We will hold UNIQUE VISION Programming Contest 2023 Christmas (AtCoder Beginner Contest 334).
- Contest URL: https://atcoder.jp/contests/abc334
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20231223T2100&p1=248
- Duration: 100 minutes
- Writer: yuto1115, cn449
- Tester: leaf1415, math957963
- Rated range: ~ 1999
- The point values: 100-250-350-400-450-550-650
We are looking forward to your participation!
Are Japanese editorials for Beginner contests still translated just by using GPT-4 to English?
I do think so!
hope to solve ABCD.
good luck!
excited
Hope everyone has good grades！
Why this submission gives wrong answer for E? Any HELP is appreciated.
Better to use $$$i * W + j$$$, Fixed
PS: I think in $$$i * H + j$$$ sometimes this value may clash
Such a beginner's mistake, thanks for your help.
https://atcoder.jp/contests/abc334/submissions/48792186
Can you help with this submission for problem E please? Even a counter test case where this might be failing would be helpful, thanks in advance.
today,i fuc**ed up
yeah,only solved AD...
https://atcoder.jp/contests/abc334/submissions/48798563 my code for B is giving wrong for 2 testcases.What's wrong in this code?
A < D < B < C
E < B
no lol
But I don't think so lol
Nice contest as usual, managed to solve 6 problems.
is $$$F$$$ dp?
This?
Does anyone accepted G with Dynamic Connectivity?
Mine got 7 tests TLE
Mine got last 5 tests TLE Submission
I think I found it.
Here it is: noimi
Update: Managed to fit my solutions just under TL by optimizing it and using C++23 Submission
I added hints for E — Christmas Color Grid 1 on CF Step
I created video editorial for F: Christmas Present 2. I also created 1D version so that you can submit $$$O(N^3)$$$, $$$O(N^2)$$$ and $$$O(N Log(N))$$$ solution.
Is this the right direction to approach G?
Only articulation points affect the answer. Hence, for each articulation point, come up with an algorithm to find out the increment in connected components after its deletion
Yeah, and the standard approach for checking articulation points gives you their exact count lol. I just built the graph and ran the cp-algo implementation on it.
In that implementation, the number of times IS_CUTPOINT($$$v$$$) is called is the number of resulting components (except for the root where its actually “children — 1”).
If you’re familiar with DFS tree, the intuition is pretty obvious, each child $$$v$$$ which doesn’t have a backedge that crosses over $$$v$$$ will be disconnected when $$$v$$$ is removed.
Is it intended that offline dynamic connectivity solutions receive TLE on problem G?
Solved A,B,D
D<B<<C
A- If conditions
B- Tricky implementation + Maths
D-Sorting+Binary Search+ Prefix Sum
Hardest B ever, lol. I spent around 1 hour for B but failed. D, E are quite trivial though.
The time is over right after solving E. Fuck
+Fuck B
and C
Why does this code get WA in E?
https://atcoder.jp/contests/abc334/submissions/48798560
DSU is quite overkilled though
I'm familiar with it :)
line 101:
line 106:
res (number of components) can be upto $$$5 \cdot 10^5$$$
so ans can be upto $$$(5 \cdot 10^5)^2$$$.
meaning ans*lt(dv, mod-2, mod) can overflow 64-bit ints
Thanks, it is such a stupid mistake.
Horrible contest. Problem B very to hard for this division and no editorial. What else ?
I solved F just because I saw this before (The problem called Mowing the Lawn in USACO 2011 Open Gold Group)...
Speedcoder........
Why and where does this code give WA for problem E? any help ?
My submission
Can you please elaborate on the idea behind your submission? I have solved the problem, but apparently with a different approach.
Yes sure, I actually gave ids through string concatenation of row and column index, after that I performed merging of grid cells which are green, now so i will have components of green through dsu, then on every red cell I will check it's neighbor and identify what are the different component numbers in all 4 directions, then just adding the components we can achieve for every red cell to green. Then dividing by number of initial red cells which have equal probable to choose.
seems to be wrong. Row 1 Column 11 has the same string as Row 11 Column 1.
Yes got it thanks, will add space lol
How can i download the testcase.txt in Atcoder?
https://atcoder.jp/contests/abc334/submissions/48787539 Can anyone tell me what's wrong with my code or help me download the testcase??? Thanks a lot!
Weird round, I solved ADE.
Now I've just finished upsolving B. Being that there's no editorial for the problem yet, I'll give an explanation of what I've done:
What we are being asked is easy to understand: we want the amount of numbers y in the range [l , r], such that y % m == a % m. One way to go about this is to subtract a from l and r, now the problem reduces to finding all y such that y % m == 0 in the range [l , r]. Good, this is easier because we can subtract (l — 1) % m to r % m and that will give us the amount of multiples in between... right? Well, the intuition is correct but there is a slight flaw: when l is negative the above formula doesn't work because of rounding. So what we can do is add m * (abs(l)/ m) to l and r and then apply the formula. Ta-da! Problem solved
please give me a little hint for problem b
for problem B why this logic is wrong
I am checking at l and r and between l to r
Logic is simple. typical arthimetic sequence
A<D<<B<<C,I hate problem B
I made video editorial for F: Christmas Present 2
B is hard,the hardest B I've ever done.(AC 15,WA 5) I think this is also the reason it gets 250 points but not 200 points.
Can anyone explain my question on B. For 0 2 1 10, why the answer is 5??
f***! My brain must have been damaged.