### dzy493941464's blog

By dzy493941464, 10 years ago,

The editorial is updated.

### 447A - DZY Loves Hash

We just need an array to store the numbers inserted and check whether a conflict happens. It's easy.

### 447B - DZY Loves Strings

Firstly the optimal way is to insert letter with maximal wi. Let {wi}. If we insert this character into the k'th position, the extra value we could get is equal to . Because of wsi ≤ num, when k = n + 1, we can get the largest extra value.

So if we insert the k letters at the end of S, we will get the largest possible value.

### 446A - DZY Loves Sequences

We can first calculate li for each 1 ≤ i ≤ n, satisfying ai - li + 1 < ai - li + 2 < ... < ai, which li is maximal.

Then calculate ri, satisfying ai < ai + 1 < ... < ai + ri - 1, which ri is also maximal.

Update the answer , when ai - 1 + 1 < ai + 1.

It's easy to solve this problem in O(n).

### 446B - DZY Loves Modification

If p = 0, apperently the best choice is choosing the row or column which can give greatest pleasure value each time.

Ignore p first,then we can get a greatest number ans. Then if we choose rows for i times, choose columns for k - i times, ans should subtract (k - i) × i × p.

So we could enumerate i form 0 to k and calculate ansi - (k - i) * i * p each time, max {ansi - (k - i) * i * p} is the maximum possible pleasure value DZY could get.

Let ai be the maximum pleasure value we can get after choosing i rows and bi be the maximum pleasure value we can get after choosing i columns. Then ansi = ai + bk - i. We can use two priority queues to calculate ai and bi quickly.

### 446C - DZY Loves Fibonacci Numbers

As we know,

Fortunately, we find that

So,

With multiplicative inverse, we find,

Now,

As you see, we can just maintain the sum of a Geometric progression

This is a simple problem which can be solved with segment tree in .

### 446D - DZY Loves Games

Define important room as the trap room. Let w(u, v) be equal to the probability that DZY starts at u (u is a important room or u=1) and v is the next important room DZY arrived. For each u, we can calculate w(u, v) in O(n3) by gauss elimination.

Let Ai be equal to the i'th important room DZY arrived. So Ak - 1 = n, specially A0 = 1. Let ans be the probability for DZY to open the bonus round. Easily we can know . So we can calculate ans in (a is equal to the number of important rooms) by matrix multiplication.

So we can solve the problem in . we should optimize this algorithm.

We can find that each time we do gauss elimination, the variable matrix is unchanged. So we can do gauss elimination once to do preprocessing in O(n3). Then for each time calculating w(u, v), the only thing to do is substitute the constants. In this way we can calculate w(u, v) in O(n3).

In this way, we can solve this problem in

### 446E - DZY Loves Bridges

Let n = 2m. For convenience, we use indices 0, 1, ..., n - 1 here instead of 1, 2, ..., n, so we define a0 = an.

Obviously this problem requires matrix multiplication. We define row vector , and matrix , where bii = 1, . The answer is row vector .

Since n can be up to 3 × 107, we need a more efficient way to calculate. Let denote the matrix when m = k. For example,

Define , then we can easily find that

where denotes the identity matrix.

For an n × n matrix and a constant r, we can prove by induction that

Let α1, α2 be two 1 × n vectors, then we have

This result seems useful. Suppose we want to find , where , we have

so we just need to find , which is a self-similar problem. By recursion, it can be solved in time T(n) = T(n / 2) + O(n) = O(n).

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By dzy493941464, 10 years ago,

Hello everyone! Codeforces Round #FF(255) is coming soon! We invite you to participate in this round, the round will be held in both divisions.

In this round, the boy DZY appears again! As we all know, he loves many things. This time he also brings us many interesting problems, which are easier than the problems in last round, but he still needs you help. In return, he will present rating to you.

Many thanks to Gerald for giving us much advice and helping us to prepare the round. Also many thanks to MikeMirzayanov, who created such a wonderful platform.

The problem setters are jcvb, jiry_2 and me. This is our first Codeforces round :)

Come and join us in helping DZY again, and you will take the high rating home!

Good luck and have fun! :)

UPD

In Div. 1, scores for each problem will be 500-1500-1500-2000-2500.

In Div. 2, scores for each problem will be 500-1000-1500-2500-2500.

UPD

The contest is over. Thanks for participating.

Congrats the winners.

Division 1:

Division 2:

You can find editorial here.

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