* Four, the sequence $1,1,3,3,3^{2}, 3^{2}, \cdots, 3^{1992}, 3^{1992}$ consists of two 1s, two 3s, two $3^{2}$s, ..., two $3^{1992}$s arranged in ascending order. The sum of the terms in the sequence is denoted as $S$. For a given natural number $n$, if it is possible to select some terms from different positions in the sequence such that their sum is exactly $n$, it is called a selection scheme, and the number of all selection schemes with the sum $n$ is denoted as $f(n)$. Try to find:
$$
f(1)+f(2)+\cdots+f(s) \text { . }
$$
For each $k(1 \leqslant k \leqslant 1992)$, it is easy to see that the sum of the first $2k$ terms of the sequence is less than $3^k$, so terms of the form $3^k$ must be chosen from two $3^k$ terms. Thus, selecting one $3^k$ has two methods, and selecting two $3^k$ has only one method.

For each number $n$ in the set $A=\{0,1,2, \cdots, S\}$, it can be expressed as
$$
n=a_{0}+3 a_{1}+3^{2} a_{2}+\cdots+3^{1992} a_{1992},
$$

where $a_{i} \in\{0,1,2\}(i=0,1,2, \cdots, 1992)$.
That is, $\overline{a_{1992} a_{1991} \cdots a_{1} a_{0}}$ is the 1993-digit ternary form of $n$.
If $a_{0}, a_{1}, a_{2}, \cdots, a_{1992}$ have exactly $k$ ones (the rest being 0 or 2), then $f(n)=2^k$.
The set $A$ can be decomposed as
$$
A=A_{0} \cup A_{1} \cup A_{2} \cup \cdots \cup A_{1993},
$$

where each number $n$ in $A_{\star}$, when expressed in the above ternary form, has exactly $k$ ones among the digits $a_{0}, a_{1}, a_{2}, \cdots, a_{1992}$. Therefore, the set $A_{k}$ contains $C_{1993}^{k} \cdot 2^{1992-1}$ numbers, and the sum of the $f$ values of the numbers in $A_{k}$ is
$$
\sum_{n \in A_{k}} f(n)=2^{k} \cdot C_{1993}^{k} \cdot 2^{1993-k}=2^{1993} C_{1993}^{k} .
$$
$$
\begin{array}{rl}
\sum_{n \in A} & f(n)=\sum_{n \in \Lambda_{0}} f(n)+\sum_{n \in \Lambda_{1}} f(n)+\cdots+\sum_{n \in 1993} f(n) \\
& =2^{1993} \cdot\left(C_{1993}^{0}+C_{1993}^{1}+\cdots+C_{19993}^{1993}\right) \\
& =2^{1993} \cdot 2^{1993}=4^{1993} .
\end{array}
$$

Thus, $f(0)+f(1)+\cdots+f(S)=4^{1993}$.
Noting that $0=1 \cdot 0+3 \cdot 0+3^{2} \cdot 0+\cdots+3^{1992} \cdot 0$, i.e., no terms are selected in the sequence, the number of methods $f(0)=1$.
Therefore, $f(1)+f(2)+\cdots+f(S)=4^{1993}-1$.

Since the sets $A_{0}, A_{1}, A_{2}, \cdots, A_{1993}$ are pairwise disjoint, thus