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;
}