Is it possible to set problems for Div. 3 or Div. 4 while being a Pupil or Specialist? If it is possible, how can I propose my problems for any Div. 3 or Div. 4 contest? Can you help me by providing some guidelines?

Let’s define a Binary Contest: a Codeforces contest whose number consists only of 0s and 1s. Sounds cool, right?

The Binary Contest Timeline So far, 9 Binary Contests have been held: 1, 10, 11, 100, 101, 110, 111, 1000, 1001

And today is the 10th one: 1010!

I've participated in two Binary Contests so far: 1000, 1001

If I keep up this streak, the maximum contest I might participate in is 1111. Why? Because after that, the next Binary Contest will be 10000—which is about 8889 contests later!

That’s practically a lifetime in Codeforces terms! Now the real question is—can I make it to 1111? Or better yet, can I set a legendary record of attending every Binary Contest from now on? Or maybe... if I don’t, my son or grandson can. :)) The Binary Legacy must live on.

UPD (22/05/2026): So the Binary Contests are finally here. The next four are: 1100, 1101, 1110, 1111. And realistically, these could be the last Binary Contests I ever get the chance to participate in.

UPD 2 (18/07/2026): Today is the last Binary contest '1111'. The next one will be 10000, and unfortunately, none of us will be able to participate.

Here are my two submission for problem : 1158A - The Party and Sweets

1. using accumulate function: 300047140

2. using loop: 300047827

1st one giving wrong answer. Previously, the same thing happened with sqrt, and then I used binary search to find sqrt. So after this WA, I used a loop to get the sum, and now it's giving the correct answer. So where has this accumulate function gone wrong?

I was looking for a solution to a problem. I saw this in a function of that solution:

bool good(ll mid) { ll x = sqrt ( mid ); while( x * x > mid ) x-- ; while( ( x + 1 ) * ( x + 1 ) <= mid ) x++ ; return mid -x >= k; }

how is it possible that: sqrt ( mid ) * sqrt ( mid ) > mid ? or ( sqrt ( mid ) + 1 ) * ( sqrt ( mid ) + 1 ) <= mid ?

all this sqrt has floor value.

That problem link: https://codeforces.me/contest/2020/problem/B