Yes, I believe this is the mirror.

Let A[i] be the input array, and let B[i] be an array satisfying B[i] = (B[i-1] + B[i+1]) / 2 + K. (There's a pretty simple pattern we can use to generate B[i].) Then, let C[i] = A[i] — B[i]. The key insight is that setting A[i] = max(A[i], (A[i-1] + A[i+1]) / 2 + K) is equivalent to setting C[i] = max(C[i], (C[i-1] + C[i+1]) / 2). Then, we can use convex hull to compute the final values of C[i]: it's fairly well-known that the final value of C[i] will be the y-value at x = i on the convex hull of the points (i, C[i]). Afterwards, we can add B[i] to C[i] in order to find the final values of A[i], and we print the largest one.

Do what Geothermal said, but also note that you have to output floats in decimal notation rather than exponential which is the default for cout. (Even though they make us suffer through reading in exponential notation...)

Edit: Based on what geothermal said, the problem may have just been cout not printing enough digits for doubles. Guess I can't blame the problem writers for that one (:

Yes, there will be overview slides for each of the new problems (B,E,G,H). We're a bit behind putting the slides together but expect we'll release them in the next day or so.