Explanation : Using concepts of Dynamic Programming to find
the maximum possible weight of a subsequence
of X, we will have two alternatives.
1. Do not include a0 in the subsequence: then
the maximum possible weight will be equal to
maximum possible weight of a subsequence
of {a1, a2....an} which is represented by Y.
2. Include a0: then maximum possible weight will
be equal to a0 + (each number picked om Y
will get divided by 2) a0 + Y/2. Here you can
note that Y will itself pick optimal subsequence
to maximum the weight.
Final answer will be Max (Case1, Case2) i.e. Max
(Y, a0 + Y/2). Hence B.