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359A - Table

If there are some good cell, which is located in the first row or in the first column, the answer is two. Similarly, if If there are some good cell, which is located in the last row or in the last column, the answer is two. Otherwise, the answer is four.

Авторское решение: 4968279

359B - Permutation

The answer is a slightly modified permutation 1, 2, ..., 2n. Let's reverse numbers 2i - 1 and 2i for each 1 ≤ i ≤ k. It's not hard to understand, that this permutation is good.

Авторское решение: 4968385

359C - Prime Number

Obviously, the answer is x^v. Let sum = a₁ + a₂ + ... + a_n. Also let s_i = sum - a_i (the array of degrees). After that let's find value v by the following algorithm: Let's consider a sequence of degrees as decreasing sequence. Now we will perform the following operation until it's possible to perfom it. Take the minimum degree v from the array of degrees and calculate the number of elements cnt, which have the same degree. If cnt multiples of x, then replace all cnt elements by cnt / x elements of the form v + 1. Since the sequence of degrees is a decreasing sequence, we can simply assign them to the end. If cnt is not a multiple of x, then we found the required value v. Also you need to check, that v is not greater then sum. Otherwise, v will be equals to sum.

Авторское решение: 4968346

359D - Pair of Numbers

Quite simple note: if the pair (l, r) satisfies the condition 1 from the statements, then min(l, r) = GCD(l, r), where min(l, r) is smallest number a_i from the segment (l, r) and GCD(l, r) is a GCD of all numbers from the segment (l, r). Calculate some data structure that will allow us to respond quickly to requests GCD(l, r) and min(l, r). For example, you can use Sparce Table. Solutuions, that uses segment tree, is too slow. So I think, you should use Sparce Table. So, now our task quite simple. Let's use binary search to find greatest value of r - l:

lf = 0;  //left boundary of binary search
rg = n;  //right boundary of binary search
while (rg - lf > 1) {
  int mid = (lf + rg) / 2;
  if (ok(mid))   //ok(mid)
    lf = mid;
  else
    rg = mid;
}

ok(mid) is the function, that determines, is there some segment where min(l, r) = GCD(l, r) and mid = r - l (mid — is fixed value by binary search). If there is some good segment, you should update boundaries of binary search correctly. After that, it's very simple to restore answer.

Some information about Sparce Table

Авторское решение: 4968587

359E - Neatness

You should write recursive function, that will turn on the light in all rooms, where it's possible. Also this function will visit all rooms, which it may visit. Let this function is called paint(x, y), where x, y is the current room. Paint(x, y) will use following idea: Let's look at all neighbors. If there is a light in the current direction (rule 3 from the statement), and the room (nx, ny) (current neighbor) has not yet visited, we will call our recursive function from (nx, ny). Also, we will turn on the light in all rooms, were we were. If some room is not visited by paint(x, y) and lights is on in this room, the answer is "NO". Otherwise, the answer is "YES". After that let's calculate value dist(x, y) by using bfs. dist(x, y) — is a minimal possible distance from the start to the current position (x, y). It's possible to use in our bfs only rooms, where lights is on. After that we will write the same function repaint(x, y). Repaint(x, y) will use following idea: Let's look at all neighbors. If there is a light in the current neighbor (nx, ny) and dist(nx, ny) > dist(x, y) ((x, y) — current room), let's call our recursive function from (nx, ny).After that we will come back to room (x, y). If there is no such neigbor (nx, ny), turn off the light in the room (x, y). Also you should look at my solution for more details.

Авторское решение: 4968657