Tommyr7's blog

By Tommyr7, history, 2 years ago, In English,

Hi, Codeforces!

I'm quite excited to invite you to Codeforces Round #439 (Div. 2) which takes place at 16:35 MSK on 6 October and lasts for two hours.

One of the five problems is created by quailty, and the others are created by me. This is our first round here.

The round couldn't have been realized without efforts of KAN and gritukan. Besides, I also want to say thanks to our testers: cyand1317, visitWorld, Nisiyama_Suzune, cdkrot and gainullin.ildar. Thanks for your help to the contest. Also, thanks to MikeMirzayanov with the fantastic Codeforces and Polygon platforms.

The contest will consist of five problems and it is rated for Div. 2 contestants. The same as before, Div. 1 contestants can take part out of competition.

The problems will feature... Well, let's wait and see for another time ;) As per Codeforces tradition, moderate length text and no spoilers about the actual plot.

I hope you can have fun during the contest. Good luck and have fun, wish you high rating!

The scoring distribution will be announced later.

See you tomorrow!

UPD1 : The scoring distribution is 500-1000-1500-2250-2500.

UPD2 : The contest is over. The editorial is available.

UPD3 : Congratulations to top five coders!

Division 2 :

  1. 1heart2plans (solved all problems!)
  2. fatice
  3. LuckyPants
  4. xza
  5. T404

Division 1 + Division 2 :

  1. 1heart2plans (solved all problems!)
  2. unused
  3. fatice
  4. nuip
  5. chemthan

Have fun! See you next time! (Let's wait and see...)

 
 
 
 
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2 years ago, # |
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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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    2 years ago, # ^ |
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    The E's data has some weak,because many people violence passed,but if Op 1 is 50000 and op 3 is 50000 many people will be TLE.

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      2 years ago, # ^ |
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      Well... It's my issue... In fact, the generator for div2. E isn't easy to code... I added 20 tests after randomly getting the ractangles... and I didn't consider this situation... Maybe I will do better next time... Thanks for pointing out!

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Tommyr7's round, must be very nice to perticipate.

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Another chinese round. Hope for short problem statements.

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Not only Chinese round,but Shanghai Round as well.It must be a fantastic one as the ones written by cyand1317!

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WOW Shanghai Round OvO

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zici!

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%%%%Q神

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Wow, another round with special starting time. I can't stop craving those thought-provoking problems.

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Very good! Hope the statement will be short.

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Nope

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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I did terribly today and cannot wait until tomorrow to try to restore my confidence.

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gl hf every one

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good luck!

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2 years ago, # |
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Hope this contest problems also have a short statements like previous. Wish everybody luck! YA!

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Its not.

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    2 years ago, # ^ |
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    If there written "Codeforces Round #***", it means that it is rated. But not always, for example.

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      2 years ago, # ^ |
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      Right. In addition, "Codeforces Round #***" can be unrated if there occurs any unexpected technical problem (though those contests are declared as rated, later they're declared as unrated in those cases).

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Wish that this contest announcement will be, as short as, statement)

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Yeah, This contest will held on my Dad's birthday.The bad thing that I will be doing this contest during my family сelebrating dad's birthday(((

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    2 years ago, # ^ |
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    I really think you should celebrate your dad's birthday in case contest clashes with the party time :)

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Finally came to a Chinese Round!Fantastic! Zici Tommyr7's first round. And ORZ Giant Red Q ! I think at Tommyr7's and quailty's level , it's not difficult at all to prepare for Div1!

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    2 years ago, # ^ |
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    Yes, and they (classified information) with (classified information).

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    Well, it's our first round. So I think it more proper to prepare for just one Div. 2 round.

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      2 years ago, # ^ |
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      proper uh... how to understand this words ? my English is not so good sorry ... proper means suit? maybe... QAQ

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Chinese Round,fighting!

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Congratulations....hope your 1st contest will be very interesting and expecting high rating :)

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So before every contest, I get an email announcement for the upcoming contest, and every time it states that the contest has an unusual start time.

Out of curiosity when is the usual start time?

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For quailty hit call(escape

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If you ask me zici not zici, I'm zici

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EMPTY

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As someone who is about to retire, I hope to have a happy ending.Come on!

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quailty, I hope the problem's quality is very good.

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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I wonder how Tommyr7 became rating ~2300 so fast (3-4 months) from rating 1800 (with many participation)...

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    2 years ago, # ^ |
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    In fact... I can't performance well in Div. 2 contests and I even don't know why...

    Always FST on div2AB may be one of the reasons...

    You can see that I fst problem B yesterday again... which makes me very upset...

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Let's Hack it!!!!

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Someone tell me how to solve C??

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    2 years ago, # ^ |
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    excuse me?

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    Let me bring my crystal ball... Oohhhh, I can't find it. Must have missed it on the field some day.

    Well, now you're gonna need to read the problem first.

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    2 years ago, # ^ |
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    You can do it greedily. Just sort the array and try all the numbers from smallest to largest.

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    2 years ago, # ^ |
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    You can use Brute force. Just O(n^2). n is less than 10000. So brute force can solve it.

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      But time limit is only 439 miliseconds. And implementation is pretty heavy so the program can be slow. How can you deal with it?

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        2 years ago, # ^ |
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        You can use fastIO, such as read() or fread(). :)

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      The brute force solution seems to be O(n^2), but is actually O(n*log(n)), so the solution can easily pass all tests.

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    I solved it with Convex Hull on linked list! :)

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Yay! I don't have to feed the cows this morning. Let's have fun in this contest :) !!!

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Really Pixar:)

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How to Solve problem C :D

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Today's round ID is 869, which is one of the substring of E869120. I'm lucky that I can participate.

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    The next contest which's ID will be in your username as substring will be 912. After some more contests :)

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the start round doesn't suit me, i've to run from college to home quickly and the contest has started 10 minutes :)

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It seems that chinese round is difficult as usual.

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    That's why people shouldn't vote up the announcement before they read problems.

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I am not being able to open any solution for hacks. [Locked solutions]

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Awesome Chinese names got me!! :-P Instead of "Koyomi", I wrote "Kayomi" and got pretests passed. So, I locked for hacking and now I realized that I made a mistake!! Worst feeling when you know your submission is going to fail in system testing and you cant do anything :-P

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    dont worry...have hope...it wont fail :p

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    The answer is always Karen so don't worry and you'll be accepted

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      Damn...that is great! Totally missed it.

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        I realized that when I tried to hack this code:

        int main() { printf("Karen\n"); }

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          2 years ago, # ^ |
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          It's not always :)

          Consider the case: 2 4 1 5 4

          No of pairs: 3 (1,1) (2,1) (2,2)

          Luck wasn't in my side, and contest got over before i could hack a submission.

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      Why is it always Karen? Sorry I'm kind of slow on these things.

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        2 years ago, # ^ |
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        It's easy to see, xor(X,Y) = xor(Y,X) right ? So if (X,Y) is a valid pair, (Y,X) would be as well.

        As you can see the answer comes in pairs, that's why it's always even.

        Don't forget that xor(X,X) = 0 and 0 is not allowed in the problem so you will never count this one.

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          Thanks, I get it now. I was suspecting it was the xor but not sure until now.

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    2 years ago, # ^ |
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    I guess the names are Japanese as the characters from the problems are a part of an anime known as the Monogatari Series.

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      Yes, they're Japanese names. Though Kanbaru didn't appear in problems XD

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        2 years ago, # ^ |
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        XD ? what does it mean ? it's a abbr?

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          Rotate it by 90° clockwise.

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            still can't understand it! sorry... Please forgive my ignorance

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502 Bad Gateway

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C is savage :/

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    Why ? As I am new to the world of CP. Why Chinese contests are not good ? :P

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since when codeforces started allowing Div2 participants in Div1 contests.....???

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How to solve A ? [ What is the hack case for A as I solved it using O(n * n)]

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    ans for A is always "Karen".

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    The answer will always be karen because the number of pairs will be even always. Suppose, for pair (1,2) the answer(xor of both numbers) exists in array, then obviously for (2,1) answer will also exist. Hence for x (x1,x2) number of pairs there are other x(x2,x1) pairs. Hence 2x pairs in total. so the answer will be karen.

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I am not able to see other solution for hacking after locking my solution. Also my score shows 0 in my room. Why is this happening? :/

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Problem descriptions were not that long. Thank you!

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It was a high quality round.

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what should be the output for

1 1048576 1048575

in A ? update- it was my mistake .

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    2 years ago, # ^ |
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    Karen
    
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      I tried to hack one solution which was giving Koyomi but ended up with unsuccessful hacking attempt.

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        bool vis[2000200]; will get Runtime Error, but int vis[2000200]; wiil not. I tried to hack the second twice but both failed! It is really strange…

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          int vis[2000200]; accepted ?

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    Karen?

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    Are there any test cases that don't give Karen?

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    It is always "Karen"

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Math, Math and more Math.

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I did hashing on fenwick tree for problem E. Even though I did double hashing, I'm still crossing my fingers.

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When you're < 30 seconds late from submitting a solution.......

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Is there any alternate approach to E, mine is ->

I maintain a 2D BIT, where in every cell I store hash of set of rectangles that affected it. For query, I just check if hash of given two cells is same. Query as well as updating all elements inside rectangle while adding or deleting it takes O(log(n)2) time.

My hashing technique though is very naive, first I assign unique numbers to rectangles, now I maintain few functions like sum of numbers, sum of squares, xor of numbers, xor of squares in every cell. Comparing two cell then will require to compare these 4 numbers. Using lesser and better functions should also suffice.

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A. Nice Hello World

B. Nice problem for Python

C. Nice problem for Python

D. Nice difficulty

E. Nice data structure

The difficulty distribution is 500-500-1500-3000-2000.

Please dynamic problem scoring.

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Mathematic round

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do the pretests of E contain big tst cases

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In problem B one should check for 0!/0! case answer should be 1 .

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How to solve C?

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    2 years ago, # ^ |
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    Fix the number of edges between a-vertices and b-vertices. Let there be x of them. Choose x out of a possible a-nodes and x out of b possible b-nodes and connect them in x! ways. Do same for a-c and b-c pairs. Multiply all 3 values to get final answer.

    Code

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      2 years ago, # ^ |
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      I don't get it :/. What do you mean by "Fix the number of edges between a-vertices and b-vertices" ? Can you explain with the example a={1,2} and b={1,2,3} ?

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        2 years ago, # ^ |
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        x can range between 0 and min(a, b). In this case, x can be 0, 1 or 2.
        If x = 0, number of ways = C(a, 0) * C(b, 0) * 0!
        If x = 0, number of ways = C(a, 1) * C(b, 1) * 1!
        If x = 0, number of ways = C(a, 2) * C(b, 2) * 2!
        C(n, r) is number of ways of choosing r objects from n objects.

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          2 years ago, # ^ |
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          C(a, 0) * C(b, 0) * 0!,
          C(a, 1) * C(b, 1) * 1!,
          C(a, 2) * C(b, 2) * 2!

          Sir, can you explain why you multiplied the expression with the factorial? 0!, 1!, 2!??? satyaki3794

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            2 years ago, # ^ |
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            After you've chosen x values each from the two sets of nodes, you need to make x pairs from them. For the first a-node, you have x possible b-nodes to choose from. Make a pair. Now, for the 2nd a-node, you have x-1 possible b-nodes to choose from, and so on. Total ways is x!.

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              Thank you so much ! :) Makes perfect sense to me now!

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    2 years ago, # ^ |
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    Calculate F(a, b) independently for 3 pairs and multiply.

    if a <  = b, F(a, b) = F(a - 1, b) + F(a - 1, b - 1) * b

    Because, for 1st element of first set, if nothing is assigned in 2nd set, there are F(a - 1, b) ways left, and if 1st is assigned in b ways, F(a - 1, b - 1) ways are left.

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      2 years ago, # ^ |
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      why not so:F(a-1,b-1)*a+F(a,b-1)

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        2 years ago, # ^ |
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        The relation is symmetric, hence I mentioned a <  = b before it.

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          2 years ago, # ^ |
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          But How are they independent of each other ???

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            2 years ago, # ^ |
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            When you have joined a point in set 1, with a point in set 2, joining this point to any point in set 3 doesn't contradict any given conditions, hence you can treat them as independent.

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swap(D,E)

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I did not enjoy the difficulty this round. The difficulty was A, A, D, E, E. One small mistake on the first two problems can drop your rank by 500.

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I've been tricked After I hacked someone's code,I saw this: #define int long long int

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    That's cruel.

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    Exactly this was something I faced in a previous round. From then, I always check ;)

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      Yeah,I will check it from now on.I were so innocent...

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    I use this, It save me from overflows :D

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      but long long int is time consuming.

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        Codeforces is not so strict with time limits :D until you are writing some sub-optimal solution in which case you need to edit this #define int long long

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wanna cry, come to the solution of C in the last ten minutes but fail to submit it.

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x[i]^y[i] Test 23 A

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Answer of A is Always Karen ????

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    Yah

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    2 years ago, # ^ |
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    Yes. Because, if A^B = C then B^C = A and A^C = B both are true. When you find a pair A and B such that A^B = C and C is one of the numbers of your list, then it's guaranteed that you will find another pair looks like A^C (Which will give B) or B^C (Which will give A). So for one successful XOR another successful XOR is guaranteed!

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    2 years ago, # ^ |
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    ordered pairs was the key word.

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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2 years ago, # |
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Oh my god, I made an obvious mistake and I passed the pretest of problem A..... Farewell my rating.

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2 years ago, # |
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using map in A gets TLE :(

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    2 years ago, # ^ |
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    Solution was O(1) answer is always Karen :D

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      2 years ago, # ^ |
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      Too lazy to think for problems which can be solved with brute force approach :p

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        2 years ago, # ^ |
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        Same happened with me in B i just looped from a+1 to b without noticing constraints lost 50 points :(

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          2 years ago, # ^ |
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          I did the same mistake too.

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    2 years ago, # ^ |
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    Same here

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    2 years ago, # ^ |
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    no it doesn't my submission

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      2 years ago, # ^ |
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      but why it's with me:31069010

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        2 years ago, # ^ |
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        It's giving TLE with [] operator but passing with find() but both are logarithmic in complexity .

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          2 years ago, # ^ |
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          Using [] operator adds keys to the map(if it doesnt exist), thus making it too large for further operations. So find() runs fine but [] gives TLE. Made the exact same mistake !

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            2 years ago, # ^ |
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            I had a small doubt, if [] is making keys then when it is called for the second time why doesn't it return true.

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              2 years ago, # ^ |
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              Map stores key value pairs. Sure it adds new keys, but maps them to false value.

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    2 years ago, # ^ |
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    Lol, never use a map just to map something to bool, you can always just use set in such case instead.

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2 years ago, # |
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O(N^2*6) DP getting TLE in C but O(N^2) Pascal's trangle passing :( too strict time limit

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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2 years ago, # |
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CodeForces runs 10^8 operations in 1 second? correct me

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2 years ago, # |
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In problem C, it is Impossible to solve the problem in Java using O(N2) Dp. However the same solution, in c++ passes like a wonder.

Remember that I used my mind to come up with an idea for a good problem, I type the Code in Java, TLE on pretest #7. The same code I type in C++, AC 740 MS. So, U should also prefer a better language, not only use ur brain to solve problems.

I have been facing disappointments like these from time to time.

I'm just going to stop using Java anymore, because I feel a lack of respect towards it in the programming Community.

Some Proof :

C++

Java

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    2 years ago, # ^ |
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    It's not impossible http://codeforces.com/contest/869/submission/31077410 , there are others, you can see it on http://codeforces.com/contest/869/status/C , just set the language to Java ... I don't think there is suck lack, look to the top ones in the rating and you gonna see that some of them use Java

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      2 years ago, # ^ |
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      It's not that it's impossible in Java, it's just that you need much lesser effort (less efficient code) using c++ and much more effort (highly optimized code ) using Java

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      2 years ago, # ^ |
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      Some problems are impossible to solve in java, especially the ones using TreeMap/TreeSet. The fact is C++ has the advantage and java is at a disadvantage.

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2 years ago, # |
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how 2 solve E in 8 minutes like fatice did ?

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    2 years ago, # ^ |
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    begin with E first read problem in 1 minute think for 2 minutes code BIT in 1 minute code others in 3 minutes submit code in 1 minute

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    2 years ago, # ^ |
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    In fact, fatice ranked 1st in CNOI2017.

    He's really cool.

    /And I was shocked when I knew he took part in the round...lol

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      2 years ago, # ^ |
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      That 's amazing !!! When I was in the game, I was shocked when I looked at the rankings. Now read his code, I think in this life I can hardly write such a simple and concise code. lol

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      2 years ago, # ^ |
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      31072413 This is fatice's solution for problem C. It is really short and clean and I think it's awesome. Any idea what dp[i][j] represents?

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Perfect contest time after a long week.

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This contest is not showing up in my profile. Is there some time to wait before it is updated?

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checking an element a1 in map like ma.find(a1)!=ma.end() passes all the testcases for A but checking like ma[a1]!=0 gets TLE. why??

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    2 years ago, # ^ |
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    When you use the second option, an element with key a1 is inserted on the map, so the runtime will be slower since you gonna have more values in your map

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      2 years ago, # ^ |
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      but time complexity is same for both the cases and also atmost 4*10^6 elements will be inserted.

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Running on tc22 for 15 mins...Can someone explain why

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why in problem A mp[] gives TL ,, okay it will insert more elements in the map but still the max size of the map should be 2000+2000+(2000*2000)=4004000 so the overall complexity should be n*n*log(4004000) and this should pass in one second!

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The constraints given in problem A are wrong thats why my solution did not pass The constraints for xi and yi given are 1 — 2*1000000 whereas actually the constraints are 1-4*1000000. Please correct me if I am wrong.

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    2 years ago, # ^ |
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    You can see my submissions for problem A. The only difference between two codes is size of array 'ha'

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    2 years ago, # ^ |
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    constrains is right but A^B may be more than 2*10^6

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when the ratings will be updated? can't wait to become purple

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Am I the only one who solved problem B by finding the repeating pattern of the last digits of b!/a! ?? This is my solution.

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Is there anyone else or just me who was not able to understand the test output of Problem C :P ?

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2 years ago, # |
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ratings?

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2 years ago, # |
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What kind of joke it is to provide editorial of only 2 problems? :/

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when the rates updated??

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what is difference between finding in map and direct checking? ex. map<int,int>mp; mp.find(x)!=mp.end() and mp[x]!=0

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    2 years ago, # ^ |
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    If you access a key using the indexing operator [] that is not currently a part of a map, then it automatically adds a key for you. This makes a really large map (up to 4000000) in this problem. You should use .find() or .at() for lookup.

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Is there a good deterministic solution to E?

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How to solve D?

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Why does my Python 2.7.12 solution get a runtime error when I use xrange (on TC 21) and pass when I use range with the same arguments. This is not supposed to happen. It works fine on my laptop but fails on Python and PyPy here. This cost me

xrange: http://codeforces.com/contest/869/submission/31091691

range: http://codeforces.com/contest/869/submission/31091663

Literally the only difference is the x[range]

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    2 years ago, # ^ |
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    If you use Python 2/PyPy 2, xrange doesn't take args from 2**31 and higher (>=2147483648) (on here that is)

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      2 years ago, # ^ |
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      Same for me.. i knew this already large arguments in xrange dont work. I checked it and 10**18 was working in my pc i got runtime error while submitting. Took time to realize.

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My solution is hacked again. It's a sad story.T_T

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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2 years ago, # |
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If 1e18*9 is not allowed by long long,then I could have hacked many solutions on B:D

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Auto comment: topic has been updated by Tommyr7 (previous revision, new revision, compare).

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2 years ago, # |
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this contest was too weeaboo for me, couldn't take it :(

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2 years ago, # |
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