11. (20 points) It is known that a box contains 100 red and 100 blue cards, each color of cards containing one card labeled with each of the numbers $1, 3, 3^2, \cdots, 3^{99}$. The total sum of the numbers on the cards of both colors is denoted as $s$.

For a given positive integer $n$, if it is possible to pick several cards from the box such that the sum of their labels is exactly $n$, then it is called a "scheme for $n$". The number of different schemes for $n$ is denoted as $f(n)$. Try to find the value of $f(1) + f(2) + \cdots + f(s)$.
11. Arrange the cards in the box in ascending order of their labels:
$$
1,1,3,3,3^{2}, 3^{2}, \cdots, 3^{99}, 3^{99},
$$
where items with the same value are of different colors. For each $k(1 \leqslant k \leqslant 100)$, the sum of the first $2k$ terms is less than $3^{k}$. Therefore, terms of the form $3^{n}$ must be selected from the two $3^{n}$ (the sum of any other terms does not equal $3^{n}$). Thus, selecting one $3^{n}$ has two methods, while selecting both $3^{n}$ has only one method.

For each number $m$ in the set $A=\{0,1, \cdots, s\}$, it can be represented as a 100-digit ternary number:
$$
m=\overline{a_{99} a_{98} \cdots a_{0}},
$$
i.e., $m=a_{0}+3 a_{1}+3^{2} a_{2}+\cdots+3^{99} a_{99}$,
where $a_{i} \in\{0,1,2\}, i=0,1, \cdots, 99$.
If exactly $k$ of $a_{0}, a_{1}, \cdots, a_{99}$ are 1 (the remaining $100-k$ numbers are 0 or 2), then $f(m)=2^{k}$ (since each 1 has two selection schemes: red or blue).
Now, decompose the set $A$ as
$$
A=A_{0} \cup A_{1} \cup \cdots \cup A_{100},
$$
where each number $m$ in the set $A_{k}$, when represented in the above ternary form, has exactly $k$ of its coefficients $a_{0}, a_{1}, \cdots, a_{99}$ as 1 (the remaining $100-k$ numbers are 0 or 2). Therefore, the set $A_{k}$ contains $2^{100-k} \mathrm{C}_{100}^{k}$ numbers (since selecting $k$ of $a_{0}, a_{1}, \cdots, a_{99}$ to be 1 has $\mathrm{C}_{100}^{k}$ ways, and the remaining $100-k$ numbers can each be 0 or 2, with $2^{100-k}$ ways).
Thus, the sum of the $f$ values of the numbers in the set $A_{k}$ is
$$
\sum_{m \in 1_{1}} f(m)=2^{k} \times 2^{100-k} \mathrm{C}_{100}^{k}=2^{100} \mathrm{C}_{100}^{k}.
$$

Since the sets $A_{0}, A_{1}, \cdots, A_{100}$ are pairwise disjoint, we have
$$
\begin{array}{l}
\sum_{m \in A_{1}} f(m)=\sum_{m \in A_{0}} f(m)+\sum_{m \in A_{1}} f(m)+\cdots+\sum_{m \in A_{00}} f(m) \\
=2^{100}\left(\mathrm{C}_{100}^{0}+\mathrm{C}_{100}^{1}+\cdots+\mathrm{C}_{100}^{100}\right) \\
=2^{100} \times 2^{100}=2^{200}.
\end{array}
$$

Notice that $0=0+0 \times 3+0 \times 3^{2}+\cdots+0 \times 3^{99}$, i.e., no number is selected in the sequence, and its scheme count $f(0)=1$. Therefore,
$$
f(1)+f(2)+\cdots+f(s)=2^{200}-1.
$$