Codeforces Round 364 (Div. 2)

There is still no formal EDITORIAL for Codeforces Round #364 (Div. 2) yet ,so I write a informal one.

Sorry for my poor English. 701A - Карты

You can calculate the sum (x) of a pair of cards.

x=2*(a1+a2+...+an)/n

For every card ai,find a card ak which ai+ak=x and k is not used.

The n is small,so you can simplely write a o(n*n) solution(also there is a o(n) one,o(n*n) is enough).

My o(n*n) solution[submission:19328905]

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701B - Клетки не под боем You can just record the number of rows and columns that is not under attack.

If r rows and c columns is not under attack there is r*c cells that is not under attack.

It's o(m).

My o(m) solution:19330540

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701C - Они повсюду

You can just use two pointers

There are x types of Pokemon

if [i,dp[i]-1] has x-1 types of Pokemon and [i,dp[i]] has x types of Pokemon

dp[i+1]>=dp[i]

so it's o(n)

my o(xlogx+n) solution:19332828

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701D - Как можно быстрее

It's a math problem.

let d=ceil(n/k)

Just look at the bus.

it goes like that:

<<<<<<<<<

>>>>>>>>>>>>

<<<<<<<<<

  >>>>>>>>>>>>

     <<<<<<<<<

     >>>>>>>>>>>>

The long lengths are the same(x) and goes to right.

The short lengths are the same(y) and goes to left.

It goes to right and put the old students down and does to left and welcome new students.

so x+y:x-y=v2:v1

it goes for d xs and (d-1) ys

so:

1:d*x-(d-1)*y=l

2:x+y:x-y=v2:v1

3:ans=(d*x+(d-1)*y)/v2

so ans=l/v2*((d*2-1)*v2+v1)/(v2+(d*2-1)*v1)

It's o(1).

my o(1) solution:19334768

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701E - Соединяя университеты

Let's take 1 as the root of the tree.

Record the depth of every vertex(depth of 1 is 0).

Record the number of schools of the subtree of every vertex.

The sum of the depth of the lca of every pair of school should as small as possible.

Consider 2x schools of the subtree of vertex y(may be 2x is not the number of all the schools of the subtree of vertex y)

every depth of lca of the 2x schools is at least the depth of y .

If a subtree of the son of y has k schools in the 2x schools(not in all the schools).

1:if the k of every son is always mink<=x

you can match the schools to make every lca y

2:if for son z,mink>x

match (2x-mink) schools of the subtree of z with the schools not in the subtree of z but in the subtree of y

just look at z like look at y

so,it's o(n)

my o(n) solution:19344918

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701F - Разрыв отношений

Any solution?

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Other solutions are welcomed.

If you find any mistakes,tell me.

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Sorry I don't know how to use format.Can you tell me how to do that?