Hello, Codeforces!
We are happy to invite you to TheForces Round #13 (Boombastic-Forces), which will take place on May/07/2023 18:10 (Moscow time)
What is TheForces Round?
You will have 120 minutes to solve 7 problems.
We strongly recommend you to read all problems.
Problems were prepared and authored by Amir_Parsa ,E404_Not_Found, Sevawy.
We would like to thank our army of testers: physics0523, Madboly, blxst, O--O, Virendra115, blxst, BallBreaker, ExpensiveAC, aphex, TheOpChicken123, leneza_41, EndlessDreams.
Also we want to thank you for participating in our round.
Discord Server (700+ people)
Nice problems. (i didnt tested yet XD)
As a tester, I will test round while listening the Mr. Boombastic song.
As a tester, I will test this round after writing this comment
As a tester, this is my first "As a tester" comment.
Auto comment: topic has been updated by E404_Not_Found (previous revision, new revision, compare).
I can't register in this contest.What's the issue it is saying that you cannot participate in group contest
If you're seeing this contest from TheForces group then you should join as a "participant" in the group to be able to register, Otherwise you can register for the round from GYM.
done!!
Educational-Forces
C was purely just annoying implementation (or knowledge about some standard functions).
I don't know how to solve D or E. I just recognised that both were standard problems (that I don't completely know how to solve) and just copied a solution from the internet.
F and G were nice though.
G was just standered $$$O(n^3)$$$ dp right? $$$dp[i][j]$$$ denotes no. of ways to remove the subarray $$$[i..j]$$$.
Transitions were $$$dp[i][j] = dp[i+1][j-1]*(a[i]>a[j]) + \sum\limits_{w=i+1}^{j-1}(2*dp[i][w]*dp[w+1][j])$$$
Where am I wrong?
Wait why is it 2*dp[i][w]*dp[w+1][j]? Wouldn't the coefficient of this term be factorial(L1/2+L2/2)*(factorial(L1/2)*factorial(L2/2))? Because we can choose the order in which we perform delete operations on these two subarrays
Aaah got it! Thanks! Yeah you are right, the two subarrays are "independent", we can freely choose the order of the operations performed inside them.
I didn't get AC though :( idk what's wrong with my approach
Now I also updated the transitions, still getting WA :(
Maybe setter's solution is incorrect (baby don't hurt me)
You should try the following test case. I tried some of your submissions and they all gave wrong (and different) answers to this:
Explanation: You need to clear the following pairs (nothing else is possible):
All of them are also independent meaning that you get $$$5! = 120$$$ removal sequences.
Here is a test case where your code fails:
Explanation: You need to clear the following pairs (nothing else is possible):
All of them are also independent meaning that you get $$$5! = 120$$$ removal sequences.
check this out: https://usaco.guide/problems/cses-1080-empty-string/solution
it's pretty much the same problem
For D task https://mirror.codeforces.com/blog/entry/99907
Solution to E: https://mirror.codeforces.com/blog/entry/90035
Auto comment: topic has been updated by Amir_Parsa (previous revision, new revision, compare).
The tests are strange? Are there any? Well, I like the problem, the rest is not important, for me. But I assume feedback to the round is ok.
I look at the code of problem D of 40th position at standings.
It got accepted.
If that code really got accepted, then the test of that problem is kinda weak. A simple test like this will make the code fail: