, 2022

In the following, we denote $d = 100$ and let $f(0) = 0$. Note that $S = f(0) + f(1) + \dots + f(10^d - 1)$. For any integer $n$ such that $0 \le n \le 10^d$, there exists unique $(a_0, a_1, \dots, a_{d-1})$ with $0 \le a_i \le 9$ ($i = 0, 1, \dots, d-1$) such that
$$n = a_0 + a_1 \cdot 10 + \dots + a_{d-1} \cdot 10^{d-1}.$$
If $n > 0$,
$$f(n) = (-1)^{a_0+a_1+\dots+a_{d-1}} (a_0 + a_1 \cdot 10 + \dots + a_{d-1} \cdot 10^{d-1})^d \quad (*)$$

(and it is obvious to check it also holds for $n = 0$). Hence, $S$ equals to the sum of $(*)$ for all $(a_0, a_1, \dots, a_{d-1})$ with $0 \le a_i \le 9$ ($i = 0, 1, \dots, d-1$).
By multinomial theorem,
$$(a_0 + a_1 \cdot 10 + \dots + a_{d-1} \cdot 10^{d-1})^d = \sum_d \sum_{b_0+\dots+d_{d-1}=d, b_i \ge 0} \frac{d!}{b_0!b_1!\dots b_{d-1}!} \prod_{b_i \ne 0} (a_i \cdot 10^i)^{b_i} \\ = \sum_d \sum_{b_0+\dots+d_{d-1}=d, b_i \ge 0} \frac{d!}{b_0!b_1!\dots b_{d-1}!} \left(\prod_{b_i \ne 0} 10^{ib_i}\right) \left(\prod_{b_i \ne 0} a_i^{b_i}\right).$$
Since $\frac{d!}{b_0!b_1!\dots b_{d-1}!} \prod_{b_i \ne 0} 10^{ib_i}$ is independent of $(a_0, a_1, \dots, a_{d-1})$, it suffices to compute, for each $(b_0, b_1, \dots, b_{d-1})$,
$$\sum_{a_0=0}^{9} \dots \sum_{a_{d-1}=0}^{9} (-1)^{a_0+a_1+\dots+a_{d-1}} \prod_{b_i \ne 0} a_i^{b_i}.$$

In the case where $b_k \ne 0$ for some $0 \le k \le d-1$, it equals zero since
$$\begin{aligned} \sum_{a_k=0}^{9} \left( (-1)^{a_0+a_1+\dots+a_k+\dots+a_{d-1}} \prod_{b_i \ne 0} a_i^{b_i} \right) &= \left( \sum_{a_k=0}^{9} (-1)^{a_0+a_1+\dots+a_k+\dots+a_{d-1}} \right) \prod_{b_i \ne 0} a_i^{b_i} \\ &= (1 \cdot 5 + (-1) \cdot 5) \prod_{b_i \ne 0} a_i^{b_i} \\ &= 0. \end{aligned}$$

In the case where $b_k \ne 0$ for all $k$, i.e. $b_0 = b_1 = \dots = b_{d-1} = 1$,
$$(-1)^{a_0+a_1+\dots+a_{d-1}} \prod_{b_i \ne 0} a_i^{b_i} = (-1)^{a_0+a_1+\dots+a_{d-1}} \prod_{i=0}^{d-1} a_i = \prod_{i=0}^{d-1} ((-1)^{a_i} \cdot a_i).$$

$$\prod_{i=0}^{d-1} \left( \sum_{a_i=0}^{9} ((-1)^{a_i} \cdot a_i) \right) = (-5)^d.$$

$b_0 = b_1 = \dots = b_{d-1} = 1$ implies
$$\frac{d!}{b_0!b_1!\dots b_{d-1}!} \prod_{b_i \ne 0} 10^{ib_i} = d! \prod_{i=0}^{d-1} 10^i = d! \cdot 10^{\frac{d(d-1)}{2}},$$
and hence
$$S = d! \cdot 10^{\frac{d(d-1)}{2}} \cdot (-5)^d = 100! \cdot 2^{4950} \cdot 5^{5050}.$$
This concludes that the answer is $\left\lfloor \frac{100}{5^1} \right\rfloor + \left\lfloor \frac{100}{5^2} \right\rfloor + 5050 = 5074$.