Consider the equation $$x_1 + \dots + x_m = N$$ where $x_1,\dots,x_m \ge 0$ and under the additional constraints $x_k \le a_k$ for $k=1,2,\dots,m$.
I'm interested in knowing whether the number of solutions to (1) can be found in sub-exponential time w.r.t. $m$.
One way to solve it would be to recursively lower the number of constraints by subtracting the number of solutions with constraints $0 \le x_1 \le a_1, \dots, 0 \le x_{m-1} \le a_{m-1}, x_{m-1} > a_{m-1}$ from the solutions without the constraint on $x_{m-1}$. Since constraints of type $x_j > a_j$ can be dealt by substituting $x_j + a_j + 1$ for $x_j$, and because there is a closed formula for the case without constraints at all, we'll get the job done in about $2^m$ steps. This is basically the principle of inclusion and exclusion.
You want the coefficient of $y^N$ in $$(1+y+\cdots+y^{a_1})\times\cdots\times(1+y+\cdots+y^{a_m})$$ Rewrite the product as $$(1-y^{a_1+1})\times\cdots\times(1-y^{a_m+1})(1-y)^{-m}$$ Expand $$(1-y)^{-m}=\sum_0^N{n+m-1\choose m-1}x^n+{\rm\ higher\ order\ terms}$$ Now multiply in turn by $1-y^{a_1+1},\dots,1-y^{a_m+1}$. You can always discard terms in $y^s$ with $s\gt N$ along the way.