Согласен.

Traditionally, the above comment is destined to be negatively rated.

Why don't you simply take a positive look at successful hacking? It gives people chances to correct their mistakes :)

it's just a small joke to entertain before the contest. the example of being optimistic. besides, it is possible to happen, so who knows.

oh c'mon, again

Wrong output for 2snd sample. Problem D We have 1 in each vertex after increasing. But we should have 0!

Ask the admin please :)

n ^ 2 * log (n) случайно проходит претесты, ждите TLE.

Придумал решение но не успел написать. Заметим, что все числа мы можем разложить по степеням двойки с показателем не превышающим 20(2^20 > 1e6). Тогда будем хранить в корнях дерева отрезков степени двойки, дающие нужную сумму на отрезке. Таким образом запрос модификации сведется к покраске вершины дерева отрезков в нужный цвет i, где i это разложение X по степеням двойки. Если текущая вершина покрашена и поступил вновь модифицирующий ее запрос, то номер покраски будет номером разложения X1 xor X2(следует из ассоциативности операции xor — (a xor x1) xor 2 = a xor (x1 xor x2) ) по степеням двойки.

Pretests seldom check for TLE.

seems that any attempts of brute-forcing the xor function work for the pretests...

расскажи поподробнее.

Я одно решение такое же ломал, тоже неудачно. Потом поставил запросы суммы — сломал. Сумма считается дольше ксора.

They add "hacks" to test cases before final judgement.

I solved E using Segment Tree.

Me too!

I got very amused when I realized that TRICK. :D

My god, I thought they were repeating the same sentence below, that the "total number of segments" was <= 10^5 =.=

Could you tell me why?

You make a segment tree for each bit.

если у нас есть клетка (X,Y) запишем ее как (S=X*10^10+Y) тогда из клетки (X,Y) можно переходит в клетки:

1)S+10^10; =(X+1,Y);

2)S+10^10+1;=(X+1,Y+1);

3)S+10^10-1;=(X+1,Y-1);

4)S+1; =(X,Y+1);

5)S-1; =(X,Y-1);

6)S-10^10; =(X-1,Y);

7)S-10^10+1;=(X-1,Y+1);

8)S-10^10-1;=(X-1,Y-1);

все "хоршие" клетки запишем в MAP затем пустим БФС из клетки S0=X0*10^10+Y0 и посчитаем минимальное количество ходов чтобы прийти в клетку S1=X1*10^10+Y1;

вот мой код : 2538101

just wait,may be will not accepted with strong tests :)

I think bfs is perfect to solve it. there are at most 100000 points you may reach

When you define the structure i think you should define the members before initializing them.

Mee too! :( , i initialized with 10^7 for max calculation, it should have been 10^9. :(

Я, когда читал задачи, пропустил этот момент. Без этого условия задача похожа на усложненную версию этой задачи...

I offer my condolences :(

bettered experience

I am also waiting for that and nothing yet.

So slow... slow... slow... ZzZZZZzzZZZzzz...

Well, I'm aware that CodeForces team work hard to maintain the system and sometimes something wrong happens and we can't handle all of them instantly. But I think it would be better for us if we get an update explaining the reason (or not) and/or just a short note, "Due to some unavoidable circumstances, rating will be updated after X hours." It will save a lot of F5 pressing and a lot of contestants won't need to sleep on the keyboard (Later one is not assured, as a lot of programmers like to sleep on the keyboard/desk first and then roll to the bed :P)

It is not gonna be rated for you, for sure :)

You can't make update segment, get sum on segment at same time using only BIT. It can be done with Segment Tree

There is always an answer, I'm not sure how to prove this. The solution is BFS. Put all vertex that have value == a to the queue. Then on each step we take vertex, increase it's value and values of it's neighbours and add to queue neighbours which values == a. Just look at my code.

I'm not sure, but I think, that when you increment the value of the neighbours you forgot that their values can become a.