I have seen solutions that show how this is done for amounts such as $1. Namely I consulted this webpage's explanation--
https://www.maa.org/frank-morgans-math-chat-293-ways-to-make-change-for-a-dollar
--which I found helpful in understanding the general method. However, I am unsure of how to scale this up for an amount so large.
It's the coefficient of $x^{2018}$ in $\dfrac1{(1-x)(1-x^5)(1-x^{10})(1-x^{25})} $.
Stick that in your symbolic algebra systen and see what you get.
Alternatively, Wolfy says that this is $\dfrac1{\left(x^{41} - x^{40} - x^{36} + x^{35} - x^{31} + x^{30} + x^{26} - x^{25} - x^{16} + x^{15} + x^{11} - x^{10} + x^6 - x^5 - x + 1\right)} $ which would allow you to setup a recurrence (of order 41) to generate the successive coefficients.