We will hold AtCoder Beginner Contest 358.

- Contest URL: https://atcoder.jp/contests/abc358
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240615T2100&p1=248
- Duration: 100 minutes
- Writer: yuto1115, cn449
- Tester: sotanishy, MtSaka
- Rated range: ~ 1999
- The point values: 100-200-300-350-475-525-550

We are looking forward to your participation!

exited

excited

excited

Only 550 pts G!

Hi, i am new to competitive programming and new to AtCoder. I am curious about the rating score of ABC questions, what are their equivalent codeforces rating scores? Is their a formula that can do the conversion? Thx

https://silverfoxxxy.github.io/rating-converter

Thanks very much!

I don't have any recollection of there being official ratings for AtCoder problems; Scores/Points are relative and do not capture ratings particularly.

You can however refer to the following two in conjunction to get an idea:

Kenkoooo:Atcoder-Problems and Silverfoxxxy:Rating-Converter

Thanks very much!

I refuse to write solution for F even if i'm getting paid.

Exactly. I had solution after 1 minute of thinking but just implementing it seemed messy so I gave up on it.

that's not even the main thing , thing is output format could have been better , they just could have asked to output pair of cells which have wall between them

Yeah that's why I said implementing seemed messy. Format is weird. I didn't want to generate sequences of characters.

What the fuck is the output format of problem F ???

What's the meaning of

and the connected component containing the vertices (1,M) and (N,M) consists???onlyof this path.just for harder implementation ???

For human-readable

My post contest discussion stream for all problems

26 correct 2 WA on C any idea submission

Just go through every possibility $$$O(2^N)$$$

your way of computing minima is not correct

ex- 001 ,010,110

you only need shop 1 and 3 to fill the mask but your approach is greedy picking which is wrong and you will end up taking 1,2,3

hint-use bitmask

any idea for g?

you only stay in only one cell

You try to go to a certain point on a board then just spam staying in that point.

Because of that, you can try to dp here, like dp[u][T] is the maximum value you will receive when you use T turns to go from start to u (all points here are converted into 1-D coordination).

is $$$T$$$ fixed for certain $$$U$$$?

I considered for max value cell only, will fix it

But how do you calculate T ? I was lucky to get AC doing similar thing using BFS but just kept increasing turns till I got AC.

T <= H * W I suppose so

Yep but is there any proof on why T <= H * W. Is it because the number of unique paths from a cell to any other cell bounded by H * W ?

Try 1968D - Permutation Game first. You can use almost same arguments for this problem.

Alright , Thank you !

Same argument.. what do you mean, I still can not understand why the maximum number of steps to check is H*W and I can not find any formal proof, this drive me crazy.

f i x,y means when Takahashi goes i steps and stops at (x,y), how many fun value can he get.

Takahashi may walk for some steps and stopped at a position forever.

It's easy to solve f i x,y when i<=n*m. And it's can be proved that Takahashi may walk at most n*m times, so the answer is MAX (f i x,y + a x,y * (K-i))

what's wrong in my solution of G?

It's the same mistake that I did, and wasn't patient enough to even debug it after the contest so went through the editorial and just noticed that the dp table is going to have 3 states and I got what I was doing wrong, hope it provides you a hint :9

yo same case , I also wasn't patient enough to debug and watched the editorial and yeah missed the crucial no of turns state

Shortest path to reach final cell is not enough.

Consider this testcase -

SpoilerAnswer is —

`99989`

but A[i][j] > 0 (never mind)

Thanks a lot!

This is somewhat that is in testcase9

how E?

Problem F is frustrating. I accept the other six problems quickly with no penalty, but I had wrong F for 5 times.

[deleted]

Why does the C++ solution in the problem E's editorial unavailable?

Oh,sorry,maybe my internet has some problems.It is now available.

I think, I solved problem like today's F not long time ago, but don't remember, where. Anyone remember?

Also, what happened with problem G? I've never seen anything such simple on this position.

G could have easily been a D/E imo.

Anyone could hack my F please? I have been testing on it for about 1 hour but I don't know why it gets WA*3.

Your code fails when $$$n=3,m=5,k=13$$$. I am extremely sorry that you were disgusted by the authors!

output:

Thank you. I didn't notice that I can turn to the other side. And to be honest, I'm not disgusted XD anyway thanks a lot.

3 5 13 Yes +++++++++S+ +o.o.o.o.o+ +.+-+-+-+-+ +o|o.o.o.o+ +.+.+-+-+.+ +o.o|o|o|o+ +++++++++G+

It seems that you filled the path in maze one row by one row, however, when n is odd, you can also fill the last row but your code did not.

I also WA 3 for 5 times because of it, and it made me angry. However, after coming up with this situation, I accepted at once.

I have a friend who is rank 45, he didn't passed the following testcase but also accepted F in 67 minutes.

It is unfair, isn't it?

3 3 9 Yes +++++S+ +o.o.o+ +.+-+-+ +o|o.o+ +.+.+.+ +o.o|o+ +++++G+

You know that, I could pass the hack easily but I always got 1 RE.

Luck is always unpredictable

https://atcoder.jp/contests/abc358/submissions/54597481

Could someone please tell why my D fails ???

don't use it only stores unique element

let say a= 2,2,2,4,4,4

and b= 2,2

your approach will result in 2+4=6

but optimal one is 2+2=4

Use Multiset because element may repeat

Can someone say why my Problem D is TLE Submission

You should use

`ms.lower_bound(b[i])`

can you please explain why does it make so much difference?

`set:: lowerbound`

is a binary search for red and black trees, with a complexity of O (log n).`std:: lower_bound`

for non randomly accessed containers, it is an iterator sequential search with a complexity of O (n).Can someone please debug my Problem G(18xAC,18xRE)Submission

`1≤K≤10^9`

You cannot open a vector whose size is related to K.

In problem F, O(n^2) is enough guys; all you have to do is do very little HOLY CASEWORK!!!

OTZ

Do they make AI more difficult to understand the statement by modifying important information in Problem F many times?

That's right!

I don't think so. I had never seen the input format or the output format when I was trying to solve Problem F because it's too long, but I still understand what the problem want me to do. I think AI is smart, and it may easily understand what did the problem mean. If AI can solve Problem F with correct information, it can also accept Problem F with wrong format.

ABC is for more than ten thousand people. If the question surface is wrong, it will make many people feel confusion. So I think it's no need to modifying importand information in order to make AI more difficult to understand Problem F.

E was such a good problem.

How to do it ?

It's DP+Combinatorics.

If you have n items, among which m are unique with quantities(q1, q2, q3... qm), the number of distinct ways to arrange these items is:

Now for DP, there are two states

[i,len](i denotes the alphabet we're currently at and len denotes the length of our sequence so far).Now, at each step, we attempt to take as many elements of that alphabet as possible. Therefore, the recurrence relation is:

and base case is

For each state, by using mod_inverse at each step of calculation we've got denominator of

`eqn (1)`

, to get the whole expression multiply it by len!.Summation of dp[0][len] is your answer for len=1 to k.

Here's a link to my submission.

Thanks mate.Amazing Solution.I appreciate your effort into writing this.

yep it's good

Why does this TLE ??

Precompute the mod inverse for each $$${factorial_i}$$$.

thx

Gis a nice follow up to the recent Atcoder problem : ABC344F : Earn to Advance. I also have a blog, hints and practice contest for the concepts used in the first problem.your articles are good.

I dont see any relation between G and Time travel technique, can you help me to elaborate more?

It makes the observation trivial that there will only be one cell where you stay multiple cells. And this cell would be retrospectively fixed as the maximum valued cell that you encountered in your journey. Just like we did in

ABC344F: Earn to Advance.Can anyone tell me why am I having a TLE XD My code

OK. I'm turning to C++.Same code in python and C++ led to 2209ms and 64ms.Almost got me crazy trying to optimize further in python :(

F is surprisingly neat if figure out the pattern, very fun to upsolve.

I don't like how I can go from TLE to a relatively fast AC in E simply by precomputing nck values.

you don't need NCK values , you just have to calculate factorials and mod inverse of it

OMG i don't know why i did not try to solve G, got stuck in F, when i read G i thought some matrix exponentiation stuff so i left it. i should have tried G

Can problem G be solved using Dijkstra's Algorithm?

No shortest path is not enough, look at this comment

I see, thank you!

The G is pretty simple than usual rounds:D

you should swap between c and d

When would the official editorial of the round be released on the website?

I think G is easier than E.

ok

Who has stronger data to help me hack my code for F question please?I have tried all the known strong data.submission

Found a test that fails the fastest solution for task G.

1 9 34

1 4

80 28 4 98 84 2 67 57 100

The optimal solution is to take 98 * 34 = 3332, but merhorn's solution outputs 3330