488A - Гигабашня

The answer b is very small (usually no larger than 10), because one of a + 1, a + 2, ..., a + 10 has its last digit be 8.

However, b can exceed 10 when a is negative and close to 0. The worst case is a =  - 8, where b = 16.

Anyway b is rather small, so we can simply try b from 1, and check whether a + b has a digit 8.

488B - Коробки конфет

Let's sort the four numbers in ascending order: a, b, c, d (where x₁, x₂, x₃, x₄ are used in problem statement). So .

With some basic math, we can get a: d = 1: 3 and a + d = b + c.

Solution 1:

If n = 0, just output any answer (such as {1, 1, 3, 3}). If n = 1, just output {x, x, 3x, 3x}, where x is the known number. If n = 4, just check whether the four known numbers meet the condition.

If n = 2, let x, y denote the known numbers (x ≤ y). No solution exists if 3x < y. Otherwise we can construct a solution {x, y, 4x - y, 3x} (certainly other solutions may exist).

If n = 3, let x, y, z denote the known numbers (x ≤ y ≤ z). No solution exists if 3x < z. Otherwise the solution can only be {x, y, z, 3x}, or {x, y, x + z - y, z}.

Solution 2:

The known numbers are no larger than 500, so all numbers are no larger than 1500 if solution exists. We enumerate x from 1 to 500, y from x to 3x, then {x, y, 4x - y, 3x} is a solution. For each solution, check if it matches the known numbers.

Solution 3:

If n = 0, just output any answer (such as {1, 1, 3, 3}). If n = 1, just output {x, x, 3x, 3x}, where x is the known number. If n = 4, just check whether the four known numbers meet the condition.

Otherwise, we can enumerate the 1 or 2 missing number(s), and check if the four numbers meet the condition.

487A - Бьём чудовище!

It is no use to make Yang's ATK > HP_M + DEF_M (Yang already can beat it in a second). And it's no use to make Yang's DEF > ATK_M (it cannot deal any damage to him).

As a result, Yang's final ATK will not exceed 200, and final DEF will not exceed 100. So just enumerate final ATK from ATK_Y to 200, final DEF from DEF_Y to 100.

With final ATK and DEF known, you can calculate how long the battle will last, then calculate HP loss. You can easily find the gold you spend, and then find the optimal answer.

487B - Лента

We can use dynamic programming to solve this problem.

Let f[i] denote the minimal number of pieces that the first i numbers can be split into. g[i] denote the maximal length of substrip whose right border is i(included) and it satisfy the condition.

Then f[i] = min(f[k]) + 1, where i - g[i] ≤ k ≤ i - l.

We can use monotonic queue to calculate g[i] and f[i]. And this can be implemented in O(n)

We can also use sparse table or segment tree to solve the problem, the time complexity is or (It should be well-implemented).

For more details about monotonic queue, you can see here

487C - Последовательность частичных произведений

The answer is YES if and only if n is a prime or n = 1 or n = 4.

First we can find . If n occurs in {a_1,…,a_{n-1}} in the prefix product sequence 0 will occur twice which do not satisfy the condition.

So a_n must be 0 from which we know a_1a2... a_n - 1 = (n - 1)!. But for any composite number n > 4 we have (See the proof below). So we can know that for all composite number n > 4 the answer is NO.

For n = 1, 1 is a solution.

For n = 4, 1, 3, 2, 4 is a solution.

For any prime number n, let a_i be . If there are two same number a_i, a_j. Then we get i / (i - 1) ≡ j / (j - 1) which leads to i ≡ j, which is a contradiction. So all n numbers will occur exactly once. And this is a solution.

Also, we can find a primitive root g of n and $g^{0}, g^{1}, g^{n-3}, g^{3}, g^{n-5}, \cdots } is also a solution.

Proof:

For a composite number n > 4 it can either be written as the products of two numbers p, q > 1.

If p ≠ q, then we immediately get pq|(n - 1)!.

If p = q, note that n > 4 so 2p < n, we have p²|(n - 1)!

So n|(n - 1)! always holds which means

487D - Ленты конвейера

This problem can be solved by classic data structures.

For example, let's try something like SQRT-decomposition. Let's divide the map horizontally into some blocks. For each grid, calculate its destination when going out the current block (or infinite loop before going out current block).

For each modification, recalculate the affected block by brute force. For each query, we can just use the "destination when going out the current block" to speed up simulation.

Let S be the size of a block, then the time for each modification is O(S), for each query is O(nm / S), since at most O(nm / S) blocks, and at most 1 grid of each block are visited.

The total time complexity is O(nm + qnm / S + pS), where p is the number of modifications. Let , the complexity can be the best: .

This task can also be solve by segment tree. The time complexity is , or , depending on implementation.

487E - Туристы

First we can find out all cut vertices and biconnected components(BCC) by Tarjan’s Algorithm. And it must form a tree.

From the lemma below, we know that if we can pass by a BCC, then we can always pass any point in the BCC.

We use a priority queue for each BCC to maintain the minimal price in the component.

For each modification, if the vertex is a cut vertex, then modify itself and its related BCCs’ priority queue. If not, modify the priority queue of its BCC.

For each query, the answer is the minimal price on the path from x (or its BCC) to y (or its BCC). We can use Link-Cut Trees or Heavy-Light Decomposition with Segment Trees.

To be more exact, we can only modify the father BCC of the cut vertex in order to guarantee complexity(otherwise it would be hacked by a star graph).When querying, if the LCA of x and y is a BCC. Then the father of the LCA(which is a cut vertex related to the BCC) should also be taken into account.

The time complexity is or .

Lemma: In a biconnected graph with n ≥ 3 points, for any three different vertices a, b, c, there is a simple path to from a to b going through c.

Proof: Consider a biconnected graph with at least 3 vertices. If we remove any vertex or any edge, the graph is still connected.

We build a network on the graph. Let's use (u,v,w) to describe a directed edge from u to v with capacity w. For each edge (u,v) of the original graph, we build (u,v,1) and (v,u,1). Build (S,c,2), (a,T,1) and (b,T,1). For each vertex other than S,T,c, we should give a capacity of 1 to the vertex.

In order to give capacity to vertex u, we build two vertices u1,u2 instead of u. For each (v,u,w), build (v,u1,w). For each (u,v,w), build(u2,v,w). Finally build (u1,u2,1).

Hence, if the maximal flow from S to T is 2, there is a simple path from a to b going through c.

Now we consider the minimal cut of the network. It is easy to find that minimal cut <= 2, so let's prove minimal cut > 1, which means, no matter which edge of capacity 1 we cut, there is still a path from S to T.

If we cut an edge like (u1,u2,1), it is equivalent to set the capacity of the vertex to 0, and equivalent to remove the vertex from the original graph. The graph is still connected, so there is still a path in the network.

If we cut other edges, it is equivalent to remove an edge from the original graph. It is still connected, too.

Now we have minimal cut > 1, which means maximal flow = minimal cut = 2. So there is always a simple path from a to b going through c.