I think it's first time, when educational round become rated. Hope problems will have short statements not as this announcement. Wish everybody luck and high rating...!

I hope not, so that this contest would really feel different.

No, the rules will be completely the same as on the previous Educational Rounds. After the contest we will build standings with Div. 2 participants only and update ratings according to taken places.

Yes!

Does the number of successful/unsuccessful hacks made during the hacking phase affect your rating like in normal Codeforces rounds?

A couple of proposals:
You may divide all rating changes by 2 (do it "half-rated") due to specificity of edicational rounds.
Add small score bonuses for successful hacks (hacking is useful for community, so it should be rewarded)

yes it is said in the blog post.

I think, someone doesn't understand trolling

Actually, we add hacks into system tests in each ER.

Of course rating change after 24-hour hacking phase, because final system test occur after 24-hour hacking phase. Hack means a wrong solution, so rating must be affect with hack.

Can you read announcement?

:p

It's experiment

Any educated person can read this blog. Sorry to say, you can't!

NO, It isn't

First, you have to beat tourist in a contest

Watch your goddamn mouth

yes, penalty = sum_time + (amount of unsuccessful submits) * 20

Long queues always sucks !!!

I think long queues are not cause of round type.

It's not our problem. So what do you want?

The answer is indeed 2. I am not sure whether I am allowed to give you an explanation in the middle of the contest.

Optimal:
              1   2   3   4     5 
Mornings      0  +5   0   0    +11
Day Balance   0  0    0   10   10
Evenings      -5  0   10  -11   0

In check-days we have non-negative balance, balance always less or equal than d.Answer: 2 deposits.

i was a bit confused too, but then i got that we can put some money to account before 0 operation comes in the evening

first preprocess all the lowest prime factors, factorials and their inverses up to 1e6+5. Then for each x find the powers of each prime in x. You can need to distribute each power in y buckets. (basic stars and bars problems) Also to accommodate negative integers see that you can have even number of negative numbers. So final answer will be:

Calculate 2^y - 1 by fast exponentiation. So time complexity: O(t * log(max(x, y)))

So don't fail the pretests

With such two diagnostic launches, the performance of the program suffers tremendously (the program is executed 5-100 times slower and requires more memory), but often it's worth it.

Please, read my post http://codeforces.net/blog/entry/55902

You must not have seen the blog post. "GNU C++17 Diagnostics" runs much slower than the standard compiler as it checks for certain errors in your code. Check http://codeforces.net/blog/entry/55902

I think it maybe a Div1+Div2 round and only to rate for Div2

diff here 32594374 and here 32596916

my idea was to use two limits mn - mx which the balance should be one of, I missed the case I meet a 0 and not update my mn to at least max(mn, 0) because if that value is below zero, some cases might add to mn which would fail my condition of failing which is mn > d it means you just can't achieve the goal!

Hope this helps!

While propagating the extra money that you are adding you need to keep in mind that you do not end up decreasing sum below 0.

Go through this once.

Let me know if you are unable to understand something

I honestly don't know I got it in the last 3 minutes so after that I started changing small things in my code (I didn't even know if the changes were going to do something or not) and then submitting the code I sent like 5 submissions in 5 minutes I was like crazy.

Anyway a few of my submissions got AC but I don't know if it's going to stick.

Try LONG LONG INT

you havent considered the smallest one

I think you've missundestood the problem.

you can go throught this ,

hope it helps

if(f[i]==0 && cur+mar>=0) cur=0,mar-=cur;
mar-=0 does not update mar.

I solved using segment tree lp. And why this line? "if(f[i]==0 && cur+mar>=0) cur=0,mar-=cur;

Here is my submission: 32585910.

First, I took the prefix sums of all a[i] in the array I called day. I also saved which days a[i]=0 in the array check so I know which days the balance needs to greater than 0. Notice that the balance on day j is day[j].

Since the balance on day j is day[j], we need to make sure that day[j] >= 0 n days that a[j] = 0. If day[j] < 0, we want to deposit money that morning. Since we want to add money for the minimum number of days, we want to add as much money as possible when we are depositing money. If we add x money to day i, the amount of money on days i...N all increase by x. Therefore, The maximum amount of money we can add on the morning of day i is D-( max(day[j] for j = i...N) ). To calculate max(day[j] for j = i...N), create an array mb where mb[i] is max(day[j] for j = i...N). Loop backwards to create mb. Now, if on day i a[i]=0 and day[i] < 0, we can deposit D-mb[i] that morning. We now need to store that days i...N have D-mb[i] more money. Create a variable add which stores how much money has been added in the mornings.

If at any point day[i]+add > D, print -1 because the account has more than D burles. If a[i] = 0 and day[i]+add < 0, deposit D-(add+mb[i]) but deposit at least 0-(day[i]+add), otherwise the account will have negative balance. So add += max(D-(add+mb[i]), 0-(day[i]+add)).

start with 2 people (1 and 2), 3 is sitting out. then start a loop and check if 1 wins then swap 2 with 3, so now the players become 1 and 3. 32591094 will tell you the logic. it becomes obvious once you do it by hand.

You just have to simulate the game, Alex and Bob start playing and Carl is spectating. So if Alex wins ( if the number is 1 ) you change Bob who lost with Carl , now Alex is playing with Carl if the number is 3 means that Carl wins, so you have to change the players again....

Do this until you find some absurd ( the person who wins is the spectator ) or until the movements finish.

for starters, your array should have a size of at least 100001, because you're storing everything 1-based.

In general, it's a good practice to add 5 elements more to your limit to avoid these problems.

I explained my solution here.

Hope this helps!

Yes, because people's solutions can still be wrong.

It depends on your rating. In your case (pupil), you won't get any points. For example, in my case (newbie), I will get exactly more than 2500 points.

Build a merge sort tree on the <value, depth> of the tree in DFS order. When merging, remove elements so that value is non-decreasing and depth is non-increasing. Then each query becomes over this range, what is minimum value if depth is maximally (depth(x) + k). Solve this by having a fenwick tree or doing a binary search.