wxhtxdy's blog

By wxhtxdy, 5 years ago, In English,

Hello everyone! Codeforces Round #268 is coming soon! We invite you to participate in this round. It will be held on September 20th at 17:00 MSK.

Problems have been prepared by me. Thanks xyz111 and dhh1995 for giving me the idea of some problems. Thanks vfleaking, foreseeable, xiaodao, ruchiose and xllend3 for testing.

I also want to thank Gerald for helping to prepare this round, Delinur for translating the statements, and also MikeMirzayanov for Codeforces and Polygon.

It is my first round on Codeforces. Hope you will enjoy this round.

You'll help a boy named Little X in this round. Good luck and have fun :)

UPD

Round has finished. Thanks for participating.

Congratulations to the winners.

Div. 1

  1. PavelKunyavskiy
  2. Kostroma
  3. HellKitsune
  4. Baz93
  5. DemiGuo

Div. 2

  1. manman
  2. topcodecheforces
  3. mhy12345
  4. sk_aswd
  5. z.last

Congratulations to ecnerwal, the only participant to solve the problem D!

Unfortunately, no one has solved the problem E in both divisions.

Editorial for the round will be added tomorrow.

UPD

Editorial is here.

 
 
 
 
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5 years ago, # |
Rev. 2   Vote: I like it +10 Vote: I do not like it

Looking at the level of the setters, I would say a Div 1 round would have been possible :( Edit: I thought today's round was their round, don't mind :P

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another china round! good luck everyone :)

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5 years ago, # |
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Why "Little X", but not simply "x"? :)

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5 years ago, # |
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It's the first time to see contest announcement before the start of previous contest :D

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5 years ago, # |
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Why are there so many numbers in the handles of the problem setters ? =D

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    5 years ago, # ^ |
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    No, many handles for testing not problem setting

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    5 years ago, # ^ |
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    Because 233 is a special number in Chinese... It means "LOL"(Laugh Out Loud) in English which shows a man bursts into laughter. And we use repeating "3" many times strengthen the funny meaning. 233 is a photo id in a famous forum of China. here is the photo.

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      5 years ago, # ^ |
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      Link is broken.

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        5 years ago, # ^ |
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        poor link...

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            5 years ago, # ^ |
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            what is APEX KEK means

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              5 years ago, # ^ |
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              Oh right, I should've included a

              see more on Know Your Meme

              and apex = top (it's made using rarely used words, like nay = no). I just found the "poor link" reaction seriously funny :D

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                5 years ago, # ^ |
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                Oh...I thought LOL is the short from Lots Of Laughter... After clicking your URL, I have just known LOL is short from Laugh Out Loud......

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                  5 years ago, # ^ |
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                  It's hard to say what people want it to be (and it doesn't really matter), it could even be short from "Lots of LOL" :D

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          5 years ago, # ^ |
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          fix the photos.

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5 years ago, # |
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Yessssss

Chinese round -> more math

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    5 years ago, # ^ |
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    There is actually going HackerRank contest "Ad Infinitum — Math Programming Contest September'14 ". It lasts 2 days :). Give it a try.

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    5 years ago, # ^ |
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    First of all, before this commented is downvoted, I would like to point out that codeforces is for developing your algorithmic knowledge. Depending on your definition of "algorithmic", math could be considered an algorithm; and this is a perfectly valid opinion. However, what is the point of competing in coding contests if all you want is math all the time. Every time you post, it's about math; nothing else. Coding contests are meant to introduce you to varied ways of thinking, not limited to math.

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      5 years ago, # ^ |
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      I'm agree with you and I don't know why many people downvoted you. Someone like math, someone not. But everyone love coding. I think we shouldn't care about kinds of problems.

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5 years ago, # |
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Oh, god, I think I can't participate any more Chinese round, because it is 6:00 am here.

Btw, I guess more testers you have then more difficult this round will be.. let's see if it is correct.

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    5 years ago, # ^ |
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    early sleep and early wake~ After brushing teeth and washing face, you can participate the Chinese round relaxedly

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    5 years ago, # ^ |
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    Have fun getting used to the U.S :)

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5 years ago, # |
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nice

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    5 years ago, # ^ |
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    newSolar BigGod is here! You are a Yellow! But you say you are only div2.....

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      5 years ago, # ^ |
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      When did I say that? I didn't use this Id for more than half a year.

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So, its going to be a typical chinese round . good luck everyone :)

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My last contest in summer! I'm going to school on Tuesday! Hope a good contest! Good Luck!

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I have registered for the contest, but due to time constraints (I have to participate in a qualifying round of some other contest), I will not be able to participate for the entire 2 hours. Can I appear as a virtual participant (If I participate for the first hour of the contest)?

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    5 years ago, # ^ |
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    No, if you submit during contest, it will be counted in your actual participation and rating changes will also apply. So it is better to take the entire contest as a virtual contest at some suitable later time.

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I will try to help Little X as much as possible but can Little X tell us score distribution ?

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Attention please,it's a Chinese round!There is reason to believe that it's impossible for the konjacs(its Chinese name ‘juruo’ means the people with low level like me)to improve rating.Good luck to every super cattle(you may be able to understand it)! TAT

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Many GranMaster & Master for prepare this round. My premonition told to me that "be careful with tricky test"

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Ehm, score distribution?

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It seems that I will participate in the next contest (Div2) Officially :D

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5 years ago, # |
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500 — 1000 — 2000 — 3000 — 3000 :(

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B(div1) 4 7 9 2 4 5 7 If someone need.

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    5 years ago, # ^ |
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    The answer is "YES 1 1 1 1", right?

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    5 years ago, # ^ |
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    would it be correct if ans is YES 1 1 1 1 ?

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      5 years ago, # ^ |
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      Yes. ) But any people have answer : "NO".

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    5 years ago, # ^ |
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    Or
    3 5 8
    2 3 5

    (sometimes they remove numbers from first set without checking if first set stay valid)
    Edit: the answer is "NO", but some were giving for output "YES 0 1 1"

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what was the hacking case in div-1 B...my solution got hacked :(

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How to solve C ??

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    5 years ago, # ^ |
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    I took the following greedy approach:

    for n <= 8, run recursion
    else
    1 * 2 = 2
    3 * 4 = 12
    2 * 12 = 24
    5 - 6 = -1
    7 - 8 = -1
    -1 - -1 = 0
    for i > 8 && i <= n ( i * 0 = 0 )
    24 + 0 = 24
    
    
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    5 years ago, # ^ |
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    I used the following :

    for n<4 print NO
    for n=4/n=5 find solution by hand
    for n>=6
    n - (n-1) = 1
    1 - 1 = 0
    for all i = 5 to i = n-2
    i * 0 = 0
    
    you are left with {0,2,3,4} - just do 2*3*4+0
    
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How do we solve Div2 Problem C?

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    There are several ways to solve it. You can see that with N < 4, there is no answer. With N = 4, you can easily take (1 + 2 + 3) * 4 = 24. N = 5, 2 * 4 = 8, 3 * 5 = 15, 8 + 15 + 1 = 24. From N = 6, 4 * 6 = 24, 3 — 5 = -2, -2 + 2 = 0, 0 * 1 = 0, then for i from 7 to N, 0 * i = 0, and finally, 24 + 0 = 24.

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    5 years ago, # ^ |
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    I took special cases for 4,5,6 and

    and for n>=6

    used pre-created sequence for 5 and 6 based on even or odd n,

    made every element n and n-1 as (n)-(n-1)=1

    and multiplied all the ones in the end.

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Can anyone share his/her approach in Div2 C problem ?

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    5 years ago, # ^ |
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    If n is even and >= 4: 1 * 2 * 3 * 4 = 24 n-(n-1) = 1, (n-2) — (n-3) = 1 and so on. So just multiply the 24 with the ones.

    If n is odd and >= 4: 5 * 4 + 2 + 3 — 1 = 24 n-(n-1) = 1, (n-2)-(n-3) = 1 and so on. Same.

    There is no answer if n < 4.

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    5 years ago, # ^ |
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    for n >= 6 erase all numbers except 2,3,4 using: (5 — 6 + 1) * z
    for n = 4 and n = 5 — hardcoded answer

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      5 years ago, # ^ |
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      Yes got the idea now couldn't think about it in contest :(

      Thanks :)

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How to solve Problem D of Div. 2 ?

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    5 years ago, # ^ |
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    My solution was, to just use pair(value,pos) in a set, and modify the set comparing function to consider only the first value. Then just keep checking if each element has its match in either A or B. hope it passes final tests :)

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    5 years ago, # ^ |
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    Your graph is inspirational !!!

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    It's a bipartite matching problem ;)

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    5 years ago, # ^ |
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    Hm, i had two maps, one <int,bool> where i stored if there exists some number, and one where do i keep position of that number in initial array. Then, for every number, check map1[a-array[i]], map1[b-array[i]], if there is both do sth, if there is only one put it into the set, if there is none return -1. See my code for details UPD: It failed on systest, so this approach may be wrong

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    5 years ago, # ^ |
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    Pick a number in the list, say x. Then we are interested in whether a-x and b-x are also in the list. If

    1) Neither are in the list, then partitioning is not possible.

    2) If only a-x is in the list, then both x and a-x go into set A

    3) If only b-x is in the list, then both x and b-x go into set B

    4) If both are in the list, then this is inconclusive. We need to further check b-(a-x) and a-(b-x) according to the same rules (for example, if only b-(a-x) is in the list then all of a, a-x, b-x, and b-(a-x) go into set B).

    The implementation of this is pretty straightforward, though in contest I mishandled the a=b case (which requires special handling as a-x=b-x=b-(a-x)=a-(b-x)=...). See 7882478.

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      Why do we need to check further ??? Can you give test ??

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      i got it on contest , but i missed when cout no

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    hopcroft-karp bipartite matching

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      5 years ago, # ^ |
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      Hii, palmerstone, I like your approach using Hopcroft karp, but I am having trouble understanding the case when both a-x and b-x (along with x) are present in given array. Here's a part of your code

      if (!c && !d){
            flag = false;
            break;
      }
      /*...*/
      if(!flag) puts("NO");
      

      how did u came to such conclusion that when both a-x and b-x are not present, there can't be any possible matching?

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        5 years ago, # ^ |
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        What do you mean? x is an element of the given set. If both a-x and b-x are not present then it is certainly not possible to include x in any of the sets, is it? Hence not possible.

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          Oh yes. Thanks, I get it now... By the way can you please tell what does array L and R store in your implementation of hopcroft-karp ??

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    My solution for div2 D/div1 B:

    Assume that b>a because we can switch them anyway and b=a is trivial case. Let x be the smallest value which hasn't been assigned yet to either of the sets. If there exist a value b-x, then we have to pair values x and b-x into set B because value b-x can't be paired into set A because a<b and therefore a-(b-x)<x and there exist no values smaller than x. So we know that if there exist value b-x, we have to pair it with x. If there doesn't exist value b-x, we have to pair x with a-x into set A. If we can't do that, the answer is NO. This approach can be implemented just by iterating over elements which haven't been assigned to either set in ascending order. 7874034

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What's the solution for problem C? Is it some kind of greedy/constructive algorithm based on number theory, or is it dynamic programming? The numbers are from 1 to N, so I assumed it was a number theory problem but couldn't come up with a solution.

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    5 years ago, # ^ |
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    For numbers up to 3 answer is 'NO'.

    Solution for 4:
    1 * 2 = 2
    2 * 3 = 6
    6 * 4 = 24
    
    Solution for 5:
    5 - 3 = 2
    2 + 1 = 3
    2 * 3 = 6
    6 * 4 = 24
    
    Other solutions:
    N - (N - 1) = 1
    (N - 2) - (N - 3) = 1
    ...
    1 * 1 = 1
    ...
    <- solution for 4 if N is even or 5 if N is odd
    
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    5 years ago, # ^ |
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    Just notice this if(n<4) no soln if(n==4) 1*2*3*4=24 if(n==5) 4*5+3+2-1=24 and for all other numbers, if n is even then you do the 4 thingy and keep doing i+1-i=1 for all the numbers, and at last do 24*1=24 as many times is you need same thing for odd n, just you would use 5 thing

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    5 years ago, # ^ |
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    You can get the 24 by using 2, 3, 4 or 1, 2, 3, 4. For the neighbor number, say, i and i + 1, we can get 1 by using - operator, and 1 can be eliminated by * with other number. Then I think you get the answer:-)

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    5 years ago, # ^ |
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    if n is less than 4, obviously no solution. Then 2 cases, if n is even, we can reduce the sequence to 1,2,3,4 (after which we just multiply each element one at a time). By subtracting n and n — 1, multiplying 1 and 1, then n — 2 and n — 3 and so on.. if n is odd, you can similarly bring it to 1,2,3,4,5, we can reduce to 2,3,4 by doing the following: 1 + 5 = 6, 6 — 3 = 3. After this we just multiply each one again. Hope its clear :)

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    5 years ago, # ^ |
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    A much easier solution is to get 1 by doing n-(n-1) and then 1-1=0. Now do i*0=0 for all i from 5 to n-2. And then do 2*3*4 = 24. For n<=6 pre calculate the ans. I hope you get it.

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This contest reminds me of the Codeforces Round #146 (Div. 1) which was prepared by WJMZBMR... Codeforces Round #146 (Div. 1)

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To those who have problems with div. 2 C here is my approach:

First print out the answer for cases n <= 5. Then for n > 6 you realize that you can make 24 by multiplying 6 and 4. Then you are left with the burden of creating n — 2 more operations, thus, you subtract adjacent numbers so you get a long chain of 1's, be careful that you cannot use 6 and 4 again. Then you simply keep alternating between: 1 — 1 = 0 1 — 0 = 1

until you are left with 1 or 0 then you can just do either 24 + 0 = 24 or 24 * 1 = 24

I did this approach during this contest and it seems to work. I was a bit dumb for the first hour because I forgot to print out "YES" at the beginning >.<

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    5 years ago, # ^ |
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    Haha I forgot to print "YES too :P

    My solution was as follows: the n = 4 is easy, because we can just multiply 1 * 2 * 3 * 4 = 24. The n = 5 is a tiny bit trickier, but we have 4 * 5 = 20, then 20 + 2 + 3 — 1 = 24. Now, if n > 5, we can take the two largest numbers, n and n-1, and subtract n-1 from n to get 1. Now, our list is 1, 2, 3, ..., n-3, n-2, 1. If we multiply the two 1s, we are left with 1, 2, 3, ..., n-3, n-2, which is the list for n-2, a smaller case! This means that n = 6 will work, since we can reduce it to the n = 4 case, n = 7 will work, since we can reduce it to the n = 5 case, and so on.

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    5 years ago, # ^ |
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    my answer is 1*2*3*4*(6-5)*(8-7)*...*(n-(n-1))=24 for n is even 5*4+3+2-1*(7-6)*(9-8)*...*(n-(n-1))=24 for n is odd

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    Our Little X is in upper case. Too big!!

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Fuck math with no programming!

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    Got butthurt? Go away.

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      Why so serious?

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        Because he was rude, used swear word and continues math-hate here which I'm completely fed up with.

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    5 years ago, # ^ |
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    I am not sure, but only the last problem seems to be mathematical one.

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      First one was also mathematical more than programming I'd say. But, a major chunk of programming is maths and logic, so it is only expected.

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        5 years ago, # ^ |
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        Of course all the problems are about math here(discrete math mostly), but here, I believe people mean complicated math by saying "fu**ing math again", but noticing that 6*4=24, a*0=0 and a*1=a.

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That was one fast system test!

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Why so weak pretests???
I guess lots of Bs will fail.
Also there was too many WA2 in problem A... For constructive problems, first sample should be more difficult. :|

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    5 years ago, # ^ |
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    IMHO, pretests complexity is completely up to the Problem Setter.

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I spent all 2 hours on problem D but still cannot finish coding it...

Calculating "lexicographically smallest" one is very confusing, and I couldn't make solution without very complex iteration. Doesn't there exist any simpler solution?

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    5 years ago, # ^ |
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    Indeed. If it weren't for this "lexicographically smallest" than it would be doable with centroids and that kind of stuff, but it is a significant obstacle :/.

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      5 years ago, # ^ |
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      Yes. Thinking about centroid is fun, but I think main part of this problem is more complex part, because only 1 people solved it despite thinking about centroid is not so hard...

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5 years ago, # |
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Like 90% of 469D - Two Sets has failed tests. Lol!

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Wow, system test over in 10 minutes this time!

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thanks for the fast system test~ I think I could back to blue.

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5 years ago, # |
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problem A:

for n=1,2,3: no

for n=4 : 1*2*3*4

for n=5 : (3*1+5-2)*4

for n>=6 : 2*3*4+(6-5-1)*7*8*...*n

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for div 2 B And D

what is the bug in my solutions ??

Code_B

Code_D

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    5 years ago, # ^ |
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    For D:

    1 2 2
    1
    
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    5 years ago, # ^ |
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    about D:

    |cpp| if (n % 2) { puts("NO"); return 0; } <||

    it will fall on the case

    3 2 6 1 2 4

    because 2-1=1.

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    5 years ago, # ^ |
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    about B: Your check for intersection doesn't look right. See if you can find your error here: if ((x >= a[j].first && x <= a[j].second) || (y >= a[j].first && y <= a[j].second))

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      5 years ago, # ^ |
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      can you please explain why this is wrong?

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      4 years ago, # ^ |
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      Same question, please explain what's wrong here?

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        4 years ago, # ^ |
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        Seriously... 17 months ago

        Anyway, look at the conditions inside the if statement. What is it doing? Is it checking that the two segments overlap, or if one segment is completely within the other segment? What we want is for the two segments to intersect, one segment does not necessarily have to completely contain the other.

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5 years ago, # |
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Isn't this code wrong 7871552 ??

Test: 2 1 0 0

1 5

2 4

7 7

UPD:The inputis illegal sorry.

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    5 years ago, # ^ |
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    I think it is wrong. The answer should be 0, but it is giving 1. oh! Inputs are not valid.

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5 years ago, # |
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Ehh..

if (n % 4 == 1) { ... cout<<"3+1 = 4\n"; ... }

There wasn't any pretest for n % 4 == 1? It's really disappointing :\

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5 years ago, # |
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My solution for div-1 B using dinic's algorithm for maxflow TLEd :(

I thought it would run in O(E * sqrt(V)) time as all edges are of unit length. Can anyone please let me know why dinic would be suboptimal? Isn't it the same as Hopcroft-Karp for unit graph?

My Submission

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5 years ago, # |
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By the way, somebody solved problem E in Division 1...

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5 years ago, # |
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printf("24 * 1 = 1\n");

and Failed A...

What a sad story :((

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5 years ago, # |
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Great test for problem D Div 2. Lots of Failed System Test

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5 years ago, # |
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Can you give me an example on which my code doesn't work ? ( Sorry for my poor English) http://codeforces.com/contest/468/submission/7883022

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    5 years ago, # ^ |
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    Why would everybody use tonns of #define? It puts my bum on fire.

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      5 years ago, # ^ |
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      Because it's easier to write MP than make_pair or something else... You write faster

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        5 years ago, # ^ |
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        It makes code harder to read in my opinion. Especially when I see that 50% of code are definitions for "for" and other shit.

        If you use snippets you can write fast enough. As for me is more important to think fast, not write.

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          5 years ago, # ^ |
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          FOR0(x[i],8) vs. for(x[i]=0;x[i]<8;x[i]++) not only saves keystrokes, but also reduces probability of typing errors.

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            5 years ago, # ^ |
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            Ok-ok. I understand, I am bad programmer if I don't use #define.

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    5 years ago, # ^ |
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    2 8 10
    4 6
    
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      5 years ago, # ^ |
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      It's well... output : YES 1 1

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        5 years ago, # ^ |
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        Sorry. Try this.

        3 12 14
        5 7 9
        
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          5 years ago, # ^ |
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          Thank you very much for this example :). Now, I understood why my code doesn't work

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5 years ago, # |
Rev. 2   Vote: I like it +1 Vote: I do not like it

I can not understand the reason why many of the chinese authors make hard and mathy problem set. In my opinion, contest's easy problem should be easy. Otherwise it is not motivating enough for people to participate.

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5 years ago, # |
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I've entered for the next div2 contest.The konjac(its Chinese name ‘juruo’ means the people with low level like me) has no human rights.

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5 years ago, # |
Rev. 5   Vote: I like it +5 Vote: I do not like it

winter is coming

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5 years ago, # |
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when will the ratings be updated?

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5 years ago, # |
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About Div1 B. Does anybody have a case with an odd n and positive answer? Or could explain why this is possible? I assumed this wasn't possible during the contest, but apparently it is. Maybe I misunderstood the problem.

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5 years ago, # |
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can someone explain why this solution is incorrect? http://codeforces.com/contest/469/submission/7867443

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    5 years ago, # ^ |
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    you should use resize not reserve

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      5 years ago, # ^ |
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      reserve only reserves space but the size of your vector stays the same. it is only useful for example you read numbers and push them back of a vector. the vector resizes itself when you push back sometimes. if u know how many numbers u are going to push back then u reserve some memory and when u push a number back of the vector it doesnt allocate new memory

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5 years ago, # |
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How to solve Div 2D/ DIV 1B ?

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5 years ago, # |
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Task D div 2 (Two Sets) Test Case 9:

How could the correct answer be YES 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ?

when it is stated in the text: If there is a way to divide the numbers into two sets, then print "YES" in the first line.

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    5 years ago, # ^ |
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    You should know that {} is a set. it is an empty set.

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      5 years ago, # ^ |
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      It is like sharing 5 apples between you and me. You take all of the apples and I get none of them. But what we just did was sharing :D

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    5 years ago, # ^ |
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    Note

    It's OK if all the numbers are in the same set, and the other one is empty.

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    5 years ago, # ^ |
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    You Didn't read this Note:

    "Note It's OK if all the numbers are in the same set, and the other one is empty."

    By the way, I got WA too in this case :(

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Can any one please point out the mistake in my code. I am a newbie and am unable to find the glitch in CODE 469D. Thanks in advance

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5 years ago, # |
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When will they update ratings?

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5 years ago, # |
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Chinese Round again, so difficult problems again...

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5 years ago, # |
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Can anyone tell me what's wrong with my code for problem D, please?

7882442

Thank you!

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    5 years ago, # ^ |
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    Check this test:

    10 12 14

    3 2 10 9 4 9 3 2 5 10

    Answer is:

    YES

    0 0 1 1 1 0 0 0 1 0

    Edit: Sorry. Test is incorrect.

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      5 years ago, # ^ |
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      Explanation: Some numbers may occur several times and you should put some of them in set A, and some other in set B.

      Edit: It's not true. My mistake.

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        5 years ago, # ^ |
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        That's not true, the problem statement specifies that the numbers are distinct.

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5 years ago, # |
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very interesting and enjoyable problem indeed..i enjoyed this very much :D

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5 years ago, # |
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I have a question regarding this submission 7868179 Div.2 problem A .. I unsuccessfully tried to hack the solution using a testcase where the algorithm will try to access h[100] while the size of the array is actually 100! so the last element i can access is h[99], but the hack was unsuccessful .. any explanation ??

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    5 years ago, # ^ |
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    When you access beyond the range of an array in C/C++, it gives an undefined behavior. This undefined behavior may or may not throw an exception.

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My first time doing the first three problems . Best wishes.

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So much math, it's awesome round!

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2333333333333 Thanks! It was a great Round!

Could anyone clarify on grading policy?

I've successfully submitted solutions for 2 tasks without penalties or resubmissions but my final round score is negative why it could be so?

Thanks!

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5 years ago, # |
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5 years ago, # |
Rev. 4   Vote: I like it +34 Vote: I do not like it

I just wanted to leave this here: 7878473

In case you're too lazy to open, here is the whole accepted code to C in Python (authored by ZhouYuChen)

m = int(input())
x,t=10**100-1,m-100*45*10**99%m
print(t,t+x)

I have to say that I'm impressed :P

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    5 years ago, # ^ |
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    A really neat/impressive solution!!

    What is the logic behind the solution?

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      5 years ago, # ^ |
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      Let S(l, r) be the sum of digits of numbers form interval [l, r]. Take big k and see that S(1, 10k) = S(2, 10k + 1) - 1 = S(3, 10k + 2) - 2 = ...

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        5 years ago, # ^ |
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        Here we have with x=10^100 — 1 and t=m-(450*10^100)%m = m-r , where r<m

        S(t,t+x) = S(1,1+x)+(t-1) = S(1,10^100)+(m-r-1)

        Above must be equal to 0 when taken mod m

        (S(1,10^100)+(m-r-1))%m should be = 0

        S(1,10^100)%m + (-r-1)%m

        If the above steps are accurate, I cant follow the proof after that.

        Can you please explain?

        Thanks.

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          5 years ago, # ^ |
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          You are taking a hard way to understand this problem. I used the same idea and I think I can explain more clearly.

          There are 4 things to observe:

          1) S(1,10^k) = 45 * k * 10^(k-1) + 1

          You can try to convince yourself doing some mathematical proof or just run a script that calculates in a brute force fashion when k is small.

          2) S(1,10^k) = S(2,10^k)-1 = S(n+1,10^k+n)-n

          This is the trickiest part! In the first case, l = 1 and r = 10^k. What happens if we increase them both? l = 2 and r = 10^k+1. In this case, we "removed" one element whose function value equals to 1 and "added" another one whose value is 2 (10^k+1 has two 1's and a lot of 0). You can observe (by induction) that the same holds for any value we add to both l and r. Therefore, this formula I wrote here is correct

          3) S(1,10^k) % M = x if and only if S(M-x+1,10^k+M-x) % M = 0

          Well, if you add M-x to both boundaries of S(1,10^k), you will have S(1,10^k) = S(M-x+1,M-x10^k) + x — M, according to property (2). Then you take modulo M and you will have S(M-x+1,M-x10^k) % M = [ S(1,10^k) + M — x ] % M = [ x + M — x ] % M.

          4) S(1,10^17) > 7 * 10^18

          Finally, as a is at most 10^18, you have to find a power of 10 that guarantees a sum that is greater than that value. I wrote a simple script in Python that calculated S(1,10^k), according to formula (1), for some k and I found that k = 17 was enough.

          If you have any further questions, you can take a look at my last submission for this code or write a message.

          Regards

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Can someone explain the greedy approach to Div.2 D / Div. 1 B Two Sets?

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This round taught me a lesson and warned me about my deficiency in coding details. Besides I eventually understood that you may never think about getting more rating in Chinese Round. But actually it may not be the reason for not participating the contest, for the main purpose that we compete in CF isn't the for the rating. Even so, there will be a long time for me to regain my confident and to participate another round especially the Chinese Round.

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Can anybody explain what is the flaw in my approach Solution ?

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5 years ago, # |
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Thank you for the tasks, I really enjoy the round.

C Div 2 is very simple and beautiful. I wish I could find the "odd part" of solution during the round :)