all newbie people need that :'(

Well, sort of regular.

A usual round at unusual time!

For someone like you I don't think you need luck

Great way to learn. Best of luck.

9 months ago I was in your position, and now I'm here (further than I expected). Good luck!

k0sk

That's my boy, well done. Dude, wish me luck next contest.

i think that is fake xD

Thanks for your support everyone. :) ♥

What the hell, man!

When will you bring your army to take tourist's rating?

جرا ايه يا عرص؟ طلعت ال٦٧ يا حيلتها؟ ما تجيب جيشك وتيجي تنفخهم

ROFL... https://en.wikipedia.org/wiki/Abdel_Fattah_el-Sisi

Best of luck :)

No, because tourist now has a 4000+ of rating. I respect that!

Kinda related, I wonder if it would be difficult to allow for div1 users the option to register to div2-only rounds as ranked competitors. If I'm not mistaken, it's not allowed because div1 users might run out of problems to solve, but if they wanted to register as ranked competitors into div2-only rounds, they'd have to accept that possibility, yet could still get the fun of competing in a ranked round. Many div1 competitors usually can't solve any problems past the first 3 anyway (I know I can't).

Probably because the F and G problems in this round would be way too easy as Div1 D and E.

Is someone going to answer his question!?

Except this one has 7 problems and 2.5 hours and is only rated for Div.2.

Здравствуйте. На Codeforces я всего 10-11 дней,по-этому не знаком с понятием разбалловка — объясните пожалуйста.

a div1 guy with fake acc

maybe for easy problems :-"

I guess maybe they found those problems were too easy for div.1. So, 7 problems for div.2. lol Okay,frankly, maybe just for fun.

you expected rank can't be computed before the contest is finished, because it depends on the people who make at least one submission

Try the NBHEXT extension for Chrome. It is not completely accurate because systest, but you get a rough estimate.

5 min? hah

I guess it's because it was added to the contest list earlier?

Чаще всего правильных ответов много.

His submissions look suspicious though. He got B and C accepted in the same minute, and it only took him 8 minutes to solve G.

By the way, I think that the problem statements are pretty formal and the next challenge for the DeepMind team could be trying to compete in the programming contests.

Correct me, please, if I am wrong and there is something about artificial neural networks that doesn't allow them to compete here.

I wanna see AlphaGo vs tourist face-off assuming AlphaGo is AI and not a team/user.

Loop through all the cells. If k is evenly divisible by the number in the cell, and cellnum*m*n is greater than k, then do a BFS across adjacent cells that are greater than or equal to cellnum. If we have processed k/cellnum nodes, immediately print YES, print a matrix such that any cell examined in this BFS is given number cellnum and all other cells are 0, and then return. If the loop completes without finding anything, print NO.

Answer can not exceed the number of connected components. So define answer=number of connected components. If there is a cycle in a connected component then decrease answer.

Consider each connected component separately.

If the connected component is a tree, the answer is 1. This is because there are only V - 1 edges — not enough edges for all vertices. It is not more, we can root the tree and direct all edges from parent to child. Only the root will have no incoming edges.

If the connected component has cycles, the answer is 0. To see this, take an arbitrary cycle and direct the edges in it the obvious way. Now you can throw out some edges (and direct them randomly) to get a tree for every vertex in the cycle. Direct all those trees from parent to child and you'll see that all vertices have incoming edges.

So, for each connected component, check if it is a tree, and if it is, add one to the answer.

Alternate (greedy) solution:

• count undirected edges as "inward" edges
• use a min-heap to store number of incoming edges to each node
• assign "inward" edges by popping off min heap

at the end, the answer is the number of nodes without any "inward" edges (http://codeforces.me/contest/659/submission/17054791)

Consider a connected component If it has a cycle it is possible to assign directions to edges such that no vertex has zero incoming edges
If it has no cycles there will be exactly one such vertex
So just count connected components which do not have cycles

if you don't want to find cycles, then just count the number of edges in each connected component, you can do it by summing up the degrees of all nodes in that component then divide it by 2. if the number of edges is equal to number of nodes minus 1 then it's tree and there's no cycle, otherwise there's a cycle

5 2 Ivanov 1 763

Andreev 2 800

Petrov 1 800

Sidorov 1 800

Semenov 2 503

3 1
a 1 1
b 1 1
c 1 0
ans a b

3 1
a 1 1
b 1 0
c 1 0
ans ?

3 1
a 1 0
b 1 0
c 1 0
ans ?

3 1
a 1 0
b 1 0
ans a b

4 2 Ivanov 1 800 Andreev 2 763 Petrov 1 800 Semenov 2 503

some people wrote if (list[i][0] == list[i][2]) printf("?"); else print answer, which is wrong (should be list[i][1] == list[i][2])

if (a[i].size() && t == a[i].top().y){
      cout << "?" << endl;
}

Он проверяет размер очереди

I kept trying to solve D for two hours. Did you solve it? . I couldn't get my algorithm right for finding whether a line segment will properly cut any segment of the polygon when extended in the direction given.

After some minutes of thinking how to solve it in a nice way, I give up and seek for pre-written geometry library that do the job for me (I hope there's no bug inside tho).

For those who solve it without a generic written code, how do you guys managed to write decently? (e.g. without using lots of if)

chameleon

https://en.wikipedia.org/wiki/AlphaGo

By seeing submission gap between B and C, probably a team.

F is already explained, as regards G I did dp:

dp[i][0] — the number of ways where you remove something from position i and the remaining height is not greater than h[i+1]

dp[i][1] — the number of ways where you remove something from position i and the remaining height is greater than h[i+1]

Obviously if h[i] <= h[i+1] then dp[i][1] = 0;

Hope that helps.

Edit: If not, here is the submission implementing that logic: http://codeforces.me/contest/659/submission/17059428

I think it can be solved in a greedy manner, using DSUs. Start processing the cells in decreasing order of their values. When you process a particular cell, check its neighbouring 4 cells. The cells among those 4 which have their values >= value of current cell, are combined with the current one using DSU. After each combination, check whether that set has the required number of cells (if value of the cell is v, the set should have at least k/v elements). The moment this occurs, you have found the answer. Assign exactly k/v cells of this set to v, and assign all remaining cells in the grid to 0.

Lol, payback. There should be an achievement for a thing like this.

That's so me in past contests, lock one problem, read others' solutions and realize mistakes in mine...

Greedy was okay.. Can you post your code ?

You just made 100100 sized array to mark used toys, but that can be from 1 to 1e9. Your update will fail if used toy is greater than 100100.

Your code->

const int N = 100100;

int n, m;
bool toys[N];

Не нужно ничего раскладывать. Все гораздо проще.

Достаточно лишь отсортировать значения в ячейках по невозрастанию и пройтись по ним поддерживая СНМ.

Каждый раз, когда добавили чиселку x, проверяем, если k делится на x и размер текущей компоненты не меньше k / x, то выводим ответ (обойдя компоненты дфсом, чтобы получить необходимое количество клеточек).

Если не получилось найти ответ — выводим NO.

Собсна, все.

(n-4)/2

Or not. TLE lol