Calculating coefficient before $x^{100}$ can be done quite easy and quickly in this situation. I will show it for coins 5,10,20,50, because the idea is relevant and that will be faster. Denote:

$\displaystyle P_5(x)=\frac{1}{1-x^5}=1+x^5 P_5(x)$ (multiply by denominator)- generating function for changing money only with 5 cents

$\displaystyle P_{5,10}(x)=\frac{P_5(x)}{1-x^{10}}=P_5(x)+x^{10}P_{5,10}(x)$ - for changing with coins 5,10, and so on..

$P_{5,10,20}(x)=P_{5,10}(x)+x^{20}P_{5,10,20}(x)$

$P_{5,10,20,50}(x)=P_{5,10,20}(x)+x^{50}P_{5,10,20,50}(x)$

We are looking for sequence $p_n$, where $P_{5,10,20,50}(x)=\sum_{n}p_n x^n$. Denote:
$P_5(x)=\sum_{n}q_n x^n, \ P_{5,10}(x)=\sum_{n}r_n x^n, \ P_{5,10,20}(x)=\sum_{n}s_n x^n$, and by the relations with generating functions above, it follows:

$$q_n=1, \ r_n=q_n+r_{n-10}, \ s_n=r_n+s_{n-20}, \ p_n=s_n+p_{n-50}$$ so it takes 5 minutes to calculate $p_n$ fo small $n$, which is an answer. For bigger $n$ maybe better will be to solve this system of recurrences (because I'm not sure about complexity of finding $p_n$ from these recurrences, but it is unsatisfying I think) and derive closed form formula for $p_n$ but for now I can't do it.