Maybe that's because you checked it right after it was created? Your comment is just an elaborate way of saying well known "First!"

Why it is always necessary to make such standard comments?

i'm pretty sure this would classify as cheating.

i don't

Nope

How about Codezilla?

that would be called abuse

stop being a hater and get good

i call bullshit

... on codeforces

My stomach says go eat your launch.

10 min after start of contest for next 20 min

Registration closes 5 min before contest. Wait for late registration (10 min after start of contest)

This is speedforces. If you still havent registered dont even bother giving this contest.

No, it's actually good to have interactive problems, we will get used to it, but the difficulty should be not this hard for DIV 2. I personally have never solved an interactive problem here. But i think it is fun.

if(problem.interactive()) binarySearch.learn();

There were 5 problems.

I guess we have to consider first 45 fibonacci numbers but then when we find between which two fib's is A I don't know how to find it

First, find p such that 2^p < a <  = 2(p + 1).
"? 2^p 2^(p+1)" // At max 30 queries.
Let x=2^(p+1).
Binary search on (x/2,x]. // At max 30 queries.
"? m x" // At max 30 queries.

Take care of a = 1.

If you consider every (x, 2x) pair from x = 1 to x<=1e9, the answer would lie in the first pair where it will answer 'x'. This is because if a > 2x, it will always answer 'y'. This would take 29 queries.

Let's call the first found segment (X, 2X). a would lie between X+1 and 2X. So, we can binary search with low = X+1 and high = 2X asking query (X, mid) in each iteration (If a is greater than mid, we will get 'y' otherwise 'x'). This would take 30 queries at max.

Does anyone have a better solution by the way?

Well for B, implement stack, and if you erase character, it will be the value s[i] is same as top of stack.

As for D implement log(n) answer, imagine you have some x and y (x<y). If answer is x then answer is in segment [x+1,y]. Then you can notice that incrementing by power of 2 (like (1,2);(2,4);(4,8);(8,16)..) you can find if answer lies between some of these segments. And special case is when a=1. At least that's how I wrote it out.

Got 4 unsuccessful hacks because of that. People who used std::erase() got it in ~750ms on the worst case(Palindrome of length = 10^5 using 'a' and 'b').

Codeforces systems are fast, really fast :(

A lot of N^2 solutions passed B. Compiler magic works wonders.

Used binary search. Twice.

What was the logic?

a = 8

Yes, but do it until (x,y) becomes y-x=1

Yes, I got ac asking

1 2

2 4

...

2^x 2^x + 1

And if answer on query 2^x 2^x + 1 is x, then you should do binary search on [2^x; 2^x + 1].

Wont it time out because string erase is linear? I guess.

Let's run dfs. If number of leaves is at least k than we can find k cycles, otherwise we can find a path with length at least n / k.

I didn't get AC, but I think it should work

Build dfs tree from some vertex. If its depth is at least n / k then we found a path. Otherwise by pigeonhole principle tree has at least k leaves. Let's choose k leaves. Our tree is a dfs tree, so all edges not in the tree go from some vertex to its ancestor. Each degree is at least 3, so each leaf has 2 edges going up. Each of these edges forms a cycle. If both of these cycles have length divisible by 3, then we can build a cycle with both of these edges and its length will not be divisible by 3.