We will hold AtCoder Beginner Contest 467.
- Contest URL: https://atcoder.jp/contests/abc467
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20260718T2100&p1=248
- Duration: 100 minutes
- Writer: vwxyz0, Nyaan
- Tester: sheyasutaka, sounansya
- Rated range: ~ 1999
- The point values: 100-200-300-425-450-525-625
We are looking forward to your participation!
Let's see how it goes
Hope I can solve 6 problems.
me too
so strong, i think i can only solve 4~5problem
so strong, i think i can only solve 0.1 problem
Currently I solved 2 problems.
1 so far for obesity gettig wa not sure why
You mustn't post it during the contest.
What's wrong with the floating point today?
The first problem is the easiest one I've have ever done!!! (OoO)
ABC461A is easier, isn't it?
Yeah!
Have you entered ABC before
yes
Hmm,i don't think so..
Hope it's going well!
Who turned the FG problem into a data structure problem?
Problem C is more difficult than the previous ones.
Yeah. Used DP to solve it.
if ( a % 2 != current_p ) { ops ++; }
Where is the mistake in this code? Please!!!
I was only able to solve $$$A$$$ and $$$B$$$ :(
Me too.
I was able to solve A, B, and D... can someone provide a solution for C? So hard :(
https://atcoder.jp/contests/abc467/submissions/77636929
how did you even solve D dude!! I got no idea then skip to E then fail miserably 🥀
for C just case work it into if you make A[1] even, if you make A[1] odd, the the choice of how much to add to A[2], A[3], and so on is deterministic by parity.
i.e. if B[i] = 1 its neccessary that A[i] and A[i+1] has different parity
and if B[i] = 0 its neccessary that A[i] and A[i+1] has same parity
For D, just create two equations equating radii and then use det to find solve conditions...
Specifically, use (x-Px)^2+(y-Py)^2=(x-Qx)^2+(y-Qy)^2 and (x-Sx)^2+(y-Sy)^2=(x-Rx)^2+(y-Ry)^2
then use det conditions..
Yoo you solved C! great job man, so how would your algorithm work for say A = 010010 and B = 01100, just want to understand, thanks
https://atcoder.jp/contests/abc467/submissions/77588124 I hope this help, I'll try to not write long so its does not overwhelm, I case work it into two case, if we make A[0] even, if we make A[0] odd, (I use 0 base index so its start with A[0])
then we consider i=1,2,... so on iteratively
check if (A[i-1] + A[i]) === B[i-1] (mod 2) already, if not we have to do +1 right? if so we do on ith element so A[i]+=1 hornestly I think my explanation is shit, somebody could prolly do better, I'm sorry brother please consult other or just look at editorial
There can only every exist two unique sequences (considering all the elements $$${mod}$$$ $$$2$$$) which satisfy $$$(A_i + A_{i+1}) \equiv B_i \pmod{m}$$$ for all $$$1 \leq i \leq n-1$$$.
These can be found out by modifying the prior condition to:
Hence, if we know know the first element we can always re create the entire sequence. So, we first assume the first element to be
0and then1. And choose the better case of the two as final state.Thank you.
Was only able to solve A and B as well... how on earth do you solve C?? Anyway, this is my second atcoder contest so I hope to keep improving, hope you guys did well!
This is my 82nd contest,my rated is 1323.but I only solve AB because Iwant to solve Cand E together that problem look easy but I failed. 1323->1267,-56.5555555555555555555555555555555fuck. Fuck d,It looks so easy but but many detail
I only solved A, B, and D.
I think C was harder than D. :(
Me Too.
First AtCoder contest. Learned a lot today. Looking forward to improving!
pure math contest ):
The contest is like a shit!
Why the Problem D inspects the carefulness instead of the algorithm ability!
Why the Problem A inspects the precision so much that it's easy to get a penalty time.
For the problem A, you can turn the division to the multiplication to avoid the problem of precision
True, but few of the participants noticed this one and get 5min pentaly
I got 3 penalty for this one
Of course. But do you think it's suitable to put this in A?
Why don't you code like this to avoid precision?
What’s the point of constant-factor optimization in the $$$O(n\log^3 n)$$$ solution to Problem G
I think either all 3log solutions should pass comfortably, or none of them should be allowed to pass.
神秘大佬30min过G，告诉我他G的写法是暴力分块。。。
It's practically impossible to allow all $$$O(n \log^2 n)$$$ while reject all $$$O(n \log^3 n)$$$. The writer's intention of the problem seems introducing $$$O(n \log^2 n)$$$ solution, which is designed to pass comfortably, but also inevitably allow some fast variants of $$$O(n \log^3 n)$$$.
Rubbish.
Rubbish contest.
E seems so difficult but have 1000+ solves,can someone explain,i don't understand the editorial.
Atcoder Math Contest
Can anyone explain problem E i have doubt how we find r[0] as we have to sum over all r[i] and once we find r[0] all other expressions depends on r[0] only and if I want to find r[0] by traverse from 0 to 10^9(M) then it's gave me TLE also I find that we can build a function and here only one variable is r[0] and then I have to check whether this function gave us maximum but how to implement I have no idea.
how to solve E?
Trash.
Did anyone get AC with sqrt heuristics in G?
Ex:the editorial of G's first merge-sort tree is on the indices.
My account has been banned for unknown reasons. Please explain the reason and lift the ban. I did not violate any competition rules. My account: xuyifei0302.