«Надейся на лучшее, готовься к худшему!»

haha good luck

i'm looking forward to join contest . i hope i will be succesfull

Yes, I'm agree

My congratulations!

finally ,there he is

مبروك

What does rated mean? Excuse me but I'm new to this website.

Stop downvoting! I don't want to have negative contribution.

What is misleading about his response?

"Please note, that if Bob doesn't make any breaks, all the bar will form one piece and it still has to have exactly one nut."

I don't find it misleading. This sentence means that it's invalid way if we don't make any breaks and bar hasn't nuts. And this sentence would be excess if it was guaranteed that bar has at least one nut.

Everyone who asked similar question got this sentence as response. By the way, it can be with the same probability my or GlebsHP's response.

I also failed system tests because I was mislead by the exact same answer you got. I'm still not quite sure what I think about it, it would've been easy enough to just answer yes or no to the question, since people were confused and asking directly, but I guess I just also learnt from it, and from now on will always include an if statement for special case like that if I'm in doubt whether it's in the test cases or not.

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I hope you will be successful in this contest

Here you go!

The same thing. Never passed it before :D

You must be really excited about Codeforces :)

It's only a minor contest for most people

500 — 1000 — 2500 — 1500 — 2500

I think it depends on perspective. I took 18 minutes to solve C, but 50 minutes to solve D. Mainly due to my inexperience with STL. While the idea for C is a bit less intuitive than D, the implementation is relatively more simple.

Let us note by T[i] the transfer from i to i + 1 in some solution. Then the following equations must be satisfied:

A[i] + T[i - 1] - T[i] = 0.

for i = 0, 1, ..., n - 1. Suppose that we know T[0] = x then T[i] = A[i] + A[i - 1] + ... + A[1] + x.

We can manipulate x and we want the maximum possible number of T[i] to be 0. The best choice for x is the opossite of the most frequent number in the pref_sum table.

So the answer is n - k where k is the frequency of the most frequent number in the sequence A[1], A[1] + A[2], ..., A[1] + A[2] + ... + A[n].

incorect:

9
0 0 0 0 1 -1 0 1 -1

correct ans: 2
your ans: 4

test:

1 -1 0 0 1 -1 0 0 0

answer is 2

Did you see that it was a circular array? :)

Well, I passed pretest, though I don't know if my answer is correct or not. Here, at first we take cumulative sum in another array, say, cum[i], where cum[i]=summation of initial array from 1 to i. Then, we check for the number which has maximum occurence in the cumulative array.Let the number of times it has occured be max. The answer is n-max.

same here!

We know as much as you!

Welcome to Codeforces, land of the trolls.

You maybe are so excited that you have found the solution, that you don't think about tricky test cases. You just go strait into the code. It's a common mistake. Just make sure you spend 10 seconds thinking about data types that you are going to use in your code when you solve a problem. Spending very little time double-checking your solution is worth getting wrong answer, specially in CF that tricky test cases are not usually included in pretests.

Your solution is N^2 in worst case (e.g. sequence 1 2 3 4 5 ... N)