$X ( e^{i \nu} )$ represents the Fourier transform of a finite sequence $x ( n )$ with a length of 10. We aim to calculate the 10 samples of $X ( e^{i \omega} )$ at frequencies $\omega_{k} \!=\! {\frac{2 \pi k^{2}} {1 0 0}} \,, \! k \!=\! 0 \,, 1 \,, \! \cdots, 9$. When computing, we must not use a method that calculates a greater number of samples than needed and then discards some. Discuss the feasibility of using the following methods:
Directly use the 10-point Fast Fourier Transform algorithm.
Use the chirp $z$-transform algorithm.

Options:
A. $$
G_{l} ( k )=\sum_{r=0}^{4} x ( 2 r+l ) \mathrm{e}^{-i \frac{2 m^{2} r} {5 0}}
$$
$$
\sum_{s=0}^{2} x ( 4 s+l ) \mathrm{e}^{-l \frac{5 \pi l^{2} s} {2 s}}+\mathrm{e}^{-l \frac{5 \pi l^{2}} {2 s}} \sum_{s=0}^{1} x ( 4 s+2+l ) \mathrm{e}^{-l \frac{5 \pi l^{2} s} {2 s}}, \; \; \; l=0, 1
$$
B. $$
G_{p} ( n )=\sum_{q=0}^{9} x ( r q+p ) \mathrm{e}^{-i \frac{3 n^{2} q} {6 0}}
$$
$$
\sum_{t=0}^{2} x ( 5 t+p ) \mathrm{e}^{-p \frac{3 \pi p^{2} t} {4 t}}+\mathrm{e}^{-p \frac{3 \pi p^{2}} {4 t}} \sum_{t=0}^{1} x ( 5 t+2+p ) \mathrm{e}^{-p \frac{3 \pi p^{2} t} {4 t}}, \; \; \; p=0, 1
$$
C. $$
G_{m} ( j )=\sum_{n=0}^{7} x ( 3 n+m ) \mathrm{e}^{-i \frac{4 j^{2} n} {7 0}}
$$
$$
\sum_{u=0}^{2} x ( 6 u+m ) \mathrm{e}^{-m \frac{6 \pi m^{2} u} {3 u}}+\mathrm{e}^{-m \frac{6 \pi m^{2}} {3 u}} \sum_{u=0}^{1} x ( 6 u+2+m ) \mathrm{e}^{-m \frac{6 \pi m^{2} u} {3 u}}, \; \; \; m=0, 1
$$
D. $$
G_{b} ( t )=\sum_{u=0}^{4} x ( 2 u+b ) \mathrm{e}^{-i \frac{5 t^{2} u} {9 0}}
$$
$$
\sum_{y=0}^{2} x ( 5 y+b ) \mathrm{e}^{-b \frac{3 \pi b^{2} y} {4 y}}+\mathrm{e}^{-b \frac{3 \pi b^{2}} {4 y}} \sum_{y=0}^{1} x ( 5 y+2+b ) \mathrm{e}^{-b \frac{3 \pi b^{2} y} {4 y}}, \; \; \; b=0, 1
$$
E. $$
G_{n} ( l )=\sum_{e=0}^{2} x ( 4 e+n ) \mathrm{e}^{-i \frac{3 l^{2} e} {5 0}}
$$
$$
\sum_{b=0}^{2} x ( 6 b+n ) \mathrm{e}^{-n \frac{3 \pi n^{2} b} {2 b}}+\mathrm{e}^{-n \frac{3 \pi n^{2}} {2 b}} \sum_{b=0}^{1} x ( 6 b+2+n ) \mathrm{e}^{-n \frac{3 \pi n^{2} b} {2 b}}, \; \; \; n=0, 1
$$
F. $$
G_{d} ( q )=\sum_{b=0}^{3} x ( 3 q+b ) \mathrm{e}^{-i \frac{2 q^{2} b} {8 0}}
$$
$$
\sum_{w=0}^{2} x ( 7 w+d ) \mathrm{e}^{-d \frac{4 \pi d^{2} w} {3 w}}+\mathrm{e}^{-d \frac{4 \pi d^{2}} {3 w}} \sum_{w=0}^{1} x ( 7 w+2+d ) \mathrm{e}^{-d \frac{4 \pi d^{2} w} {3 w}}, \; \; \; d=0, 1
$$
G. $$
G_{q} ( m )=\sum_{k=0}^{6} x ( 2 m+k ) \mathrm{e}^{-i \frac{6 m^{2} k} {1 0}}
$$
$$
\sum_{z=0}^{2} x ( 6 z+q ) \mathrm{e}^{-q \frac{2 \pi q^{2} z} {4 z}}+\mathrm{e}^{-q \frac{2 \pi q^{2}} {4 z}} \sum_{z=0}^{1} x ( 6 z+2+q ) \mathrm{e}^{-q \frac{2 \pi q^{2} z} {4 z}}, \; \; \; q=0, 1
$$
H. $$
G_{c} ( o )=\sum_{l=0}^{4} x ( 4 l+c ) \mathrm{e}^{-i \frac{7 o^{2} l} {8 0}}
$$
$$
\sum_{a=0}^{2} x ( 4 a+c ) \mathrm{e}^{-c \frac{5 \pi c^{2} a} {3 a}}+\mathrm{e}^{-c \frac{5 \pi c^{2}} {3 a}} \sum_{a=0}^{1} x ( 4 a+2+c ) \mathrm{e}^{-c \frac{5 \pi c^{2} a} {3 a}}, \; \; \; c=0, 1
$$
I. $$
G_{j} ( p )=\sum_{o=0}^{5} x ( o+2j ) \mathrm{e}^{-i \frac{8 p^{2} o} {4 0}}
$$
$$
\sum_{v=0}^{2} x ( 3 v+j ) \mathrm{e}^{-j \frac{2 \pi j^{2} v} {5 v}}+\mathrm{e}^{-j \frac{2 \pi j^{2}} {5 v}} \sum_{v=0}^{1} x ( 3 v+2+j ) \mathrm{e}^{-j \frac{2 \pi j^{2} v} {5 v}}, \; \; \; j=0, 1
$$
J. $$
G_{o} ( b )=\sum_{m=0}^{5} x ( 3 m+o ) \mathrm{e}^{-i \frac{4 b^{2} m} {9 0}}
$$
$$
\sum_{c=0}^{2} x ( 5 c+o ) \mathrm{e}^{-o \frac{5 \pi o^{2} c} {3 c}}+\mathrm{e}^{-o \frac{5 \pi o^{2}} {3 c}} \sum_{c=0}^{1} x ( 5 c+2+o ) \mathrm{e}^{-o \frac{5 \pi o^{2} c} {3 c}}, \; \; \; o=0, 1
$$
Answer: A