С почином) Надеюсь ты сможешь найnи вдохновение для следующего контеста раньше, чем через 3.5 года)

Офигеть, когда это номер раунда уже подошел к 600! только вчера решал триста какой-то.

I was the fifth person to register for the round.

Hopefully a round that took 3.5 years to prepare is a good round :D

Евгений, правильнее будет сказать 3.5 года, а не 3.5 лет. А так, надеюсь, что раунд будет норм. А то обычно первые раунды у людей так себе. Удачи!

Its too late for me...

Please add my handle as co-author in your announcement QAQ

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UPD1: McDic is joining as a coauthor of the contest (reasons to be revealed later)!

No chance of giving the round now! ;D

I have already risen to master, fallen to candidate master, risen to master, fallen to candidate master, ...... for totally 11 times. And I just rised to master in last round, I hope I can stay in master in this round.

Ставлю бороду Снарка, что orz 300iq вылетит из топ-10 после раунда.

How to become a tester for contests and contribute problems??

I think I know the reason why McDic added as co-author. In his last round he removed a problem from the contest beacuse of no one can solve it in testing. I thing that problem will be added this round

Slow work yields fine products,it mights be a good contest.

Looking forward to my first div.1!

Надеюсь, функция качества раунда от времени примерно похожа на функцию коньяка.

Hey, can someone tell me more about the sites which actively create contests and where I can use my knowledge of Data Structures and Algorithms?

Apart from codeforces, I know about hackerearth, codechef, google competitions, atcoder, topcoder.

Recently, I came to know about leetcode through Errichto's youtube video.

Please, can someone suggest anything about it?

Last round in Codeforces Round #5xx i hope it would be a great closing

So much red color in div2 announcement makes feel worried :| . However, let's hope that this will be balanced.

In this contest you will have to help Ujan deal with his renovation issues.

Hoping that the problem statements will be short and clear!

Is it rated?do not rate me

How many problems will be in the round?

Will be there multiple-task problems?

How many problems? And what scoring?

I can already smell mathforces XD

I will come back whenever I will become pupil

Expecting speedoforces after seeing the score distribution.

I missed this contest as I thought the contest will be started at 9:35 AM (usually CF contests will start at this time) in my time zone. However, unfortunately, the start time for contest was at 9:05 AM.

I know it was my fault that I didn't double check the time. But just as a suggestion for MikeMirzayanov and Codeforces team, it would be great if they unify the time of the contests, which are close to each other, to a certain, fixed time.

That was a great contest!

Hopefully gonna be expert for the first time. Just an awesome contest

What hacks did you guys find and what were they about?

Fun fact, Div2D/Div1B's code is almost identical to the problem 920E, even though the thing they asked for is different. I've just done 920E in practice, so I recognized it and ctrl-CVed my code. I'm betting there are many who does this.

How to solve div2 D? with 920E :))

There are k boxes numbered from 1 to k. The i-th box contains $$$n_i$$$ integer numbers. The integers can be negative. All of the integers are distinct.

I understood this as "all integers in any single box are distinct" and wasted over an hour on C before realising that wasn't the case. This could have been stated much more clearly. For example, "no integer occurs twice, not even in two separate boxes".

Div2 D Spoiler

Unbalanced round ;__;

How to solve Div1 C?

My D solution:

Let's look at the index of the $$$i$$$-th sequence($$$0$$$-indexed) where a number is skipped because it appeared as a sum of some other sequence. Let's call this $$$f(i)$$$.

There are $$$2$$$ cases:

1) $$$k$$$ is even

$$$f(i)=\frac{k}{2}+i*k$$$

2) $$$k$$$ is odd

$$$f(i)=\frac{k-1}{2}+i*k+g(i)$$$

Let's write $$$i$$$ in base $$$k$$$. If there is no digit $$$\geq \frac{k+1}{2}$$$, then $$$g(i)$$$ is $$$0$$$. Else, let's say that the least significant digit which is $$$\geq \frac{k+1}{2}$$$ represents $$$k^x$$$. $$$g(i)=1$$$ if $$$x$$$ is even and all digits representing $$$k^y$$$ where $$$0 \leq y \lt x$$$ are equal to $$$\frac{k-1}{2}$$$ and $$$0$$$ otherwise.

Using this the rest of the implementation shouldn't be too hard in an awful complexity(I'm not sure if it's ever actually achieved) but the average case is really good. You can feel free to hack this if the worst case really is too slow.

How i figured this out:

Just make a brute force and notice some patterns.

Code: 64419936

There are definitely better solutions though.

Edit: my solution is in fact too slow, it will soon be hacked. There might be ways to use $$$f(i)$$$ which aren't too slow though. The reason the solution is so slow is because finding $$$g(i)$$$ takes $$$O(log n)$$$, I'm not sure if there's a better way.

Edit 2: I realized $$$g(i)$$$ can be calculated in $$$O(log log n)$$$ but that's still too slow.

Edit 3: Actually, the slowness doesn't have anything to do with $$$g(i)$$$; i actually call $$$f(i)$$$ $$$log(n)*log(n/k)$$$ times in the worst case, which is clearly too slow.

The Idea for the DIV2 — C was somewhat similar to 1245A - Good ol' Numbers Coloring !!!

hakobdilif must have been practicing really hard since Monday!

Hi , Can anyone tell me why I got wrong on pretest 6 in Problem C ?

My Code :

long long n ; cin >> n ; bool chk=0;

if(n%2==0) cout<<2 ;
else
{
    if(n==1) cout<<1;
    else
    {
        for (int i=2; i<=sqrt(n); i++)
        {
            if (n%i == 0)
            {
                cout<<i ;
                chk=1;
                break ;
            }
        }
        if(chk==0)
            cout<<n ;
    }
}

UPD1: McDic is joining as a coauthor of the contest (reasons to be revealed later)!

What was the reason?

...

My problem in this contest is Div1D. Thanks to all people who participated!