Problem A. PawnChess

Player A wins if the distance of his nearest pawn to the top of the board is less than or equal to the distance of the Player’s B nearest pawn to the bottom of the board (Note that you should only consider pawns that are not blocked by another pawns).

Problem B. The monster and the squirrel

After drawing the rays from the first vertex (n - 2) triangles are formed. The subsequent rays will generate independently sub-regions in these triangles. Let's analyse the triangle determined by vertices 1, i, i + 1, after drawing the rays from vertex i and (i + 1) the triangle will be divided into (n - i) + (i - 2) = n - 2 regions. Therefore the total number of convex regions is (n - 2)²

If the squirrel starts from the region that have 1 as a vertex, then she can go through each region of triangle (1, i, i + 1) once. That implies that the squirrel can collect all the walnuts in (n - 2)² jumps.

Problem C. The Big Race

Let D be the length of the racetrack, Since both athletes should tie DmodB = DmodW.

Let M = lcm(B, W), then D = k·M + r. None of the athletes should give one step further, therefore r ≤ min{B - 1, W - 1, T} = X.

D must be greater than 0 and less than or equal to T so  - r / M < k ≤ (T - r) / M.

For r=0, the number of valid racetracks is floor(T / M), and for r > 0 the number of racetracks is floor((T - r) / M) + 1.

Iterating over all possible r, we can count the number of racetracks in which Willman and Bolt ties:

Note that floor((T - r) / M) = v ≤ ftrightarrow T - v·M - M < r ≤ T - v·M. That means that floor((T - r) / M) = v for exactly M values of r.

We can count the number of values of r in which floor((T - r) / M) = floor((T - 0) / M) = v₁, and the values of r in which floor((T - r) / M) = floor((T - X) / M) = v₂. Each of the remaining values v_1-1, v_1-2,...,v_2+1 will appear exactly M times.

Problem D. Super M

Observation 1: Ari should teleport to one of the attacked cities (it doesn't worth going to a city that is not attacked since then she should go to one of the attacked cities)

Observation 2: The nodes visited by Ari will determine a sub-tree T of the original tree, this tree is unique and is determined by all the paths from two attacked cities.

Observation 3: If Ari had to return to the city from where she started, then the total distance would be 2e, where e is the number of edges of T, that is because she goes through each edge forward and backward

Observation 4: If Ari does not have to return to the starting city (the root of T), then the total distance is 2e - distanceofthefarthestnodefromtheroot.

Observation 5: In order to get a minimum total distance, Ari should chose one diameter of the tree, and teleport to one of its leaves.

The problem is now transformed in finding the diameter of a tree that contains the smallest index for one of its leaves. Note that all diameters pass through the center of the tree, so we can find all the farthest nodes from the center...and [details omitted].