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
- Problems were prepared by Amir_Parsa, O--O, and E404_Not_Found
- Would like to thank our testers: merlin_, Psychotic_D, k1r1t0, psychobot, and iliya_mon
- Also we want to thank you for participating in our round
Thanks for participating.
Winners are:
Solutions are opened.
Here is a group archive of our previous contests
I am waiting for nice problems!
[Deleted]
Is it rated?
is it rated? or just for practice purpose.
It's just for practice
thanks for helping the community
i am not able to register
Maybe something went wrong. Everyone can register.
am I alone who received browser alert with this blog entry?
I also did
I did too.
what's the expected complexity of the round? like div2?
you will see...
It was most likely div 2.5, what do you think?
Did you like the contest?
Yes. How to F though ??
ya i liked the contest
How to solve F?
Hint: 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19 * 23 has 512 divisors
How to solve Problem B Cube Sum for Perfect 100 points
Prefix sum + Binary search after precalculation upto 1e6.
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
$$$1^3 + 2^3 + ... + n^3 = (\frac{n * (n + 1)}{2}) ^ 2$$$
Yeah I have also come up with this formula but it was running for only samples :(
It works!
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 :
Why Pratice Mode isn't enabled? I had to give virtual contest to submit $$$G$$$. I failed G by few minutes :(
Anyways... Nice Problems. Enjoyed solving G :D
how to solve F
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.
Submission
Got it thanks
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.
Auto comment: topic has been updated by E404_Not_Found (previous revision, new revision, compare).
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
Done.
It is very hard to participate in your rounds i am very active on codeforces but still could not see your blog.
Very interesting gym, enjoyed it. Thanks to the organizers. If they need tester, I would help them with pleasure (never was tester before).