Hi to all!

215A - Bicycle Chain

Because of small constraints we can iterate i from 1 to n, iterate j from 1 to m, check that b_j divide a_i and find max value of . Then we can start this process again and find amount of the pairs in which is max value.

215B - Olympic Medal

Let amagine we have values of r₁, p₁ and p₂. Then:

Now it’s easy of understand, we must take maximum value of r₁ and p₁, and minimum value of p₂ to maximize r₂. You could not use three cycles for iterating all values, but you could find property above about at least one value and use two cycles.

215C - Crosses

Let’s iterate n₁ = max(a, c) and m₁ = max(b, d) — sides of bounding box. Then we can calculate value of function f(n₁, m₁, s), and add to the answer f(n₁, m₁, s)·(n - n₁ + 1)·(m - m₁ + 1), where two lastest brackets mean amount of placing this bounding box to a field n × m.

Now we must calculate f(n₁, m₁, s). At first, if n₁·m₁ = s then the result of the function is 2·(⌊ n₁ / 2⌋ + 1)·(⌊ m₁ / 2⌋ + 1) - 1 (you can prove it to youself).

If n₁·m₁ > s then we must to cut 4 corners which are equal rectangles with square . We can iterate length of a one of sides, calculate another side, check all constraints and add 2 to the result of the function for all such pairs of sides.

The time of a solution is O(n³), but it’s with very small constant, less then 1.

215D - Hot Days

You can use only two features about this problem: the solution is independenly for all regions and there are only 2 possible situations: all children are in exactly one bus or organizers must take minimum amount of bus such no children got a compensation. It’s bad idea to place some children to hot bus and to place some children to cool bus simultaneously.

For solving you must calculate this 2 values and get minimum.

215E - Periodical Numbers

Will be soon…