Question: my problem is:

A real sequence $C$, the length is $N$ is multiple of 32, the pattern is as below:

$$C =\left(a\_0,\, a\_1,\, \ldots,\, a\_{\frac N2-1},\,\, b\_0,\, b\_1,\,\ldots,\, b\_{\frac N2 -1}\right)$$

the other two sequence is $A$, $B$, length is also $N$, as below:

$$\begin{align}
A &=(a\_0,& a\_1,& \ldots,& a\_{\frac N2-1},&&0, &&\ldots&0&)\\
B &=(0,&& \ldots,& 0,& b\_0,& b\_1,&\ldots,&& b\_{\frac N2-1}&)
\end{align}$$

you can see, $C= A+B$.

the DFT of sequence $C$ is known, but sequences $A$, $B$ and $C$ are all unknown.

I just want to know the DFT of sequence $A$ and $B$.

I know the ordinary method is

* perform a backward FFT to get sequence $C$, then
* sequence $A$ and $B$ becomes known, then
* perform forward FFT about $A$, the DFT of $B$ can be get by subtract DFT
of $A$ from DFT of $C$;

but the computaton is too complicated,

Is there a faster computation method?

Answer: 
Alas, a multiplication by a rectangular window in the time domain is a convolution in the frequency domain. Thus (assuming no other constraints on or knowledge about A or B), the DFT of A (or B) would have to be the convolution of the DFT of C with a twisty Sinc function, which is computationally far more expensive than 2 FFTs (except by fast convolution, which would require the same number of FFTs as your "ordinary" method).
If your DFT of the sum-sequence is real-valued, then the original $C$ *must* be hermitian!

Thus, if we set

$$a = \left(a\_0,\,\ldots,\, a\_{\frac N2-1}\right)\\
b = \left(b\_0,\,\ldots,\, b\_{\frac N2-1}\right)$$

$$ A + B = C \in \mathbb R \\\implies\\
a\_k = \left.b\_{\frac N2 -1 - k}\right.^\* \\ \text{with } k \in \{0,1,\ldots,\frac N2-1\}
$$

That implies that 
$$\text{DFT}\{b\}(f) = \text{DFT}\{a\}(-f)^\*$$
so, your subtraction method is actually superfluous; just calculate the transform of $a$ and read it backwards and invert the sign of the imaginary part. That is usually but a bit flip.

So your complexity reduces to the sum of complexities of

* 1x $\text{FFT}\_{N}$
* 1x $\text{FFT}\_{\frac N2}$
* $\frac N2$ bitflips