omg tourist round!

note: the unusual duration for 6 problems

omg tourist round:)

omg tourist round:)

omg tourist round

inb4 long wait for the editorial

is it rated for you?? :)

Note the unusual time (⓿><⓿)

omg lazy editorial round!

Желаю удачи финалистам

omg tourist round!

omg yet another lazy editorial round!

stO tourist Orz But I'm going to start school this Sunday.

omg lazy editorial round!

I don't know Russian but is it okay to participate and use google translation? Anybody doing it?

Unfortunately，rating should be between 1,900 and 9,999 in order to register for the contest of div1.

orz tourist

Second time seen that there is no thanks to MikeMirzayanov for Codeforces and Polygon platforms,from the same author blog,hoping it never get repeated.

I am really afraid of tourist rounds. This is where my downfall starts

God round again.

Hope problem C is solvable this time

Why this time seperated Div1,2 round though the elimination round was Div1+2? I guess Div1 and the final are the same problem set and some easier problems were added for Div2.

Hope this won't be an overrated round again from tourist

Ahh yes a short announcement concisely thanking all people responsible for the betterment of the contest and that amazing chad energy coming off from the announcement. Is it possible this is a tourist round?

Tourist round = the hardest round

If no mention of interactive problems is made, does it always imply that they will not be present?

my first div2 contest. hope to solve at least 2 problems

Omg tourist round!

omg trash round!

omg editorial after 10 days round

First time to take part in div1 only round, in which problem A could be as hard as problem D in div2. Perhaps I'll not be able to solve any problem.

Do take note of the unusual time of the contest.

afk btw i'

6 problems 3 hours, so scare!

Yet Another tourist round

Will it be codeforces mode or ICPC mode?

HERE WE GO...

Finally a contest on a convenient time. It will be at 5pm for me.

I forgot the unusual start time, I guess this will be a no-sleep-forces round for me :')

omg Lantern Festival round!

what about Score distribution ?!

Note the unusual timing.It is 17:35 IST (UTC+5.5).

I don't know about tourist, but one piece is real!

The time of the round is perfect for Chinese participants. Usually the codeforces rounds start at 22:35 CST (UTC+8), and I have to stay up late to participate. This round starts at 20:05 CST, which is much better.

No score distribution again!

Hoping for 18 plus points...

Yup it's Tourist round and i have accepted my fate here (Don't have good past experiences despite of problems being too good).

Guts Feeling, It will be a hard round.

Did you do it on purpose or accidentally?

While attempting problems you can feel that this is tourist round.

Div 2 speedforces

Div 2 Speedforces

Requesting for codeforces feaure to contest as event when registered (Add to calender like leetcode,codechef)

So, what tourist round has?

• Stupid Div 2's problems that are just about reading English.
• NO thanks to MikeMirzayanov for Codeforces and Polygon platforms.
• NO noted unsual time.
• NO score distribution.

I know that sometimes being the best CPer allows you behave in such way, but please at least respect people.

I don't like the round anyway... huge work for 1B and 1C

Present passed 1A and 1B for 30 minutes, then sitting for 2.5 hours. So painful tourist round.

This round was a bit unbalanced. At least for div2

When 766 participants solved C, D was solved by 1

When 143 participants solved D, E was solved by 5

Speedforces or Implementation-forces

Spent 3 hours and only solved the problem A of div1. Game Over.

By the way, as a former HearthStone player, problem A and C are pretty familiar for me.

Update: Now I've find a easy solution for div1B/div2D (look at this comment ). Very easy to implement but I've not come up with this idea for 2.5 hours. Although it's logic is easy, I've used over 100 lines of code to implement it.

#include <bits/stdc++.h>

using namespace std;

char nl = '\n';
char sp = ' ';
struct exc{
	int p1;
	char char1;
	int p2;
	char char2;
	
	exc(){
		p1=char1=p2=char2=-1;
	}
	
	exc(int a,char b,int c,char d){
		p1=a,char1=b,p2=c,char2=d;
	}
};

string inw="inw";
string iww="iww";
string iiw="iiw";
string nnw="nnw";
string nww="nww";
string inn="inn";
string iin="iin";
string www="www";
string iii="iii";
string nnn="nnn";

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int _t;
	cin>>_t;
	while(_t--){
		int n;
		cin>>n;
		// wi stores the index of edges pointing from w to i 
		//(we note them as (w->i))
		stack<int> wi;
		stack<int> iw;
		stack<int> in;
		stack<int> ni;
		stack<int> nw;
		stack<int> wn;
		vector<exc> ans;
		for(int j=1;j<=n;j++){
			string s;
			cin>>s;
			sort(all(s));
			// edge (c1->c2) means: we need to remove c1 from this string, and add c2 to it.
			if(s==www){
				wi.push(j);
				wn.push(j);
			}else if(s==iii){
				iw.push(j);
				in.push(j);
			}else if(s==nnn){
				ni.push(j);
				nw.push(j);
			}else if(s==iin){
				iw.push(j);
			}else if(s==inn){
				nw.push(j);
			}else if(s==iiw){
				in.push(j);
			}else if(s==iww){
				wn.push(j);
			}else if(s==nnw){
				ni.push(j);
			}else if(s==nww){
				wi.push(j);
			}
		}
		//cancel pairs of (w->i) and (i->w)
		while(!wi.empty() && !iw.empty()){
			int p1=wi.top();
			int p2=iw.top();
			
//			cout<<p1<<":"<<p2<<endl;
			
			wi.pop();
			iw.pop();
			ans.emplace_back(p1,'w',p2,'i');
		}
		//cancel pairs of (i->n) and (n->i)
		while(!in.empty() && !ni.empty()){
			int p1=in.top();
			int p2=ni.top();
			
//			cout<<p1<<":"<<p2<<endl;
			
			in.pop();
			ni.pop();
			ans.emplace_back(p1,'i',p2,'n');
		}
		//cancel pairs of (n->w) and (w->n)
		while(!nw.empty() && !wn.empty()){
			int p1=nw.top();
			int p2=wn.top();
			
//			cout<<p1<<":"<<p2<<endl;
			
			nw.pop();
			wn.pop();
			ans.emplace_back(p1,'n',p2,'w');
		}
		//cancel cycles of (w->i),(i->n),(n->w)
		while(!wi.empty()){
			int p1=wi.top();
			int p2=in.top();
			int p3=nw.top();
			
//			cout<<p1<<":"<<p2<<":"<<p3<<endl;
			
			wi.pop();
			in.pop();
			nw.pop();
			ans.emplace_back(p1,'w',p2,'i');
			ans.emplace_back(p2,'w',p3,'n');
		}
		//cancel cycles of (i->w),(w->n),(n->i)
		while(!iw.empty()){
			int p1=iw.top();
			int p2=wn.top();
			int p3=ni.top();
			
//			cout<<p1<<":"<<p2<<":"<<p3<<endl;
			
			iw.pop();
			wn.pop();
			ni.pop();
			ans.emplace_back(p1,'i',p2,'w');
			ans.emplace_back(p2,'i',p3,'n');
		}
		cout<<ans.size()<<nl;
		for(auto [p1,c1,p2,c2]:ans){
			cout<<p1<<sp<<c1<<sp<<p2<<sp<<c2<<nl;
		}
	}
}

I got a thousand of WA and I wanna kill myself!!!

Is D graphs or just super ugly implementation? Or both?

How to solve Div2B? I needed to solve it to become cyan.

UPD: solved it

I hate the ROUND.... Unusual and confusing 1B and 1C...

time for tourist to settle down.

Is Div 2 D just a case work or is there any other methods?

this was the first time i saw an easy version and a hard version as a different problem code (the letter thingy)

Is Div 2 D just a case work or is there a elegant method?

wow, my first div1 round and I havent lost 100+ rating)))

btw could anybody please tell how to solve 1C & 1D?

Predictor: my $$$\Delta=-1$$$ (now). DON'T FST!!!

If my solution failed for a HIDDEN PRETEST, (like Pretest 2), how much time after the contest can i see that testcase? Because the contest has ended, but I cant see the testcase which went wrong.

Does the problem 2D involves case work on bitmask? or this is graph problem?

Really interesting div 1 contest, thanks!

Why did you put a trivial 12D DP (= easy idea, looooong implementation) as Div1E? It's not funny.

Was Div1C a seg tree + binary search (based on the idea that after some time some numbers are fixed)? I could not come up with easier solution.

C was too hard for me ;)

Please give some hint how to solve problem 2D.

my suggestion after this,1782,1561: avoid vk cup rounds for boring implementation problems and random performance

I just wrote the ugliest code of my life for problem D (B for div1): 192346636 Didn't really like this problem, and C was too easy, it should have been B. Other that these, I enjoyed the contest, although I would have loved to have time for E :(

My username is pretty much how i felt during the entire 3 hrs.Last tourist round for me.

So when will system test begin?

C, D and F are interesting, E is quite standard, and B is too hard for me.

If problem setter is tourist, We expect better Contest, but I can say, this is the worst contest, I have ever given. Respect to tourist but Worst Contest ever seen... Mainly Problem D

where editorial?

Can anyone explain to me how to solve problems E and F on Div2? Thank you so much.

is this approach correct for B? align the first cake with the first dispenser, such that a1-w=b1-h, then for the rest of n-1 cakes check if you can move them to the left by less or equal to (a1+w)-(b1+h) so that they are aligned with their dispensers, if ever needed to move the cake to the right the answer is "no" otherwise "yes"

Was div2D ugly implementation?? There was a huge gap no of solves for b and d. I have no idea how to even start approaching D