Да.

ДА, если DDOS-еры не проснутся

We never sleep.

once again a required hope prevailed UPD : And a few unspoken : extended queues, clear stmts, etc.

Now it is in English!

Wasn't it because of the temporary rating change rollback?

Must be some bug, usually only standalone Div 2 rounds are rated for <= 2100. Endagorion, is this normal that I could register for both Div 1 and Div 2 rounds?

Registered before becoming expert

Div. 1 and Div.2 editions are open and rated for everyone

Is that a threat?

Hope not, I kinda like the problems

2 hour hacking phase for DDosers only

I believe one possible reason could be codechef lunchtime today?

Of course

Why do you ask such question,DST？

It was mentioned in blog.

Пожалуйста, подвиньте раунд на 30 минут, т.к. пересекается с уроком физики в СУНЦ МГУ, мой преподаватель не увидит меня на уроке и обидится:(

Now, it has become one of the cliche comments. Every one of us wants the round to be safe from DDoS attacks.

Не думаю, что можно зарегаться на оба раунда одновременно.

Same here. I know a good input to hack, but unable to submit it :(

I agree. I wanted to submit a solution in the very last minute but couldn't because site went down :(

What's the solution for C?

same, here

6 2 1 2 4 8 2 4

I would be very sad if that happens :(

Try for smaller P?

5 2

I thought that answer is always in a small range, I could be wrong though...

I SOLVED IT OMG I AM SO PROUD

ok anyways so basically if you write n as sum of k (2^i+p), then

n = (2^a + p) + (2^b + p) + ... + (2^x + p) right

n = (sum of k "power of 2s") + kp

sum of k "power of 2s" = n — k * p

So what remains is to check whether n — k * p can be written as k "power of 2s" (with duplicates). This is kinda well known lmao. But basically

def check(x, k):
    if k < x: return False
    if bin(x).count('1'):
        # counting the number of set bits in binary of x
        # This is because the binary expression is the unique way of representing x as sum of power of 2s with minimal number
        return False
    return True

Sorry if you don't code in python but I'm sure you can code the "count set bit"

Now loop through all k (while n — k * p >= 0) and check that

Code: https://mirror.codeforces.com/contest/1247/submission/63459624

(still proud plz upvote minecraft is the best)

have u declared an array for counting in every test case??

Firstly,we use backet sort and convert the input to backet[number]={count}.

And do next for all $$$i(1 \le i \le 10^5):$$$

• find the minimum $$$p$$$ that satisfies $$$i*p = x^k$$$(we can use prime factorization for this.)
• count pair $$$(i,p * v^k)(1 \le v \le 500$$$ thanks to $$$k \ge 2$$$ and $$$N \le 10^5)$$$

We have to use valid pre-calculation for doing this.

9 4. ans will be -1 and not 2

Something like 11 5.

Basically, checking if N — i * p <= i.

How to solve DIV1 — C?