Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

i lost the rated contest again :weary:

As a tester, you know ... what I want :)

As a tester, I would like to say "Good Luck everyone. Have Fun!!"

As a tester, I would like to say that I haven't seen the problem yet.

Notice the unusual time !
Good luck for your maiden contest !!

The blog is written $$$4$$$ days before the contest -> bad contest

As a tester, I would like to say, good luck and have fun!

What time is it? 18:35 or 21:05?

Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

Auto comment: topic has been updated by flaviu2001 (previous revision, new revision, compare).

It seems the blog text does not use the "Contest time" feature of the blog editor, like [contest_time:1421]

I expect that this contest can be held smoothly.

As a Romanian, I am excited to participate in a round made by Romanians!

Good luck to everyone

if someone knows why there is no announcement of the Codeforces Raif Round 1??

What is "Raif" in Codeforces Raif Round 1 (Div. 1 + Div. 2)?

Clash with codechef cook-off, but do we even care?

What a great way to promote a youtube channel :D, JK,

stefdasca orz

I am a newbie so should i participate in div2 or div3 contests Please help. i want to increase my rating seriously

This contest clash with the cook-off.....They should prepone or postpone the contest whichever is possible

A lots of thanks to MikeMirzayanov and other involved organisers for rescheduling the round:)

Why can't CodeChef reschedule for CodeForces?

Great another 2AM contest

As a tester, I can say that the problems are really interesting!

stefdasca Can you leave a dot in comment section? So that we can upvote :p

That's a good time for Chinese competitor!Thanks flaviu2001!

what will be the scoring distribution? or is there a fixed scoring distribution for div2 rounds and I'm just not aware of it?

Whoah that's really early round, gives me no time to have dinner. Anyways glhf

15 min to be ready (:

Good time for a participant from India. Just after lunch full of Energy.

In question B what does "adjacent by side" means does that exclude diagonal cells?

all problems have nice music :) thanks <3

Well, cool round! Problem C was sickkkk

https://www.youtube.com/watch?v=ebgjlXyH9s8

Do watch the video editorial here

Sorry, but it was a very stupid problem B.

The Kind of day i want to wake up to almost everyday during lockdown!!!

Afternoon — CF round

Evening — Kickstart

Night — CC round Cook Off

HAPPPYYYYYYYYYYYY!!!!!!!!!!!!!!!!!!!!!!!!!

Gap between D and E is very Huggggggggge.

stefdasca, flaviu2001, koala_bear00 You guys surely have a great taste in music:) I had never imagined that I would be introduced to new music via a contest!

Am i the only person who find C much easier than B and regret for wasting too much time on B instead of solve C. Or i overcomplicated B?

Solution for C is simply,

R n-1

L n

L 2

Which takes about 15 minutes to come with idea and solve.

For B my idea is

Let b1 and b2 two adjacent cell of F and c1,c2,c3 be adjacent cell of b1 and b2.

Then i checked condition for possible cases which takes 1 hour to implement.

ConstructiveForces...xD....nice problems. Here are my submissions..hope to pass ST. Still 20 short of Specialist tho :(

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
const ll maxn=100005;
const ll mod=1e9+7;

int main(){
	fast;
	ll t,a,b,x;
	cin >> t;
	while(t--){
		cin >> a >> b;
		ll x = (a^b);
		cout<<x<<"\n";
	}

}
 

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
const ll maxn=100005;
const ll mod=1e9+7;


int main(){
    fast;
	int t,n,i,j,k;
	cin >> t;
	while(t--){
		cin >> n;
		vector<vector<char>> gr(n,vector<char>(n));
		for(i=0;i<n;i++){
			for(j=0;j<n;j++)
				cin >> gr[i][j];
		}
		int n1 = gr[0][1]-'0';
		int n2 = gr[1][0]-'0';
		int n3 = gr[n-1][n-2]-'0';
		int n4 = gr[n-2][n-1]-'0';

		if(n1==n2 && n3==n4){
			if((n1==0 && n3==1) || (n1==1 && n3==0)){
				cout<<"0"<<"\n";
			}
			else{
				cout<<"2"<<"\n";
				cout<<"1 2"<<"\n";
				cout<<"2 1"<<"\n";
			}
		}
		else{
			if(n1!=n2 && n3==n4){
				if(n3==1){
					cout<<"1"<<"\n";
					if(n1==1)cout<<"1 2"<<"\n";
					else if(n2==1)cout<<"2 1"<<"\n";
				}
				else if(n3==0){
					cout<<"1"<<"\n";
					if(n1==0)cout<<"1 2"<<"\n";
					else if(n2==0)cout<<"2 1"<<"\n";
				}
			}
			else if(n3!=n4 && n1==n2){
				if(n2==1){
					cout<<"1"<<"\n";
					if(n3==1)cout<<n<<" "<<n-1<<"\n";
					else if(n4==1)cout<<n-1<<" "<<n<<"\n";
				}
				else if(n2==0){
					cout<<"1"<<"\n";
					if(n3==0)cout<<n<<" "<<n-1<<"\n";
					else if(n4==0)cout<<n-1<<" "<<n<<"\n";
				}

			}
			else{
				cout<<"2"<<"\n";
				if(n1==0)cout<<"1 2"<<"\n";
				else if(n2==0)cout<<"2 1"<<"\n";
				if(n3==1)cout<<n<<" "<<n-1<<"\n";
				else if(n4==1)cout<<n-1<<" "<<n<<"\n";
			}
		}
	}
	
}

 

//Think simple yet elegant.
#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll  long long
#define all(v) v.begin(),v.end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define REP(i,n) for(int i=0;i<n;i++)
const ll maxn=100010;
const ll mod=998244353;


int main(){
	fast;
	string s;
	cin >> s;
	int n = s.length();

	cout<<4<<endl;
	cout<<"L"<<" "<<2<<endl;
	cout<<"L"<<" "<<2<<endl;
	cout<<"R"<<" "<<2<<endl;
	cout<<"R"<<" "<<2*n+1<<endl;
}

 

Any ideas on D?

ppl solved D with O(1)?? I keep getting wrong answer on testcase 3 too. Is O(1) method correct way? find enclosing axis then max three ways to reach point??

I just dont get why authors dont get it .. Giving such stories just confuses us ... It does not make any thing "cool about your set". It makes it worse.

Man on a serious note, I know coming up with a interesting plot is definetly tough but I dont want to know that what the hell your character wants to achieve then first grab the info of what is doing what .. and not to mention the names u come up on

I tried to hack the following solution for A by generating 1e4 test cases where 1e8<=a, b<=1e9. This solution should time out but I wasn't able to hack it. Can someone please help me why it didn't TLE ?

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    cin>>t;
    while(t--){
        int min=0;
        long long a,b,temp;
        cin>>a>>b;
        for(int i=0; i<=b; i++){
            temp = (a^i)+(b^i);
            if(temp>min){
                min=temp;
            }
        }
        cout<<temp<<endl;
    }
return 0;
}

B was so stupid because it was all about typing, annoying question imo.

Unless there's a better way of solving it..

C was just solving this test case — "abc"

Is it only me, who did B in 15 mins but struggled in C for an hour? C definitely was a good problem, but required some thinking, and took me an hour to come up with a 4 line solution :(

I missed C by one second. As soon as I submitted, the contest ended:(

we got speedforces again lol

Ohhhhhhh as a rock fan I looooove these discription! C and D were so cute！hard E,really. Like this contest!

C is one of the worst Adhoc ever, you have to keep trying combinations until you get it, no clever obervation, no analysis, just have pattern aSb and keep trying.

B is equally annoying problem, so many if conditions, pure implementation

Really dissappointed

I did two bfs on B. smh

Thanks for Tzuyu's reference in Problem A

Is it just me or was this contest too much ad-hoc?? A,B,C,D all can be implemented in O(1), it was only about solving various test cases until you can see the pattern. I may be very wrong here but I think not much real programming or barely any Data structures and algos were used in this contest :/

Damn man, why all Constructive. Imma lose a 100 rating points, where gaining 14 would've put me over to CM. It was so hit or miss.

In problem B, is diagonal movement allowed?

Well... Instead of CM i I failed system test for problem B.
Granted, I did implement it in a disgusting way, bit it passes pretests :/

Solution for C was so easy :(

crying

Ad-hoc Missiles. I'm not complaining though :))

Problem B: Consider the grid as:

A B C ...
D E F ...
G H I ...

A = Start Case 1. if B=D=0 and C=E=G=1 then STUCK. Case 2. if B=D=1 and C=E=G=0 then STUCK.

We have to make anyone of the two cases in atmost two steps.

Is this logic correct?

I think the Problem BCD are boring

Great contest, great songs. However, I should not have listened to them during the contest.

This was a total disappointment since every question was a case work, I came here to solve the problems hoping to use my implementation and coding skills and not using my case work skills. I don't know about problem E but other 4 questions were total case work. Were you guys expecting us to write only if-else statements rather than using some great concepts out there?

Was this contest unrated??

Problem E is so hard, so interesting, so mysterious. Thanks for the author's ideas and efforts!

As soon as I read the first line of problem A, I was like — Aaaah... Imma kill this round... sad not all the problems had "twice" background.

Note for C/C++ users: I lost problem D because I was doing negative % positive in the indexes of the c[] array. We know that (-1) % (6) should give 5 but in C/C++ when you do that you will get -1. So I wrote my own modulo to overcome this and I said it would be good to share it with you people:

#define mmod(a,b) ( (a >= 0) ? ((a%b)%b) : ( ( ( a + ( ((abs(a)+b)/b) * b ) ) % b ) % b ) )

Hope it help you and myself get out of Gold Nova! I mean Expert!

Problems B, C, and D were just if-else...Not so great round, Should rather call it speedforces, implementationforces, etc.

I have a Rating of 369 and I had solved 1 problem in this contest. Will my rating change as this is a Div 2 contest?