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.

don't cook soups with water, water doesn't belong in cooking, it's not that tasty

i see this kind of memes after every contest...

Should have warned they upgraded their servers ;D

mine actually got hacked with that time limit?

The questioner should read How not to read problem statements — Um_nik.

string::erase uses a __builtin_memmove function call (as defined in basic_string.h, char_traits.h), which is just a well optimized function. For example, check vincentgable.com/code/memcpy_memove_lab.c code which memmoves an array of size 4 million 100 times in 0.03s, so can do 1.4e10 operations in one second, which contradicts the standard rule of "1e9 operations in a second" but does not mean the answerer is wrong...

I think this gives worst case.

char s[100000];
for(int i=0;i<50000;i++) {
    	int ii = i%26;
    	s[i]=(char)('a'+ii);
    	s[100000-1-i]=(char)('a'+ii);
}

erase doesn't probably shift all elements ahead of a deletion. And it gives much better than O(n)

I couldn't pass pretest. I used the graph of x mod a. start with x = 1. x = 1, y =2. check if and y lie on the same bar i.e. x and y are less than a. keep doing this with x = 2, y = 4 then x = 4, y = 8... . The moment y reaches next bar, there are only two possible conditions: x%a = y%a or x%a > y%a ( I'm not sure about this one. I used graph to get this). In second case, if you go from x to y, then X%a first increases, becomes 0, again starts increasing. So apply binary search in this region to get the value at which X % a is 0 and this value of X is ans as you are at end of first bar. In first case, ans will be y — x. I don't know what's the mistake in it.

Ask for {(0,1), (1,2), (2,4)... (x,2x)} until x % ans > 2x % ans, now you know x < ans < 2x. Binary search between x and 2x can do the rest.

codeforces system doesnt allow you to send same code as previous

Usually it's done to disallow the hacking attempts by throwing loads of random test cases into solutions that "just don't look right but I'm too lazy to create a hack".

But yeah, there are also some disadvantages of this rule (your argument being probably the most important).

Maybe the setter just wants to let slow solutions pass, with the time limit there just so you don't hog CPU time. Consider that problems with a limit on queries correspond to situations where you have plenty of computing power for yourself, but communicating with the oracle is costly.

UPD: I forgot about non-divisibility by 3, which of course breaks this solution. Dealing with it isn't hard, but we need the leaves — or more specifically, we need to encounter at least K vertices with at least 2 backlinks.

Also, I got AC despite a small bug.

Also, this restricts you from hacking solutions which fail on subsequent test cases usually because global variables are not reset.

We have to match same characters similar to as in balanced paranthesis. Stack can be used for that purpose. Along with that keep count of number of matches. If number of matches are odd then first player is the last one to make a move otherwise second player is the last one to make a move.

Never mind found my mistake.

Sorry i explained div2 c problem

This happened with many times and really made me scared. This happens with me when I have ranklist or status page and contest ends.

Yes, this has also happened to me when I have a solution opened and the contest ends. It's just that the message needs to have "or the contest has ended" appended to it.

can you please explain how you arrived at this solution?

My solution was: Place horizontal tile on (1,1) and (3,1) alternatively. Place vertical tile on (3,4) and (1,4) alternatively.

MadLad sighted

This is hackers profile Mr_Emrul