G: Every number can be represented as 2^x*3^y*z, group them by z and let (x,y) be its coordinate, then it is a standard mask-DP problem on the grid(you can't choose a shape like 'L').

D: first for each query calculate minimum of "when this segment started to accumulate snow after clearing last time" using segment tree or other data strycture. Next, you need to answer queries of type "how much snow was there from moment x to t (current moment)". It will have several full blizzards (use prefix sums for that) and 2 partial blizzards (it's 2 quadratic functions)

L: Use inclusion-exlusion formula. We choose some subset of rows, columns and diagonals such that all cells there are equal and add  ± 2^{number of cells not covered by subset + number of connected components in subset} to the answer. Without diagonals it is easy to do in O(n²). With diagonals when we choose rows and columns we need to keep track of cells on diagonals that are covered by some row and some column. We can do it with dp.

M: First kick out all points that can't reach (1,1) or (n,m). What we need to do is to count the number of points each point can reach in the graph. Note that it is a planar graph, so for a point, lets track the wall by our left hand and right hand, we will get a polygon. The answer is equal to the number of points inside the polygon. We can use DP to calculate this. The complexity is linear.

I got WA18 by nested ternary search. I got AC after eliminating one TS and did some reflection trick.

I solved it only with reflection trick, but costed me lots of time. Ternary search sounds better.

Three cases: if shortest path to one line passes through other line, that's the answer. (times 2)

Otherwise, if segment between reflection of O over two lines intersects both lines and in the correct order, answer is tbat distance.

If none of these work, answer is path to intersection and back.

Any idea what is WA 115 in A?

I: Let (p,t) be points on 2D-plane, rotate the plane by 45 degree. According to Dilworth's theorem, the answer is equal to the length of LDS.

E is an extension of the 2018 USACO US Open problem Disruption. As mentioned, for edges off the MST, you do an LCA query to find the maximum path on the MST between those vertices, and the edges in the MST part can be handled in the way described in the reference solution.

I has probably appeared more than once in other contests, see for example the 2009 BOI problem Candy Machine, except you don't need to produce a proof of the answer. The reference solution here also passes once you omit the extra output.

J: Sort all numbers, the optimal way is to select consecutive numbers a[l],a[l+1],...,a[r] and change them to a[(l+r)/2]. So two-pointer iterate l and r to find the longest one.

How to write an integer as the sum of Fibonacci numbers using minimum number of terms? It's called Zeckendorf representation and can be obtained greedily.

Let f(X, K) be the number of integers up to X, whose Zeckendorf representation contains at most K terms. If F_i is the largest Fibonacci number less than or equal to X, f(X, K) = f(F_i - 1, K) + f(X - F_i, K - 1) holds. Write memoized recursion using this.