Why would this be a deterministic context free language? $L = \{a^n b^n | n>0 \}$ $\cup$ $\{ a^n b^{2n} | n < 100 \}$
Maybe you can create 100 states that push two $a$'s onto the stack for every $a$, and then pop these $a$'s for every $b$. In the event of getting to the 100th state, when you read the 101st $a$, pop 100 $b$s from the stack and continue as you would for the first language alone.