赞
踩
F 1 = Q 2 × F 2 + F 3 deg ( F 1 ) = 7 deg ( F 2 ) = 6 deg ( F 3 ) = 5 F_{1} = Q_{2} \times F_{2} + F_{3}\ \ \ \ \ \ \ \ \ \ \deg\left( F_{1} \right) = 7\ \ \ \ \ \ \deg\left( F_{2} \right) = 6\ \ \ \ \ \ \deg\left( F_{3} \right) = 5 F1=Q2×F2+F3 deg(F1)=7 deg(F2)=6 deg(F3)=5
∣
S
′
∣
=
F
1
F
1
F
2
F
2
F
2
F
2
F
2
F
2
F
2
F
1
F
1
F
1
F
1
F
1
F
1
∣
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
∣
=
F
1
F
1
F
2
F
2
F
2
F
2
F
2
F
2
F
2
F
3
F
3
F
3
F
3
F
3
F
3
∣
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
∣
\left| S^{'} \right| =
对应子结式 S 5 S_{5} S5:
S
5
=
(
−
1
)
(
m
−
5
)
(
l
−
5
)
d
e
t
p
o
l
(
F
1
F
1
F
1
F
0
F
0
(
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
a
8
a
7
a
6
a
5
a
4
a
3
a
2
a
1
a
0
0
0
a
8
a
7
a
6
a
5
a
4
a
3
a
2
a
1
a
0
)
)
=
(
−
1
)
(
m
−
5
)
(
l
−
5
)
d
e
t
p
o
l
(
F
1
F
1
F
2
F
2
F
3
(
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
)
)
S_{5} = ( - 1)^{(m - 5)(l - 5)}detpol
F 2 = Q 3 × F 3 + F 4 deg ( F 2 ) = 6 deg ( F 3 ) = 5 deg ( F 4 ) = 4 F_{2} = Q_{3} \times F_{3} + F_{4}\ \ \ \ \ \ \ \ \ \ \deg\left( F_{2} \right) = 6\ \ \ \ \ \ \deg\left( F_{3} \right) = 5\ \ \ \ \ \ \deg\left( F_{4} \right) = 4 F2=Q3×F3+F4 deg(F2)=6 deg(F3)=5 deg(F4)=4
∣
S
′
∣
=
F
1
F
1
F
2
F
2
F
3
F
3
F
3
F
3
F
3
F
3
F
2
F
2
F
2
F
2
F
2
∣
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
∣
=
F
1
F
1
F
2
F
2
F
3
F
3
F
3
F
3
F
3
F
3
F
4
F
4
F
4
F
4
F
4
∣
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
0
0
0
0
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
0
0
0
0
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
0
0
0
0
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
0
0
0
0
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
∣
\left| S^{'} \right| =
对应子结式 S 4 S_{4} S4:
S
4
=
(
−
1
)
(
m
−
4
)
(
l
−
4
)
d
e
t
p
o
l
(
F
1
F
1
F
1
F
1
F
0
F
0
F
0
(
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
a
8
a
7
a
6
a
5
a
4
a
3
a
2
a
1
a
0
0
0
0
a
8
a
7
a
6
a
5
a
4
a
3
a
2
a
1
a
0
0
0
0
a
8
a
7
a
6
a
5
a
4
a
3
a
2
a
1
a
0
)
)
=
(
−
1
)
(
m
−
4
)
(
l
−
4
)
d
e
t
p
o
l
(
F
1
F
1
F
2
F
2
F
3
F
3
F
4
(
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
b
7
b
6
b
5
b
4
b
3
b
2
b
1
b
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
c
6
c
5
c
4
c
3
c
2
c
1
c
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
d
5
d
4
d
3
d
2
d
1
d
0
0
0
0
0
0
0
e
4
e
3
e
2
e
1
e
0
)
)
S_{4} = ( - 1)^{(m - 4)(l - 4)}detpol
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