We will hold AtCoder Beginner Contest 353.

- Contest URL: https://atcoder.jp/contests/abc353
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20240511T2100&p1=248
- Duration: 100 minutes
- Writer: MMNMM, nok0
- Tester: physics0523, kyopro_friends
- Rated range: ~ 1999
- The point values: 100-200-300-400-500-550-550

We are looking forward to your participation!

wow

I think the problem G is easier than usual because of the point values.

hope get a positive delta

Hope to Solve ABC

ABC 353 may include:

Guessing problem from contest number is very useful for ABC.

I think the really useful thing for ABC is to improve your skill instead of make some "luck based" guesses.

True, tourist is just insanely good at guessing

What do you mean by that ?

Problem B is not clear of todays AtCoder Contest.

After reading the example, it will clear.

obviously primes which are palindromes

hfhf

mAThCoder

how to do E?

hinttrie

E using hashing.

https://atcoder.jp/contests/abc353/submissions/53356096

Use hashing and count for each prefix length upto the highest length. Remember to optimise checking only those strings which have length greater than equal to the current length you are checking for.

Here is the code for the same — https://atcoder.jp/contests/abc353/submissions/53371485

Nice C, D, E

太难了！！！

認同

Please talk in English（请使用英文对话）

oh my i don't understand English!

Can someone help me find out this code get WA for problem G:

https://atcoder.jp/contests/abc353/submissions/53380216

Changing solve() to the following makes it accepted:

The changes are that the for loop bound is n instead of m, and at the end take the maximum between 0 and the dp maximums to handle the case if it is better to not participate in any market at all.

Thanks, maybe today is my unlucky day.

c d e just a little more difficult as it should be in that place.

Why this gives WA for F? Code

There is a corner case for K=2

Example

Answer is 3

Hi VLamarca,i've updated the code for k=2,still it gives WA for 6 test cases,Code ,any help?

Your update is wrong, you can make assert(k!=2) to confirm that the 6 cases you are still failing are for k=2

Thanks for the help,the mistake i made was printing ans one loop earlier for k=2,this is beyond stupid.Code AC

Hi Mr VLamarca，I still feel confused about what is special about the case when k = 2, and what ideas should we use to handle this case. Would you like to talk more details about it? Thank you so much.

can someone please explain what should i do to stop tle in problem C .What should i study to no get tle in next rounds?

your solution might be O(n * n) just sort the array and use binary search

https://atcoder.jp/contests/abc353/submissions/53347370 how i binary search and sorting helps in this ?

for a particular A[i] break the cases in 2 parts A[j] < 1e8 — A[i] and A[j] >= 1e8 — A[i] and then you will see the soln

i didnt .-.

I added hints for

G: Merchant Takahashion CF StepHi, amazing solution and explanation. I have one doubt though. While choosing the optimal position to the right of current index (say t), why are you checking for the max value in range [t,n-1] ? Shouldn't it be [t+1,n-1] ?

$$$[t + 1, n - 1]$$$ will also work. But then you need to divide into 3 cases,

Left,Rightand the third case is when you go to a market from town $$$t$$$ to town $$$t$$$.But you can take advantage of the fact that distance between town $$$t$$$ and town $$$t$$$ is zero. So it can either be clubbed with the left half or right half. So both of these are valid.

Understood. Thanks a lot!

How to G ? I tried to come up with CHT but it may not work cuz i don't know how to deal with t[i]-t[j] :(

Define $$$dp[i]$$$ to be the profit when you end your journey at index $$$i$$$.

Then, define $$$ldp[i] = dp[i] + c \cdot i$$$ and $$$rdp[i] = dp[i] - c \cdot i$$$.

Then, $$$dp[t]$$$ would take contribution from prefix maxima of $$$ldp$$$ and suffix maxima of $$$rdp$$$. So a

Segment TreeorFenwick Treesuffices.Whoa, so many people know about segment trees, it's crazy... after seeing so many ACs on G I expected it to have a simpler solution, but nope...

Tbh G was simpler than C to me. It could have easily been an E, considering it was quite literally just about implementing a segtree.

Isnt segment tree somewhat basic? Specially without lazy propagation, it is a very common topic, a little bit more advanced than binary search. You can just learn to use it as a black box, it is easier to learn than dynammic programming, about the same level as dijkistra or fast pow IMO.

Agreed, especially with Atcoder providing its AC library, they do expect the participants to be aware of its usage as a black box.

If you recall, there have been problems at position

E(out of 7 problems) in past ABC where the intended solution was segtree. See ABC340E : Mancala 2I reached high purple without ever using a segtree, and now many cyans casually use it. Blackboxing data structures in later problems is not the direction I want competitions to go in.

ABC is literally built for the purpose of making participants aware of these data structures. What do you expect?

By the way, it's not like CF hasn't had such problems before.

This old thread shares my sentiment and has indeed established that it is better to perceive ABCs as AtCoder

EducationalContests with no specific limitations regarding beginner-friendliness of involved concepts. It's a pity I wasn't aware of it at the time of writing my previous comment.When I didn't AC Problem C,my brain be like:

Same, lol

Am I crazy or is double hashing really meant to fail on E? I solved it using trie in the end, but struggled with 3 tests failing my double hash.

Got AC using double hash

C >> D , how to solve C ?

comment

## include<bits/stdc++.h>

using namespace std;

int N = 1e+8; int main() { ios::sync_with_stdio(0); cin.tie(0); int n; cin >> n; vector vec(n); for(int i=0;i<n;i++){ cin>>vec[i]; } vector vec1; for(int j=0;j<n-1;j++){ for(int i=j+1;i<n;i++){ vec1.push_back(((vec[j])%N+(vec[i])%N)%N); } } long long int sum = accumulate(vec1.begin(),vec1.end(),0); cout<<sum;

}

i wrote this code but i m getting tle but this should be o(n^2) time complexity so is there any way to optimise this code

why is using 1E8 (link) instead of 100000000 (link) giving wrong ans. I submitted 6 different solutions for prob c all of them correct but used 1E8 and none passed

If you convert to int as in (int)1e8 it should work. It's the double comparison that is giving you troubles most likely.

Can someone tell me why problem F got

Corner case for K=2 probably?

Answer is 3

thanks, although my code gives the correct answer for this case :(

My mistake is when a point is in a $$$1 \times 1$$$ matrix, i will let it go to a $$$K \times K$$$ matrix. But two points in the same $$$K \times K$$$ matrix like below case, it will give WA.

If you want to build intuition for today's

G(specifically the trick of splitting the $$$|.|$$$ operator in DP problems), you can attempt ABC334F : Christmas Present 2. I also created a video and practice contest for the same.sigma problemsetCould someone tell why WA in C? T_T

Code

Join me laughing at myself for what is probably the most overengineered E solution possible boop

Really liked F, really disliked G.

Hi,this Code for F gives WA in 6 testcases,can you help?

Dunno but they will publish testcases on dropbox soon

same

You can check video editorials of E and G here

I wrote a solution of problem D in 50+ lines. Implemented Binary Modular Exponentiation, Grade School Addition and Suffix Sum. I was happy to get AC. Then found out that these guys are writing 10 lines solutions with no fancy algorithms.

Its giving me TLE for C. PLease help me optimize the code

## include

using namespace std;

int main() { int n; cin>>n; long long a[n];

}

it's O(n^2) and n <= 1e5. btw you could also write this soln with 2 for loops why did you complicate it

I wrote a 2 for loop approach...thought this approach would give me O(n)

Dislike F.

in question D it is giving wrong answer only 2 test case passed .

## include <bits/stdc++.h>

using namespace std;

## define int long long

## define fastio \

int32_t main() { fastio; int n; cin>>n; int mod=998244353; vector v(n);

for(int i=0;i<n;i++) cin>>v[i];

int ans=0,s=0,m=0;

for(int i=n-1;i>=0;i--){

}

cout<<ans%mod;

}

unable to find the error

For problem D: I am not able to find the bug, it passes for the example test cases but fails for the rest. Here is my submission for the problem, Can someone please help me debug this?

can someone please debug my submission for G? I'm not able to pinpoint where i'm going wrong, everything seems right on a high level. I'd really appreciate you taking your time for this

Maybe your mistake is when there are 2 markets in the same location one after the other, are you handling this correctly?

yessir that seems fine as well

When calculating $$$c*(y+1)$$$ you may have an overflow.

maybe not, because $$$c \leq 10^{9}$$$ and $$$y \leq 10^{5}$$$

Buth both c and y are int

i've defined a macro

`#define int long long`

You are mixing up the zero based indexing and one based indexing. Here's your almost correct solution, where I used zero based indexing for segtree as well as $$$|.|$$$ operator. After the fix, your code passed 55 testcases instead of just 14 testcase.

The rest 18 testcases fail due to a completely different issue, which is obvious to me, but I'll let you figure out and debug on your own. If you are still not able to, feel free to ping on the thread.

well i changed -inf to $$$-10^{18}$$$ and changed final answer to $$$max(0, max(dp))$$$ now it fails on last 2 tests only xD, no clue as to where they are failing

Upd : Got AC now, changed $$$dp[y] = max({left, right})$$$ to $$$dp[y] = max({left, right, p + dp[y]})$$$. Thank you so much for helping me out!! (also i love cf step, great initiative!!!)

Can someone please help- me understand my the below solution for C is failing https://atcoder.jp/contests/abc353/submissions/53411613

Can someone please help with problem E? My solution uses trie and got 41 AC, 7 WA

https://atcoder.jp/contests/abc353/submissions/53412393

Use

`long long`

instead of int. I changed your code and submitted. Now it is AC.Thanks a lot!

Ha the code for the editorial of F is relly strange to me

I thin E is more easier than D. D->60min, E->15min

why no english editorial atcoder_official

## include<bits/stdc++.h>

using namespace std;

## define int long long

const int mod=998244353;

const int inf=1e18;

int bin_exp(int n,int x,int mod) { if(x==0) { return 1; }

}

int32_t main()

{

ios_base::sync_with_stdio(false);

cin.tie(NULL);

cout.tie(NULL);

}

question number 4 solution