Recently I wrote my very first problem (that was $$$>$$$ D2A and not just a standard application of an algorithm) that I both wrote and solved myself.
Link: https://codeforces.com/gym/104669/problem/L
Editorial: Let $$$R$$$ be the sum of all the elements.
Denote the sum of the first set to be $$$s1$$$ and the sum of the second set to be $$$s2$$$. Notice that the gcd has to be a factor of $$$R$$$ because if $$$\text{gcd} | s1$$$ and $$$\text{gcd} | s2$$$ then $$$\text{gcd} | (s1 + s2)$$$. So we can check every factor of $$$R$$$, see if it possible achieve this factor as the $$$\text{gcd}$$$, and then take the max factor that works.
A factor $$$f$$$ works if there is a subset $$$s$$$, $$$|s| < b$$$ that adds up a multiple of $$$f$$$ because the sum of the numbers from $$$a$$$ to $$$a + b - 1$$$ not in $$$s$$$ will also be divisible by the $$$f$$$. We need to find some way to optimize knapsack dp using the fact that the numbers are consecutive.
Considering fixing the size of the subset to be $$$s$$$.
Claim: There exists a subset of $$$a, a + 1, ..., a + b - 1$$$ of size $$$s$$$ that adds up to every number from the sum of the first $$$s$$$ numbers to the last $$$s$$$ numbers of the subset.
Proof: We can prove this with induction. Our base case is the subset containing the first $$$s$$$ numbers. Notice for a given subset, we can increasing the subset sum by $$$1$$$ by increasing by $$$1$$$ the greatest element that can be increased. A number cannot be increased if it would become $$$> a + b - 1$$$ or if increasing it by $$$1$$$ would give you a number already in the subset. The only case where we can't increase the subset sum by $$$1$$$ is when we have the last $$$s$$$ numbers as our subset. Thus we can represent all subset sums we can hit as a series of $$$b - 1$$$ intervals (we don't consider interval of length $$$b$$$ because if we choose this for one set, the other set would be empty).
We can check in $$$O(1)$$$ if a multiple of $$$f$$$ is in each interval by checking either if the interval is of length $$$\ge f$$$ or if the smallest subset sum mod $$$f$$$ is larger than the largest subset sum mod $$$f$$$. In both cases it is guaranteed there is some subset sum which is $$$0$$$ mod $$$f$$$. So this is a total of $$$O(m)$$$ for each factor.
The time complexity of the solution is: $$$O(R^{\frac{1}{2}} + R^{\frac{1}{3}}b) = O((ab + b^2)^{\frac{1}{2}} + (ab + b^2)^{\frac{1}{3}}b)$$$
We take $$$R$$$ to the power of $$$\frac{1}{3}$$$ because the number of factors of a number $$$n$$$ is bounded by $$$n^{\frac{1}{3}}$$$. You could also check this oeis sequence to confirm that the number of factors is small: https://oeis.org/A066150
