(0,0)`

then compare it with`X`

. If`DP(0,0)`

is more than`X`

, then the answer must be >`X`

, vice versa.Nice!

Thanks a lot.

I accept that if we put

Xas the optimal answerR, then, indeed,dp_{0}_{0}will have the value ofRafter the recurrence. But how does one prove that by assigningX=Ysuch thatY>Rwon't makedp_{0}_{0}≥Y?First, assume the expression is simple, and doesn't contain

minoperations. So, basically it'll be likedp_{00}=a+bX.Of course this can be easily solved. But I want you to observe that in this special case,

dp00 <Y.Proof:dp00 -R=b(Y-R)0 ≤

b< 1|

dp00 -R| < |Y-R|,hence if

Y>R,dp00 <Y.Try to generalize this proof to include

minoperations. If you couldn't, I can elaborate more.Got it. Thanks for the help!

D: I passed pretest using segment tree (greedy approach).

What is the solution for "Buy high sell low"? This looks like a very well known problem with failing greedy... or not?

I did a simple greedy that passed but no proof

What is your solution?

well i think its easier to read the code : https://ideone.com/giXvIl

It is like that http://codeforces.com/problemset/problem/391/F3 but different a bit.

It is like searching for "max-cost matching" with edges of weights c_i-c_j (where i > j) in complete graph. However, I don't know how to do it effectively enough...

I passed pretests using this. I always choose the lowest available stock. Then buy the highest possible "valid" stock. To check valid, I maintain prefix sums in segment tree, then I execute a range max query on some suffix of the array to find the selling point.

UPD: Got AC. :D

Code

Does it passes :- 4 1 10 5 12 by your algo u should pair 1 and 12?

Yes, it passes. Answer is 16.

Yes, I pair 1 and 12 first. Then I buy 5 but after that, 10 is valid selling point as after buying 5, the prefix sum array is [1, 1, 2, 1].

sorry i dont understand how are u choosing the range for valid stock. I understand that it will be some suffix of array but how to find where does that suffix starts?

I maintain prefix sum in a segment tree. I find leftmost index such that no point to its right has prefix-sum 0, using binary search on segment tree. When I buy a stock, I do +1 range update and when I sell a stock I do -1 range update.

I have pretest passed with this greedy algorithm: https://ideone.com/HPBkr4

I iterate from the last object to the first, and push to the priorityqueue all the items with a flag if it was sold to a item with higher price (and greedily adding this difference). If this item appears at the top of the pq it means that if I bought it at Y and sold at X, and now I'm buying at Z with Z < Y < X, then I can change it to buying Z and selling at X, and being able to sell at Y when I encounter another item.

How to solve Div2 D ?

div2 B is div1 A !!!!!

My approach for C was to sort the array by a'i and b'i then to search for number of pizzas of type 1 with ternary search and greedy. But I could not debug for two hours! Please tell me that is wrong approach or I' gonna cry :D

I tried problem E with a greedy + segment tree based solution but was getting runtime error in pretest 3?

Can anybody tell me the mistake in algorithm or code?

Code — https://pastebin.com/MPQHs8eb

i too tried using seg tree + greedy ...but it TLE'd on test 3

Can you please attach your code?

Thank you all for participation, I hope you enjoyed the problems!

The system testing will be delayed by about one hour, because onsite finalists need to have some rest and talk before they will be ready for the results revealing. Hope for your understanding!

Can we at least look at other people solutions?

`onsite finalists need to have some rest and talk before they will be ready for the results revealing`

no we do not

systest already please

Agree — I also do not understand wtf?

How systest is preventing them from talking and resting?

most of the time it takes more than an hour anyway

How to solve Div2-C ? I Wrong answer on pretest 7.

My friend forgot to sort an array and he failed at test 7 too.

I have done that.

My code : http://codeforces.com/contest/867/submission/30885182

What's the idea for the solution of Div 2-B?

If a = 1, just print 1 1 1, else if we have only 2 and 1 denominations, answer for some n will be n / 2 + 1 (can be easily proved), so for some a we print 2 * (a — 1) 2 1 2

Oh, I missed the case of 1 :(.

you could have used 2 * a — 1 and not have to worry about the case of 1.

I actually tried to shoot the mosquito with a cannon," I used binary search and then coin change to fix the number of ways".

You can get any number of combinations by using the set of coins

D= {1, 2}.We can use the formula for Combinations With Repetition to calculate the number of combinations.

I've read the wikipedia. But

"You can get any number of combinations by using the set of coins D = {1, 2}. ", it's still not clear to me. Can you please explain a little more? :(Sure!

It's good that you've asked. I've read my comment again and I see that it's not good =)

Yesterday I have started with a Diophantine equation

x_{1}+ 2·x_{2}=kand came to 2·a- 1. This equation just stuck in my head and was thinking that there is a small mental step tomultichoose(n,k), but looks like I was wrong =)Do you understand how to get to 2·

a- 1?No, it's still not clear to me. I've just started combinatorics from a few days ago. I know the basics. But I am not able to relate those with this. I can catch up only if you explain. Would you please? Thank you.

Suppose we have to make k cents from a set of 1 cents and 2 cents. let k=1,then possible states are (1,0) (1 coin of 1 cent and 2 coin of 2 cent) i.e 1 way if k=2 then states are (2,0),(0,1)---2 ways if k=3 then states are (3,0),(1,1)---2 ways if k=4 then states are (4,0),(2,1),(0,2)--3 ways if k=5 then states are (5,0),(3,1),(1,2)-- 3 ways if k=6 then states are (6,0),(4,1),(2,2),(0,3) --4 ways so,there for k=n, there are (n/2)+1 ways. In the question,we are given ways and we have to find k. So,it will be 2*a-1 where a are given ways

Okay, it's clear to observe now. Thanks a lot :)

But did it came just from observation finding pattern or there's a theory that helped you which can be used for a different variation of this type of problem?

[user:codeforces.com/profile/Pixar]helped me to understand.

I drew it out and saw basically:

1 = 1 = 1 combo

2 = 1+1 or 2 = 2 combo

3 = 1+1+1 or 2+1 or 3 = 3 combo

4 = 1+1+1+1 or 2+2 or 2+1+1 or 4 = 4 combo

5 = 1+1+1+1+1 or 2+1+1+1 or 2+2+1 or 5 = 4 combo

6 = 1+1+1+1+1+1 or 2+1+1+1+1 or 2+2+1+1 or 2+2+2 or 6 = 5 combo

You see a pattern that for each combo amount, the number that makes the combo amount is floor(num/2)+2 (ex: floor(5/2) + 2 = 4). So you "reverse" this because they give you the combo amount, making it (combo-2)*2 (ex: (5-2)*2 = 6). This way, you can make the given combo amount made up of numbers 1, 2, and (a-2)*2.

The Div2C problem left the same impression as GCJ problems. I like it a lot =)

can u explain your solution please

Split all the people into 2 groups:

A,B.A= { people who prefer pizzaA}B= { people who prefer pizzaB}S(A) — the number of pizza slices needed for groupA.S(B) — the number of pizza slices needed for groupB.Now the best thing to do is to feed group

Awith pizzaAand to feed groupBwith pizza of typeB.We can do that almost always.Some pizzas of type

Awill be completely eaten by groupA, but there may be needed some more pizza slices to feed the groupA, but these slices do not form a complete pizza.Consider the remainders of pizza slices:

R(A) =S(A)modS— pizza slices needed for groupA, without forming a complete pizza.R(B) =S(B)modS— pizza slices needed for groupB, without forming a complete pizza.If

R(A) = 0 orR(B) = 0 orR(A) +R(B) >S, we can buy new pizzas of correct type and give them to corresponding groups. So in that case everyone will get the pizza slices from pizza that he wants.There is only one case when we need to feed

somepeople from groupAwith pizza of typeB(or the other way around). This is whenR(A) > 0 andR(B) > 0 andR(A) +R(B) ≤S. In this case we need to buy only one more pizza and we check two cases: 1. This pizza is of typeA, 2. This pizza is of typeB. In each of these cases we do greedily calculate max pleasure and choose the largest between the two."In each of these cases we do greedily calculate max pleasure and choose the largest between the two."

How do we do that? Pixar

If you have two people that like pizza A more than B. Suppose that the first one likes A = 15 and B = 14 and the second one likes A = 5 and B = 1.

If you have to chose one of them to feed using pizza B (because you don't have pizza A anymore), which one you would you chose? It's easy to see that the first one is better because you would lose less than if you chose the second (15-14 < 5-1).

So if you have to change someone take the one whose difference between A and B is minimum.

Short but funny statements :D

DIV-1C. Why this idea is wrong (fails on sample 3)?

Let's Tprev is one of the possible expected total times to complete n-1 levels.

In order to surely complete level N in Fn seconds, required time delta is: Fn + (Sn + Tprev) * (1 — Pn) / Pn (including reset progress cost)

In order to surely complete level N in Sn seconds, required time delta is: Fn * Pn + Sn * (1 — Pn) (there is no reset progress cost)

Submission 30885018 (GetAns() function)

Fn * Pn + Sn * (1 — Pn)

The above one doesn't calculate the expected time ending at some 'X' second . If fn is chosen you complete earlier than 'X' by Sn — Fn seconds. Later this propagates wrongly for Fn case.

Thanks, got it!

Do you see any way to tune this formula? Or idea "DP by completion time from level to level" is completely wrong?

D is too hard for undestand

Div 1 D?

"The position of the boxes within the basket matters"

Didn't see this sentence in G, and gave up.

Problem C remind me my first participation of ICPC Regional in Asia Kuala Lumpur site 9 year ago... There is a problem containing a step with same idea(UVa12164).

No Hacks this time !! :)

Tonight, I'll dream of a pizza containing 100000 slices :D

How to solve Div1 E, F, G? I tried to solve G using matrix powers, but my solution is too slow.

The answer is the constant term of series

.

Let

M=maxc_{i}. So the answer is.

Subtract the remainder from the numerator, and substitute x=0. You can get the constant term of the quotient.

You can find the remainder using polynomial division only. So the time complexity is .

So this problem is only about knowledge. Once, you know, just write and get it. Not much thinking.

Yes... I thought the hardest problem in the memsql cup should be unsolvable (like 457F). So I skipped this one during the contest.

However, I think this one is much easier than E and F for me :(

I got your solution but don't know how to compute remainder. can you plz explain or provide some source where I can learn it

My solution to F (which hopefully will sneak under TL and ML):

First, don't end the game when one person eats R raw eggs. Then if one person ever gets at least R+1 total raw eggs, that person is guaranteed to lose. Therefore we can only consider the cases when each person eats exactly R raw eggs and C cooked eggs, and we want to minimize |# scenarios in which A wins — # scenarios in which B wins|, out of a total of (R+C choose R)^2 total scenarios.

Now look at a sequence of A and B as path on an (R+C) by (R+C) grid. The number of scenarios in which A wins is the area under this path (grid segments will have length not 1 but (x choose r-1) ). We can use dp with a map to keep a list of which areas under the path are possible at each points in the grid, as well as how many paths attain that area.

Finally, we need to use a meet-in-the-middle to speed this up. For me, the best choice for "middle" in the case where R = 10 and C = 20 was the first 38 turns vs last 22 turns.

To speed this up, I also precalculated the optimal difference for each R,C and hardcoded this into my program.

Close to intended solution, however if you notice that the first R-1 rows and columns in the grid have zero area, you only need to meet-in-the-middle on the remaining 2*(C+1) moves.

I think my solution implicitly uses this (which gives the choice of split at 38/22). It is very tight at the time limit, though.

E: first, let's try all possibilities for which of the positions have "carry" when the subtraction happens. Now we know for each position the difference between the numbers. The sum of all those differences must be zero (this cuts down the number of possibilities from 2**13 to C(13, 6)).

Now let's look at the cycles of the permutation of the digits. Each cycle must have at least one zero digit (otherwise we can decrease all digits to get a smaller number). So we can assume that we have just one big cycle, since they can be joined using the zeros.

Now we need to build this big cycle. Let's build it starting with its zero, and then each next digit is given by the current digit plus the difference in the current position (which we already know from the first paragraph). So we can do dynamic programming where the state is the set of positions we have already placed, and which remembers the smallest number formed by digits in those positions. In order to find the next digit to be placed, we just sum the differences of the already placed positions.

Overall running time is around C(13,6)*14*2**14 which is 400 million.

When will system test start?

Div1 D is a problem which is already online. https://www.quora.com/Is-there-an-O-nlogn-algorithm-for-maximum-profit-from-buying-and-selling-stocks

came after 1 hours to see standings! still pending system testing!

Just start the system test already. I can't wait to see whether I become blue or not

You will get 170 ratings if all 3 of your solutions pass system test.

more than 170

Actually there is an extension available for chrome that predicts the exact rating change. It said he'll get 170 ratings before system testing started

So that's why he will get more than +170 as his rank must get better because of people getting FST, assuming that his all 3 solutions passed.

He'll now get +108 as he got FST on C

I should delete the extension and let you tell me :p

how you predict rating ?

Link

actually his rank can be better

I got wrong answer on C whyyyy :(( #still_not_expert

WA on test case 60 on C. Disappointing!

sorry but i was stupid, and problemsetters were not stupid.

When will we be able to submit?

you can submit now.

Never in my life did I imagine my name will be in contest announcement. Thanks cerealguy

First one to solve Problem C. Rank 1 for 5 minutes. Great feeling :D

can u pls explain your solution, basically when p1 % s + p2 % s <= s in your code?? tia

if (p1%s + p2%s <= s) then we can make 1 pizza of remaining slices from both pizzas. So, we'll try both the cases. First add remaining slices of first pizza to second pizza starting from the slice whose difference between first and second pizza is lowest. Second is add remaining slices of second pizza to first pizza starting from the slice whose difference between second and first pizza is lowest. Then the maximum of above two cases is the answer.

Hey, Can you explain How you solved problem C ?

When will the editorial be posted ?

It have posted. Link: click here

For Div2 B / Div1 A, can someone explain tourist's code? 30874326

Why do we need to build max(1, 2(n-1)) using 1 and 2 coins to achieve n possible combinations?

Any insights would be appreciated c:

Each number has a certain number of twos. These twos can be made either of the coin value 2 or two coins value 1. So each time you increase by two, you are increasing the result by one possibility. Therefore it's 2(n — 1).

Thanks man, that's really helpful.

Welp, this made me realized I probably don't belong in codeforces contests until I practice more.

Hey Please Give me the approach to solve Ordering Pizza :(

First try to allot everyone the slices they prefer (

`max(a[i], b[i]`

), ignoring the restriction of buying minimum number of pizzas.To fit into the restriction of buying minimum number of pizzas, you will have to give some participants slices they do not prefer. To minimize this 'regret', you will want to disappoint the people with least (

`a[i] — b[i]`

or`b[i] — a[i]`

) values. In case of a tie, you want to disappoint the one who eats less slices (`s[i]`

.Create two groups, people who prefer pizza of type 'A' over 'B' (

`a[i] > b[i]`

) and others (`a[i] <= b[i]`

). Let's call these groups group-1 and group-2. Try to disappoint participants from one group by storing pairs of (`a[i] — b[i], s[i]`

) (eg for group-1) for both groups.Sort these vectors of pairs and greedily disappoint participants of both groups independently. Your final answer is (

`max_happiness_ignoring_restriction - min_regret_of_the_two_possibilities`

)You can look at my solution for details 30902595

Thanks a lot sir :) Can you also give me some suggestions on how to start dp??

Problemset — dp

Topcoder tutorial

Lecture

http://codeforces.com/contest/867/submission/30902968 plzz help, why i m getting runtime error on testcase 6?

when will the editorial be out?

The editorial is available since one hour, but the contest announcement hasn't been updated yet, so here's the link

When can we expect Editorials? Other's solutions too hard for me to grasp :/

can anybody explain me Reyna's solution in D/Div 1 ? 30879284

has it editorial or analysis?

Can anyone explain div2 C?

Here is a list of comments explaining Div2C. Probably, something will click:

http://codeforces.com/blog/entry/54821?#comment-388332

http://codeforces.com/blog/entry/54821?#comment-388391

http://codeforces.com/blog/entry/54821?#comment-388462

http://codeforces.com/blog/entry/54821?#comment-388465

How to prove that binary search can be applied to Div2D? I solved this problem but I can't prove it.

please publish the editorial

It has been available since yesterday: link. The announcement isn't updated.

Thanks !

I just see that Problem Div 2B is so entusiactic. :D

Where are editorials of this round? I am looking for the editorial of 'Ordering Pizza'.

Editorial

Why don't tag this editorial to the announcement? It's really disturbing to find the hint for a difficult problem consuming a lot of time.

I really liked your article , your article is very petrified me in the learning process and provide additional knowledge to me, maybe I can learn more from you, I will wait for your next article article. Thanks a lot for providing this much informative stuff is perfectly excellent. color switch

So are there any updates about the T-shirts(tbh, I only care about this year's...). I hope there is still hope.

:'(

In problem F constraints were

R,C≤ 20,R+C≤ 30. At the frist sight it seems that onlyR+Cshould matter, so one may think that if it is solvable at all then it is solvable also without constraintR,C≤ 20, but after some time it can be seen why that can matter. Solution provided in editorial in fact works inO(C·2^{C}), so it is fast for small values of C. My solution however works really fast if C is larger! For example, for R=20, C=10 it works ~2.4s and for R=10 and C=20 (if I change preprocessing from 25 to 15 steps) it works in 0.01s! So it seems that in fact merge of mine and original solution would lead to solution handling all possible cases with R+C<=30 and probably will handle even R+C as large as 35. However I would be reluctant to give any precise bound on running time of my algorithm, it uses some heuristics :P http://codeforces.com/contest/866/submission/31928336EDIT: Uhh, I dug a little bit and it turned out that in fact running time of my algorithm is not increasing with R, for (R,C)=(1,20) my code runs significantly longer than on (10,20) and my code can't handle (R,C)=(1,29) which I claimed earlier it will handle in no time :(.