Comments

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  bobhob314  commented on June 12, 2015, 10:02 p.m.

  Hello DMCI members,

  Is there going to be no DMOPC '15 June? I have grown to love these contests over the months and hope that they will be continued.

  Thanks!

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  GHETTO  commented on May 11, 2015, 2:40 p.m. ← edited →

  Greetings kind sirs and madams,

  I would like to inquire if there will be competitions that run in the duration of the summer times.

  Will the don mills Students be hosting the DMOPCs?

  Thanks you,

  Rabhit

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  Xyene  commented on May 22, 2015, 8:46 p.m. ← edited →

  It's not certain that we will be hosting the DMOPC during the summer months (we might, though). However, even if we don't, there still won't be a shortage of other contests for everyone to participate in!

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  belowthebelt  commented on May 6, 2015, 4:57 a.m.

  I managed to solve the first 4 out of the 6 problems completely. I liked the problems which I managed to solve. Though, I wasted a lot of time debugging the fourth one, and then had to go away for a while to submit a report so as to fail to not find time to pick up partial points in the 5th or the 6th problem. Anyway.

  1. Flare
  My solution is basically this:

    print input()*2.0/(9.8)

  Given the value v, we have to find the value t such that the value of y is equal to zero. So, we rearrange the equation, and thus, it changes to vt = 0.5*(gt²) - since g had a negative sign. After which, when the t is canceled, the equation reduces to 2v = gt, which further reduces to: t = 2v/g which is our answer.

  2. Tides
  This question was also simple, but had a lot of cases to deal with. I got 70/100 only initially, because I was not considering the case where the high tide ends up occurring before the low tide, which according to the problem statement (If you read carefully!) is also not a valid case.

  Depending on the language you choose to code this one in, it's just an implementation based problem. Checking if the sequence is an increasing one between low-tide and high-tide values is easy.

  n = int(raw_input())
  l = map(int, raw_input().split())
  mn, mx = min(l), max(l)
  mni, mxi = l.index(mn), l.index(mx)
  flag = True
  if mni>mxi:
      flag = False
  for i in xrange(mni,mxi):
      if l[i]>l[i+1]:
          flag = False
  if flag == True:
      print mx-mn
  else:
      print "unknown"

  3. Globally Unique Shells
  This problem like its name is actually ends up being a problem forcing you to think for a while, if you don't look at the constraints properly. For a while, I didn't realize that the value for the shells can range till 999999, I thought it was till N. So, I made a hash table till 60,000+1 for the value of N, which obviously resulted in a WA.

  Here's what I did:

  • I found the total number of the shells which were given to us.
  • I created a list with the union of the two halves, which thus reduced the total number of shells.
  • This number of shells reduced from the total which were given to us, if subtracted from n, gave me the number of shells which were incomplete.
  • Since the ones which are given to me are not completed anyway, and I just counted the number of ones which got completed.

  What are the other ways in which people solved this one?

  n = int(raw_input())
  ans = 0
  a, b = map(int, raw_input().split())
  c = a + b
  A = map(int, raw_input().split())
  B = map(int, raw_input().split())
  un = list(set(A) | set(B)) #Union of two lists.
  print n - ( c - len(un))

  4. Sand Triangle
  For a long while, my solution kept passing the smaller constraint case, while consistently failing on the larger one because of a silly mistake I was doing. (Counted the number of 9s in my code wrongly - shame on me!)

  I guess the general ideas, which I guess would have been used by people in this problem could be: pre-computation of certain values, then binary searching them, then finding out the required answer.

  • I calculated the starting point of all the levels of the triangle till 99999+1, because that'll be the last level to take care of 10⁹.
  • Then, using simple linear search, I found out the level a number was in.
  • We can already see that the number of elements in a level is the number of level itself.
  • Once I got the level a number is situated at, I had already pre-calculated all the sums for the levels till 10⁵, so I could just print the required value.

  Definitely the best problem I solved in this contest, required thinking, and making me use pen and paper.

  Here's my code:

  n = int(raw_input())
  l, x, y = [], 0, 99999+1
  
  for i in xrange(1,y):
      l.append(1 + i*(i-1)/2)
  
  for i in xrange(len(l)):
      if l[i]<=n:
          x = i + 1
      else:
          break
  ans = 0
  
  hashed = [0]*y
  hashed[1] = 1
  hashed[2] = 5
  for i in xrange(3,len(hashed)):
      hashed[i] = hashed[i-1] + i*(i-1) + 1 + (i*(i-1))/2
  
  print hashed[x]

  General Feedback:
  The contest in general was pretty good. The timings for me, in India were pretty odd, though. It started at 1 am and ended up at 4 am. I was sleepy by the end of it. Not that I mind. :) I would love to take part in more contests like this, and improve my level and skills. And I'm now expert: 1346. Which is the TopCoder equivalent of Division 1 in my first contest, which is... good, I suppose?

  Also, quick question to the admins: is there a way by which I could set up questions / test them / write editorials for the judge?

  Keep up the good work, guys!

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  omaewamoushindeiru  commented on May 6, 2015, 12:30 p.m.

  Interesting

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  nullptr  commented on May 5, 2015, 10:48 p.m. ← edited →

  https://ideone.com/1vT6q7

  Editorial coming later

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  LOLWHATOMGBBQ  commented on May 11, 2015, 11:29 p.m.

  "Later"

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  kobortor  commented on May 12, 2015, 12:04 a.m.

  Soon™

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  omaewamoushindeiru  commented on May 5, 2015, 10:18 p.m.

  14'? Have we traveled back in time?

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  Sentient  commented on May 5, 2015, 10:33 p.m.

  DMOPC'14 represents the 2014-15 school year.

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  omaewamoushindeiru  commented on May 5, 2015, 10:34 p.m. ← edited →

  Oh, I see.

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  kobortor  commented on May 5, 2015, 7:29 p.m.

  TSS IS READY

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  pyrexshorts  commented on May 1, 2015, 3:35 a.m.

  Can this start at 4:30 or some time around that? Some schools end later, and it's hard to compete when there's one hour less for some contestants.

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  Xyene  commented on May 1, 2015, 10:15 p.m. ← edited →

  This is the last contest of the school year. In the summer months, it is possible (and perhaps likely) that we'll run the DMOPC as a virtual contest.

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  bobhob314  commented on May 1, 2015, 3:34 p.m.

  This question is asked about once every DMOPC. Unfortunately, there are good reasons for the starting time and it will not be changed.