div2 round witih more div1 testers. not good.

Contest with short statements. I hope we'll enjoy.

Looking forward towards a positive delta! Wish Every body good luck.

Div.2 round with "3250" point's problem, quite interesting.

As a participant, I will appreciate the work of authors and testers. Thank you for the contest!

I hope being able to write short code for short statements , cuz short statement do not mean short code or being easy. However short statements is always good , good luck to everyone ! Thanks for authors )

Ha felt like forever since the last contest.

Excited to take part in this contest.

Short statements -----> Good contest o(*^ w ^*)o

Good luck everyone!

.

The author asked us to read all problems. we will do. Good luck everyone!

No tester comment? Bad round.

No 'As a tester' comment, interesting

okwedook Orz

hoping for easy A this time ;)

Good Luck or everyone

thanks for this "....to create some good problems and clear short statements"

Wish i could do will in this Round

All the best for all participants

Good luck for all of us

Let's See How It Goes

Looking forward to be sad like him

Again hard 'A' :(

Div √2

Thanks for the toughest contest I have ever seen.

In B, if 'x' can be infinitely large print 0. anything divided by infinity will give remainder 0. So why can't infinity be answer for each case?

Yeah True! Trust Noone Even the contest.

Hardest A, A took me longer than B and C combined.

Problem E — F ratings are different from the blog post. Neither that it matters to me. xD ... cos I solved none of it.

Toughest A or Easiest C?

• 13 minutes for D
• 26 minutes for C
• 10 minutes for B
• 45 minutes and 5 wrong answers for A.

Weirdest contest for me.

how to solve E?

RIP my rating. That first problem was brutal and then I struggled on the second after :(

1826B - Lunatic Never Content copied from https://codeforces.com/gym/102035/problem/I

How to optimize in Problem E

include <bits/stdc++.h>

using namespace std;

define int long long int

int32_t main() { int t; cin>>t; while(t--) { int n; cin>>n; int a[n]; for(int i=0;i<n;i++)cin>>a[i]; vectorv; for(int i=0;i<((n+1)/2);i++) { int k= abs(a[i]-a[n-i-1]); if(k==0)v.push_back(-1); else v.push_back(k); }

int ans=-1;
    int cnt=0;
    for(auto x:v)if(x!=-1)cnt=1;
    if(cnt==0)
    {
    ans=0;
    cout<<ans<<endl;
    }
    else if(cnt!=0)
     {
         vector<int>v1;
         for(auto x:v)if(x!=-1)v1.push_back(x);
        int ans=v1[0];
        for(int i=1;i<v1.size();i++)ans=__gcd(ans,v[i]); 
        cout<<ans<<endl;
     }

}

}

I am getting WA at 4 testcase please help me asap

So apparently every single one of the testers saw A and solved it and though that it's a good idea to put something like that in the contest?

Difficulty jump from D to E is too much. Up till task D it feels like div 3

Brutal Div2A

Any hints for C and D

great contest! thanks

Codeforces Round 657 vibes

Can someone tell where can I find some similar problem to practice :(

F is an absolute joke. Just query y=0, x=0, y=240x

Very cool problem

Problem A is very hard for me and i can't even solve A in the contest...

can anyone please give hints for problem E.

C is easier than A,,,Hardest A ever...

Хотелось бы поменьше задач вида А-С и побольше вида Д-Е. Вцелом контест понравился, спасибо за труд)

Was D a dp problem ? Hope to become specialist this time.

As a newbie, I started doubting myself when I was unable to solve A, but looks like it is really a tough problem. Now waiting for someone to post it's solution.

Problem B is from past contests.

Problem Link: https://codeforces.com/gym/102035/problem/I

Did anyone have the same problem as me in C? I figured it out pretty quickly and tried using a sieve to implement it. Kept on getting TLE on 4. preset tho...

Yay, Pupil!!!

For me, A was quite impossible, I think I had a decent idea, but couldn't implement it. Spent at least forty-five minutes attempting it just to get nowhere. Similar to the last contest, A was really difficult to understand resulting in A being difficult for me and I ended up spending >15 minutes re-reading it over and over. Although, both B and C were far easier, all A, B, and C seemed to be fairly tricky.

Anyways, thanks for the contest, the problems were really enjoyable. I need to think about A more!

A: Brute force for all possible numbers of liars from 0 to n.

B: To make a[1]%x==a[n]%x we must have x is a divisor of abs(a[1]-a[n]). So the answer is the gcd of abs(a[i]-a[n+1-i]) for 1<=i<=n/2.

C: If m is a divisor of n, m can remain the same after voting, otherwise m must be decreased. So if m < the minimal prime factor of n, m will be decreased to 1.

D: Let the indexes of sights we pick are i, j, k from left to right, then definately we will let L=i, R=k. So we just need to find the minimum of b[i]+b[j]+b[k]-(k-i)=(b[i]+i) + (b[k]-k) + b[j], where i<j<k. We can do this by pre-calculating the prefix-maximum of (b[i]+i) and the suffix-maximum of (b[k]-k).

E: If model u can be ordered before model v, for all i we have r[i][u]<r[i][v]. So we can build a DAG (directed acyclic graph) in O(m*n^2), and we can solve the problem by toposort and dp. Naive approach will get TLE and we can optimize this solution to O(m*n^2/w) by bitset, which could pass the pretest (and hopefully system test).

PS: Is there any O(m*n) solution of E? I think there must be something faster than O(m*n^2) but I can't come up with it.

Problem B was exanctly this problem

Is there any solution for problem E that is faster than O(n^2*m/w), like O(mn+n^2) ? I tried to find a O(m*n) solution to compute all pairs of i,j that j can go after i, but failed.

B was google-able, solved it after finding this which is the first link when you google "largest number that gives same modulo for 2 numbers"

870A — Trust Nobody with Explanation

https://codeforcer.blogspot.com/2023/05/trust-nobody-870a-solution-in-cpp.html

I hate E

You can solve problem 1826E - Walk the Runway in $$$O\left(n^2 \times m\right)$$$ in C++ without bitsets. Based on naive $$$O\left(n^2 \times m\right)$$$ approach, you need to apply next optimizations one-by-one for getting 1872 ms.

Accepted 1872 ms submission

Optimization 5 for getting 1200 ms

good luck whoever is participating.

The problem C is copied from https://codeforces.com/gym/433264/problem/F

can problem B be solved using binary search ?

I'm stupid and had to do a bunch of random tests for B to find out the set of possible answers for a mod x = b mod x are the factors of abs(a-b). From there, I didn't realize that it's literally just equivalent to GCD XD

Got TLE 8 on that sadge

It says it's rated but when I open my account it's viewed as unrated what's going on? (this is my first contest in a long time so idk how these work.)

For E), I had the following DP recurrence:

and I think it runs in

Never would have thought to use bitsets though, I was trying to optimize it to either

because it looked suspiciously similar to longest increasing sequence.

10134 — 6367 = 3767 participants (37.2%) can't solve a single problem!

In problem D, I made this 204647760 during the contest and after about 20 minutes I realized that it can be hacked with a test case like this:

1

100000

1 1 2 3 4 5 6 7 .....

The reason is that I use -1 to check if I visited a state in dp or not and the above test produces -1 a lot which mean the dp will recompute states which was computed before, so I made another submission to avoid this mistake.

But after the contest, I submitted the first solution again and it passed the tests of the problem with about 46 ms , I think that this test should be added to main tests!

I liked the contest, it was nice. But E is so bad

My E solution with bitset took 2800 ms :(

I found a tricky way to exclude those pairs without ordering fast: if a[i]<b[i] for evety i, then sum(a)<sum(b), min(a)<min(b), max(a)<max(b). Problem E, a 592 ms solution without using bitsets.

Problem B was written in GYM before:

B- https://codeforces.com/problemset/gymProblem/102035/I

Ratings updated preliminary, it will be recalculated after removing the cheaters.

problem A so bad statement

Can D be done using ternary search https://codeforces.com/contest/1826/submission/204648453 Or binary search https://codeforces.com/contest/1826/submission/204631760

good contest, problems were quite interesting, i am able to prove my solutions of a b and c!

Problem A.. Uff..

Has anyone come up with a dp solution for problem d？

My code can pass the example on my computer but failed when testing. Can somebody help me ? My code

I can't solve some problem during contest, but after contest I had solve.

All problems are Good. I loved it.