Ripatti's blog

Codeforces Round #125
By Ripatti, 12 years ago, translation, In English

Hello everyone!

There is yet another Codeforces round. Now it is the 53-th one.
The round will run for both divisions by classic rules of Codeforces format.

Points are standard: 500-1000-1500-2000-2500.

Round was prepared by Ripatti , Alex_KPR , Gerald , Aksenov239 , RAD , Delinur .

Good luck!

UPD. Round is ended.
Winners of div1:
1. Petr
2. tourist
3. SergeyRogulenko
4. bmerry
5. UESTC_Nocturne

Winners of div2:
1. gflegar
2. mylyanyk.ivan
3. arbesfeld

Petr is only who solved all 5 problems in the first division. No one solved all 5 problems in the second division.

UPD. Editorial.

  • Vote: I like it
  • +138
  • Vote: I do not like it

Comments Comments (37)
| Write comment?
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

0.0012 from 100010

→ Reply
»
12 years ago, # |
← Rev. 2   Vote: I like it -39 Vote: I do not like it

be careful! 5^3 is meaning something... maybe this is hint for some tasks ))

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it -46 Vote: I do not like it

    This is an algorithmic contest Mr. Sherlock Holmes :P

    → Reply
»
12 years ago, # |
  Vote: I like it -26 Vote: I do not like it

Can we suppose fixed scoring, as it is not specified?

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +5 Vote: I do not like it

    It is already specified that the points are standard ->500,1000,1500,2000,2500 .. So no dynamic scoring

    → Reply
»
12 years ago, # |
  Vote: I like it -68 Vote: I do not like it

但愿不坑爹

→ Reply
»
12 years ago, # |
  Vote: I like it -12 Vote: I do not like it

Very tough problem set .. I could not understand what problem B wanted . :(

→ Reply
»
12 years ago, # |
  Vote: I like it +20 Vote: I do not like it

Looks like I will get more than minus 100 rating for birthday present T_T such bad luck T_T

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +8 Vote: I do not like it

    Regardless, Happy Birthday! :)

    → Reply
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

I'm try to solve 4 problems..., and solve only one :/

→ Reply
»
12 years ago, # |
  Vote: I like it +8 Vote: I do not like it

C is very annoying :) Got 9 WA's and still don't know what's wrong with my solution.

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +18 Vote: I do not like it

    I had a WA 4 when I checked for line-circle intersection instead of segment-circle.

    → Reply
»
12 years ago, # |
  Vote: I like it +8 Vote: I do not like it

C was very fun (I didn't get it though). Thanks for a good problem set!

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    any idea for it?

    → Reply
    • »
      »
      »
      12 years ago, # ^ |
      ← Rev. 2   Vote: I like it -6 Vote: I do not like it

      the thng was tht the entire process was totally recursive..

      so if 1 value matches the sequence the rest will follow..;)

      simply simulate the expt1 till u do not exceed t..

      after tht both will take same number of steps....

      ans is n-i+1 where i is no. of steps to reach t

      P.S.: be careful not to output -ve..:P

      → Reply
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

In problem B in div 2, we just have to find number of circles(out of 4) which are not intersected by any other circle. Am I right ?

Please help me someone.

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it -11 Vote: I do not like it

    Yes, you're right, my young padawan.

    → Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    You have to check for every circle from each pair that it lies either in smaller of the other pair or outside bigger of the other pair. Checking only intersection is not enough as you can face a situation when a circle lies between 2 others ( exactly in the ring), so the color is not changed from white to black or vicer versa).

    → Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    You should find the number of circles that are not intersected by the other 'disc'. For a circle in disk 1, simply checking intersection with both circles of the other disc is not enough. For example, the 1st disc may lie entirely inside the black region of the other disc, in which case there are no circle-circle intersections, but the number of contours is 2.

    → Reply
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

can you add tutorials?

→ Reply
»
12 years ago, # |
  Vote: I like it +10 Vote: I do not like it

Today is Dragon Boat Festival. Happy Dragon Boat Festival!

Also, today is Alan Mathison Turing's 100th birthday.

→ Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    I can add to your wonderful list that today there is another one celebration in Saint-Petersburg called Scarlet Sails show ("Алые паруса" in Russian). It's the holiday for all school-leavers. However, that event gathers not only them, but a lot of citizens and visitors of Saint-Petersburg.

    → Reply
  • »
    »
    12 years ago, # ^ |
      Vote: I like it +4 Vote: I do not like it

    Strange !! People here vote for colors not for content .. Even I wished him in my blog and got a lot of negative votes .. Sad but truth...

    → Reply
    • »
      »
      »
      12 years ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      The problem with your post in my opinion is: How can Turing be happy on his birthday when he has passed away many years ago ?!! So, wishing a happy birthday to him may not make that much sense.

      → Reply
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

Editorial ?

→ Reply
»
12 years ago, # |
  Vote: I like it 0 Vote: I do not like it

It seems a spell error? Winners if div1 -----> Winners of div1:

→ Reply