11. Let $f(x)$ and $g(x)$ be increasing functions defined on the set of positive integers, and for any $k \in \mathbf{N}^{+}$, we have $f(k+1) + g(0) > f(k) + g(k-1)$. The sequence $\left\{a_{n}\right\}$ is formed by arranging all the numbers in the set $M=\{f(t)+g(s) \mid 0 \leqslant s<t, s, t \in \mathbf{N}\}$ in ascending order. Find the general term formula for the sequence $\left\{a_{n}\right\}$.
11. Arrange the terms of the sequence $\left\{a_{n}\right\}$ in ascending order into the following number table:
$$
\begin{array}{l}
a_{1}=f(1)+g(0), \\
a_{2}=f(2)+g(0), \quad a_{3}=f(2)+g(1), \\
a_{4}=f(3)+g(0), \quad a_{5}=f(3)+g(1), \\
a_{6}=f(3)+g(2), \quad a_{7}=f(4)+g(0), \\
a_{8}=f(4)+g(1), \quad a_{9}=f(4)+g(2), \\
a_{10}=f(4)+g(3), \\
\cdots
\end{array}
$$

Since $f(k+1)+g(0)>f(k)+g(k-1)$.
Therefore, in the above number table, the $k$-th row contains $k$ numbers arranged from left to right as $a \frac{k(k-1)}{2}+1=f(k)+g(0), a \frac{k(k-1)}{2}+2=f(k)+g(1), \cdots, a \frac{k(k-1)_{-j}}{2}=f(k)+g(i-1), \cdots, a \frac{k(k-1)_{-k}}{2}=f(k)+g(k-1)$.
Suppose $a_{n}$ is in the $k$-th row of the number table, because $a \frac{k(k-1)}{2}+i=f(k)+g(i-1)$,
so when $\frac{k(k-1)}{2}+1 \leqslant n0
\end{array}\right.
$$

Solving this, we get: $\frac{-1+\sqrt{8 n-7}}{2}<k \leqslant \frac{1+\sqrt{8 n-7}}{2}$,
and $\frac{1+\sqrt{8 n-7}}{2}-\frac{-1+\sqrt{8 n-7}}{2}=1$, so $k=\left[\frac{1+\sqrt{8 n-7}}{2}\right]$.
Thus, $a_{n}=f(k)+g\left[n-1-\frac{k(k-1)}{2}\right]$, where $k=\left[\frac{1+\sqrt{8 n-7}}{2}\right]$. ([x] denotes the floor function, which is the greatest integer less than or equal to $x$)