No,you are the last PP submission. 0.0

There would be people who have missed the usual timing rounds because it is "unusual" for them .

It's your choice to live in Islamic country

People believing in Jesus are free at Sundays

you're not the only one. i can relate to this.

ok :)

Indeed

for me it was a sudden decrease from 56 to 48 :(

What part of "Unfortunately, this time we weren't able to come up with a good problemset for a first division contestants" do you find hard to understand?

Oh boy, I literally dieded.

I almost reached rank 200, but my computer hung at the last second and it costed me 16 points. Shouldn't have submitted the code after the contest, that was freakin painful.

they forget it :)

no un

Типо да!

Во-первых, вы написали по-русски и англоязычные пользователи не увидят вашего комментария.

Во-вторых, зачем?

How does it affect you?

Well for B, you can use discount only if you are buying 2 that should solve it :| explanation sucks

If he should buy 3 pizzas, he can buy 2 pizzas with discount and 1 pizza with coupon.

Am I the only one feeling tired of reading repeated comments like this in each contest blog ?

Can you show the code please(I guess it is n * log(n)(like this))

May be complexity isn't n*n?)

Hint: think about DFS

dsu i think

Build a graph like this : If you have socks l[i], r[i], then add undirected edge from l[i] to r[i]. Now for each connected component in this graph, you must have 1 color. So we can solve each component independently. It's optimal for each component to change colours of all nodes in that component to the colour that appears the most in that component.

Consider socks as nodes and pairs as edges. Now in this graph,paint each component with the color which have highest frequency in that component.

For input
5 3 5
1 1 3 4 5
1 2
2 3
4 5

You can split all the socks into groups (sets):
S₁ = {sock₁, sock₂, sock₃},
S₂ = {sock₄, sock₅}

All the socks in group S₁ should be painted in a single color and all the socks in the group S₂ should also be painted in a single color (probably, different from the color of S₁).

For the group S_i we count the most frequent color among the socks and paint all the rest socks in that group to that color. In my example the socks in the group S₁ have following colors:
S₁ = {black, black, red}

The most frequent color in group S₁ is black, so we paint the remaining sock (which is red) to color black. After repainting the set will look like that:
S₁ = {black, black, black}.

Maybe you have to learn how to solve A and B, too early for F ..

I used some prefix array to know in O(1) time how many graphic cards are there with power from i to j.

Then for each available power of graphics, for example for power 2, i was checking prefix array every 2 (2,4,6,8) and i was increasing max result for this power by (cards in this interval)*(current value/power i was checking(2) ).

O(nlogn)

Consider every distinct tape as the leading tape and calculate the answer for this using sieve-like technique. Consider tape a[i], then all elements in range i*a[i] to (i+1)*a[i] will be added as i*a[i]. Number of elements between a range can be calculated using a precalculated array since the numbers are less 10^6.

I did it using bfs, i made disjoint sets, and tried to find out the most repeated color in each disjoint set, and subtracted that value by size of the disjoint set.

BFS works fine. The input is just not usual. For example, the input l_i = 10, r_i = 42 may be repeated:
10 42
...
10 42

this may lead you to insert 42 into adjacency list twice:
adj[10].push_back(42);
adj[10].push_back(42);

Just use dsu.