Zero-padding data for FFT If I take a discrete Fourier transform of $\{ c_1, c_2, \ldots, c_n\}$ where $n$ is prime, I am rather limited in the FFT algorithms available to me and their performance. Additionally, having mixed-radix FFT algorithms and prime-length FFT algorithms, plus the code necessary to determine which to use, increases my code size which is a limiting factor.
Suppose $n$ is prime and I decide to, instead, take the discrete Fourier transform of $\{ c_1, c_2, \ldots, c_n, 0\}$ so that, now, the length of the data is no longer prime. I know that the "picture" I get will be more smooth as compared to the true DFT of the original data.
Is there any way to zero-pad the original data slightly, perform a FFT, and then reverse the effects of the zero-padding afterwards? By that, I mean the length of the output list should be $n$, not $n + 1$. I do not expect to obtain precisely the same result from the $n + 1$ padded FFT as with the original $O(n^2)$ DFT. I understand it is practically impossible to reverse the smoothing affect of the padding.

Summary: I can zero-pad my data so it has a non-prime length, but then the result of my FFT has the wrong length, and the values on the indices don't match the true DFT. I can't just drop the last element of my FFT result, I need to something more "involved". What post-processing can I do to my zero-padded data so that it looks more like the original DFT, and has the same length?
 A: Actually, by padding with zeroes you are not smoothing or losing information.
The original FFT and the "padded FFT" just corresponds to sampling the same (true) DTFT $X(\omega)$ in different points; but both samplings are enought to recover (in theory) the true $X(\omega)$.
Let me change a little notation to be more consistent with common usage.
Let $N$ be the length of the original signal $x_n=x_0, x_1 \cdots x_{N-1}$ and let $X(\omega)$ be its DTFT , $X(\omega)= \sum x_n \exp(-i \omega n)$
Let $y_n$ be the same signal right padded with zeroes, so that its length is $N_p>N$.
It's clear that its DTFT is the same.
The DFT (what the FFT computes) are different, though. That's because they correspond to sampling $X(\omega)$ at different points. If $X_k$ ($k=0,1, \cdots N-1$) is the $k$ component of the DTF of $x_n$, then the correspondence is
$$ X_k \leftrightarrow X(\omega)\biggl|_{\omega=2\pi k /N}$$
They are different, but from one you can get the other. This is quite obvious if one thinks that any DFT -padded or not- is invertible, so from the padded FFT you can recover the original signal, and then you can recover the unpadded FFT.
$$X_k = \sum_{n=0}^{N-1} x_n e^{-i 2 \pi n k /N} = \sum_{n=0}^{N-1} \frac{1}{N_p} \sum_{j=0}^{N_p} Y_j e^{i 2 \pi n j /N_p}  e^{-i 2 \pi n k /N}$$
This is sort of a trigonometric interpolation. But this is surely not practical, in your scenario: instead of doing this, you'd directly perform a plain DFT on the original unpadded data.
A more practical option (but non exact) would be to do some simpler interpolation. I guess a quadratic interpolation could work decently. But don't plug blindly the $Y_k$ into some interpolation routine, you must take into account the nature of a DFT or a real signal (preserve hermitic property), and you'd respect the zero component as your true "origin", so it is strictly preserved.
