Quantum computing 'amplitudes'
The state of a qubit, or of a collection of qubits, is given by a state vector. However, most people would say that the output of a quantum computer is in fact the result of a measurement, which transforms the state into one which indicates a single outcome.

There are many ways to approach this, but the simplest one to describe mathematically is complete measurement in the standard basis. In this case, you transform a quantum state on $n$ qubits into an $n$-bit string.

Suppose that you have a state $|\psi\rangle \in \mathbb C^{2^n}$ on $n$ qubits, given by $$ |\psi\rangle \;=\; \sum_{x \in \{0,1\}^n} \!u_x \;|x_1 x_2 \cdots  x_n \rangle \;, \quad \text{such that } \sum\limits_{x\in\{0,1\}^n} \!|u_x|^2 = 1 .$$

In order to get a classical output from a quantum computer, one thing you can do is to perform a complete measurement in the standard basis. What this means is that we collapse the state to some single output vector $y \in \{0,1\}^n$. We get different bit-strings $y$ with different probabilities, governed by the formula
$$
 \Pr\nolimits_\psi(y) \;=\; \Bigl| \langle y | \psi \rangle \Bigr|^2
\;=\; \left|  \sum_{x \in \{0,1\}^n} \!u_x \; \langle y_1 y_2 \cdots y_n | x_1 x_2 \cdots x_n \rangle \right|^2
\;=\; \left|  u_y \right|^2 \;.
$$
The outcome is then just a random output string $y \in \{0,1\}^n$, where the probabilities are given by the non-negative real numbers $|u_y|^2$. Afterwards, you can compute with the measurement result $y \in \{0,1\}^n$ however you like, with a classical computer.

It's possible to describe performing a measurement only on a single bit, or with respect to bases other than the standard basis, but that's not really necessary to answer your question; measuring all of the qubits is certainly something you can do, and it's enough to get classical outputs from quantum computations.