798A - Майк и палиндром

Let cnt be the number of such that s_i ≠ s_{n - i + 1}.

If cnt ≥ 2 then the answer is NO since we must change more than 1 character.

If cnt = 1 then the answer is YES.

If cnt = 0 and n is odd answer is YES since we can change the character in the middle, otherwise if n is even the answer is NO because we must change at least one character.

Complexity is O(|s|).

798B - Майк и строки

First of all, you must notice that the operation of removing the first character and appending it to the left is equivalent to cyclically shifting the string one position to the left.

Let's denote by dp_i, j the smallest number of operations for making the first i strings equal to string s_i moved j times.

Let f(i, j) be the the string s_i moved j times,then .

The answer is min(dp_n, 0, dp_n, 1, ..., dp_{n, |sn| - 1}).

The complexity is O(|S|³ × n).

798C - Майк и GCD

First of all, the answer is always YES.

If then the answer is 0.

Now suppose that the gcd of the sequence is 1. After we perform one operation on a_i and a_i + 1, the new gcd d must satisfy d|a_i - a_i + 1 and d|a_i + a_i + 1 d|2a_i and d|2a_i + 1. Similarly, because d is the gcd of the new sequence, it must satisfy d|a_j, j ≠ i, i + 1.

Using the above observations we can conclude that , so the gcd of the sequence can become at most 2 times bigger after an operation. This means that in order to make the gcd of the sequence bigger than 1 we need to make all numbers even. Now the problem is reduced to the following problem:

Given a sequence v₁, v₂, ... , v_n of zero or one,in one move we can change numbers v_i, v_i + 1 with 2 numbers equal to . Find the minimal number of moves to make the whole sequence equal to 0.

It can be proved that it is optimal to solve the task for consecutive ones independently so we divide the array into the minimal number of subarrays full of ones, if their lengths are s₁, s₂, ... , s_t,the answer is .

Complexity is .

798D - Майк и распределение

In the beginning, it's quite easy to notice that the condition "2·(a_p1 + ... + a_pk) is greater than the sum of all elements in A" is equivalent to a_p1 + ... + a_pk is greater than the sum of the remaining elements in A.

Now, let's store an array of indices C with C_i = i and then sort it in decreasing order according to array A, that is A_Ci ≥ A_Ci + 1.

The complexity is .

798E - Майк и код перестановки

Let's consider a_i = n + 1 instead of a_i =  - 1. Let's also define the sequence b, b_i = j such that a_j = i or b_i = n + 1 if there is no such j. Lets make a directed graph with vertices be the indices of the permutation p with edges of type (a, b) representing that p_a < p_b. If we topologically sort this graph then we can come up with a possible permutation,if S is the topologically sorted graph then we can assign to p_Si number i (in other words p_Si = i).

But how we can find the edges? First of all there are edges of form (b_i, i) if b_i ≠ n + 1.For a vertex i he visited all the unmarked vertices j (1 ≤ j < a_i, j ≠ i) and you know for sure that for all these j,p_j < p_i. But how we can check if j was already marked? The vertex j will become marked after turn of vertex b_j or will never become unmarked if b_j = n + 1. So there is a direct edge from i to j if j = a_i or 1 ≤ j < a_i, j ≠ i and b_j > i.

The main problem is to extract j for a given i in a sub-quadratic time,the solution is segment tree and is based on the fact that each time we want to extract vertex j from all non-visited edges (1 ≤ j < a_i,j ≠ i) with maximal b_j,if b_j < i then there aren't remaining edges.

Complexity and memory are and O(N).