What you're asking for is related to the classic Frobenius Problem. Let me illustrate with the example you and Gerry Myerson discussed in comments.

In looking for symmetric polynomials of degree $8$ with non-negative coefficients summing to $\sigma$ (which we'll set shortly to $18$, so that Gerry's polynomial is a solution) for which $f(r)=s$ (soon to be set to $r=3$, $s=14842$), we must have $a(r^8+1)+b(r^7+r)+c(r^6+r^2)+d(r^5+r^3)+er^4=s$ with $2(a+b+c+d)+e=\sigma$. This can be rewritten as

$$a(r^8-2r^4+1)+b(r^7-2r^4+r^3)+c(r^6-2r^4+r^2)+d(r^5-2r^4+r^3)=s-\sigma r^4.$$

Now plugging in $r=3$, $s=14842$ and $\sigma=18$ (and removing what turns out to be a common factor of 4) produces

$$1600a+507b+144c+27d=3346$$

to be solved in non-negative integers $a$, $b$, $c$, and $d$ (after which one must also check that $e=18-2(a+b+c+d)$ is also non-negative). You can check that Gerry's coefficients, $(1,3,1,3)$, satisfy this equation. I'll leave it to someone else to find the rest.

Note that changing $s$ simply changes the number on the right hand side whose sum as a combination of "coins" of value $1600$, $507$, $144$, and $27$ is sought. (To be precise, $s$ can only change by multiples of $4$ in this example.) The Frobenius problem looks for the largest number that cannot be so represented, which in general seems to be difficult to decide. It's possible that the special form of the numbers here (all of the form $r^m-2r^{n/2}+r^{n-m}$) allows for a Frobenius miracle, but if they do I don't see how. Maybe someone with more knowledge of the Frobenius problem can comment or answer with greater authority.