On last div3 server works fine with 28k registration. So 25k is going to be regular participation.

Thank you, sounds like these rounds are not too harsh and beginner friendly.

One other thing, How do you hack during a contest or after its over? After some research I found out it gives you points after successful hacks. What I don't found out how you do it?

Are there points for hacks in educational rounds?

In normal round you got -50 penalty if you passed the problem from pretests

As far as I remember educational rounds also have system tests after them, after hacking phase. So #4 is only about name of verdict, isn't it?

Why does the educational round give accepted instead of pretest passed?

Normal codeforces round has 50 penalty for wrong submission afaik.

I have a doubt. If the round is rated for only people less than 2100 rating, why is the rank calculated considering all ratings. And also, is there the case with rating change as well ??

Just a doubt ...

thank..so many doubt are now clear..one last question ...is there time for hacking phase after the normal contest like div 1,2??

SIMP!

I guess it's a fake profile photo. The photo is of an actress..

we do but let's hope that in our hearts I don't think writing hope comments will actually affect the problem statements or how strong pretests are ... just an opinion maybe someone else has a different point of view

Congrats buddy

I had +88 in previous ^^ I cant imagine how myself was when I solved C accepetedly ^^

Damn ,Stealing memes even in codeforces..

thrice !! English Language experts prefer to say 'once' and 'twice' instead of 'one time' and 'two times;' but the –ice suffix ends there, –'three times' is standard expression in today's world and not 'thrice. ... To answer the question (title) of this post, 'Yes, 'thrice' is a correct English word; and yes, it is dated! i copy this from google :)

It is the "BEST".

Same here :( I didn't feel like solving any more questions after I wasted time on 'simultaneuosly'

Same for me, I noticed after like 30 minutes.

more interesting

It doesn't require much coding, but it feels like it was put there just to trip up contestants who didn't read carefully. (I'm also biased because I got tripped up as well)

Can we print the cyclic order starting from the smallest element? Thank you

I don't see how it's a "big hint" that it's greedy. Many dp solutions can be reconstructed and printed out in a batch like this too, like Thursday's Div1B/Div2D.

Contest should be unrated as problem statement was really confusing

Your example added more confusion.

How though would it make a difference if you didnt cover up that fact?

The word "simultaneously" used in this context literally (in all sense) means that going from state 0,1/1,0 to state 0,0 is not correct

You could perhaps say, at the end of all operations, $$$x$$$ and $$$y$$$ must both be $$$0$$$. The current statement wasn't optimal (the announcement could have been part of the problem statement), but I agree that the question itself wasn't ambiguous.

Thanks emma, I just saw your submission. Damn it! It didn't occur to me to use binary exponentiation to solve it. I just gave up after reaching the expression :(

To attack every empty cells, either every row or every column contains a rook (both only if $$$k = 0$$$). Now notice that , if, say, you have decided to fill every column, you will have to group pairs of attacking rooks within the different rows. A row will consist of possibly $$$0$$$, or more rooks. $$$n$$$ rooks means $$$n-1$$$ pairs, for $$$n \geq 1$$$. Now we are almost there for the reduction. There will always be $$$n-k$$$ groups of row (or "used" row if you prefer). If less, it can be shown that there are too much pairs. If more, there are not enough. The $$$n-k$$$ rows are the "bags". The position of the rooks in the columns is important. That is why you have the forumla $$$(n-k)^n$$$ at the beggining.

At the end you also need to multiply by two if $$$k=0$$$ (because of what I said at the beggining) and... you have to take into account that you also have to ditribute the "bags" ($$$n-k$$$ of them) to the different rows ($$$n$$$ of them). This is yet another binomial coefficient.

I hope this was somewhat helpful, the editorial should be better at explaining the solution anyway :)

I cross verified, there seems no variable uninitilaized. set I am using, made sure to only erase if a value exists. That makes me believe, there is no undefined behaviour. Please point out if I am missing something. Thanks Be_dos

Can place more though

Apparently the values could arrive to 0 at different steps, as well. They made an announcement stating that during the contest.

include

include

include

include

include

include

include

include

include

include

include

include

include

include

define endl "\n"

define pb push_back

define ll long long int

using namespace std; int main() { int t; cin>>t; while(t--) { ll a,b,q; cin>>a>>b>>q; if(a==b) { for(int i=0;i<q;i++) { ll l,r; cin>>l>>r; cout<<"0"<<" "; } cout<<endl; } else if(a<b) { ll k=0;

for(int i=0;i<q;i++)
    {
        ll l,r;
        cin>>l>>r;
        ll cnt=0;

       cnt=r-max(l,b)+1;
      if(b%a!=0)
       cnt-=b*((r/a)/b-(max(l-1,b-1)/a)/b);
       else 
       {
           ll g=__gcd(a, b);
            cnt-=b*((r/g)-(max(l-1,b-1)/g));
       }

        cout<<max(cnt,k)<<" ";
    }
    cout<<endl;
}
else 
{
     ll k=0;
     for(int i=0;i<q;i++)
    {
        ll l,r;
        cin>>l>>r;
        ll cnt=0;

       cnt=r-max(l,a)+1;
       if(a%b!=0)
       cnt-=a*((r/a)/b-(max(l-1,b-1)/a)/b);
       else 
       {
             ll g=__gcd(a, b);
             cnt-=a*((r/g)-(max(l-1,b-1)/g));
       }
        cout<<max(cnt,k)<<" ";
    }
    cout<<endl;
}

} return 0; }

how about now??