O--O's blog

By O--O, history, 22 months ago, In English

Hello, Codeforces!!!

Here's the invitation link of our discord server for contest discussion and announcements.

We are happy to invite you to TheForces Round #5 (PerfectForces), which will take place on [contest_time:425963]

You will have 2 hours to solve 7 problems

Thanks for participating.

Winners are:

  1. AndreyPavlov
  2. IzhtskiyTimofey
  3. siganai
  4. tyr0Whiz
  5. DrearyJoke
  6. Svyat

Editorial

Solutions are opened.

Here is a group archive of our previous contests

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22 months ago, # |
  Vote: I like it +10 Vote: I do not like it

I am waiting for nice problems!

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22 months ago, # |
Rev. 2   Vote: I like it 0 Vote: I do not like it

[Deleted]

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22 months ago, # |
  Vote: I like it +4 Vote: I do not like it

is it rated? or just for practice purpose.

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22 months ago, # |
  Vote: I like it +7 Vote: I do not like it

thanks for helping the community

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22 months ago, # |
  Vote: I like it +5 Vote: I do not like it

i am not able to register

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    22 months ago, # ^ |
      Vote: I like it +6 Vote: I do not like it

    Maybe something went wrong. Everyone can register.

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22 months ago, # |
  Vote: I like it +31 Vote: I do not like it

am I alone who received browser alert with this blog entry?

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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

what's the expected complexity of the round? like div2?

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    22 months ago, # ^ |
      Vote: I like it +3 Vote: I do not like it

    you will see...

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    22 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    It was most likely div 2.5, what do you think?

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22 months ago, # |
  Vote: I like it +19 Vote: I do not like it

Did you like the contest?

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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

How to solve F?

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    22 months ago, # ^ |
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    Hint: 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 has 512 divisors

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22 months ago, # |
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How to solve Problem B Cube Sum for Perfect 100 points

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    22 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    Prefix sum + Binary search after precalculation upto 1e6.

    My code-
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    22 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Precompute the cubes of each numbers till 1e6 and also store prefix sum of the cubes. Now for query part use binary search to find indexes of l and r in cubes array and use prefix array to find the sum.My solution

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    22 months ago, # ^ |
    Rev. 2   Vote: I like it 0 Vote: I do not like it

    $$$1^3 + 2^3 + ... + n^3 = (\frac{n * (n + 1)}{2}) ^ 2$$$

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      22 months ago, # ^ |
        Vote: I like it 0 Vote: I do not like it

      Yeah I have also come up with this formula but it was running for only samples :(

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        22 months ago, # ^ |
          Vote: I like it 0 Vote: I do not like it

        It works!

        Code
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    22 months ago, # ^ |
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    Take the cube root of l and r , let it be lc and rc respectively. Then required ans is (rc*(rc+1)//2)**2 — (lc*(lc+1)//2)**2 . As sum of cubes from 1 to n is (n*(n+1)//2)**2 .

    My Code :


    def perfect_cube(n): x = int(n**(1/3)) while True: if x * x * x <= n: break x -= 1 while True: if x * x * x >= n: break x += 1 return x - 1 def is_perfect_cube(n): x = int(n**(1/3)) while True: if x * x * x <= n: break x -= 1 while True: if x * x * x >= n: break x += 1 return x * x * x == n mod = int(1e9)+7 for _ in range(ii()): l,r=li() lc,rc=perfect_cube(l),perfect_cube(r) if is_perfect_cube(l) and is_perfect_cube(r): # lc+=1 rc+=1 elif is_perfect_cube(r) and not(is_perfect_cube(l)): rc+=1 # print(lc, rc) # lc-=1 ans = ((pow(rc*(rc+1)//2,2,mod)%mod - pow(lc*(lc+1)//2,2,mod)%mod)%mod) print(ans)
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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Why Pratice Mode isn't enabled? I had to give virtual contest to submit $$$G$$$. I failed G by few minutes :(

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    22 months ago, # ^ |
      Vote: I like it 0 Vote: I do not like it

    Anyways... Nice Problems. Enjoyed solving G :D

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      22 months ago, # ^ |
        Vote: I like it +3 Vote: I do not like it

      how to solve F

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        22 months ago, # ^ |
        Rev. 3   Vote: I like it +3 Vote: I do not like it

        I precomputed primes upto 1e7 using Linear seive. and made a set of pairs $$$($$$ no of divisors, smallest number $$$)$$$ and did a single loop [upto 1e7] and taken the number into account, whose no of divisors are greater than max no of divisors present in the set.

        Code

        Submission

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22 months ago, # |
  Vote: I like it +4 Vote: I do not like it

Thanks a lot for the contest bro, I really liked it. I'd recommend you make a facebook page where you can announce the contest date.

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22 months ago, # |
  Vote: I like it +3 Vote: I do not like it

Auto comment: topic has been updated by E404_Not_Found (previous revision, new revision, compare).

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22 months ago, # |
  Vote: I like it +3 Vote: I do not like it

can you guys please make a discord server or some other type of group that we can join so that we get notified ...as most of the times blogs don't show up and we miss these contests

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22 months ago, # |
  Vote: I like it +3 Vote: I do not like it

It is very hard to participate in your rounds i am very active on codeforces but still could not see your blog.

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22 months ago, # |
  Vote: I like it 0 Vote: I do not like it

Very interesting gym, enjoyed it. Thanks to the organizers. If they need tester, I would help them with pleasure (never was tester before).