翻了各课件，发现……
某个英文网站 是这么说的：
There is a way to calculate the GCD and resultants in
O
(
n
log
2
n
)
O(n\log^2n)
O ( n log 2 n )
a
(
x
)
=
a
0
+
x
k
a
1
a(x)=a_0+x^ka_1
a ( x ) = a 0 + x k a 1
b
(
x
)
=
b
0
+
x
k
b
1
b(x)=b_0+x^kb_1
b ( x ) = b 0 + x k b 1
k
=
min
(
deg
a
,
deg
b
)
/
2
k=\min(\deg a,\deg b)/2
k = min ( deg a , deg b ) / 2
a
(
x
)
a(x)
a ( x )
b
(
x
)
b(x)
b ( x )
a
1
(
x
)
a_1(x)
a 1 ( x )
b
1
(
x
)
b_1(x)
b 1 ( x )
a
(
x
)
a(x)
a ( x )
b
(
x
)
b(x)
b ( x )
你如果脑子一冲，傻乎乎地去直接分治一半算出线性变换，DEBUG 一下会让你冷静过来……
而 Picks 的 Introduction to Polynomials 也写的比较晦涩，因此在这里稍微详细地谈谈，以正视听……
给多项式
a
,
b
∈
F
[
z
]
a, b \in \mathbb F[z]
a , b ∈ F [ z ]
x
,
y
∈
F
[
z
]
,
a
x
+
b
y
=
gcd
(
a
,
b
)
x, y\in \mathbb F[z], ax + by = \gcd (a, b)
x , y ∈ F [ z ] , a x + b y = g cd( a , b )
由于多项式取模最坏情况下每次只让一个多项式的度数减小，这种方法的复杂度最好也就是
Θ
(
(
deg
a
)
(
deg
b
)
)
\Theta((\deg a) (\deg b))
Θ ( ( deg a ) ( deg b ) )
我们考虑求解一个中间情况叫做 HALF-GCD，设
k
=
min
(
deg
a
,
deg
b
)
k = \min(\deg a, \deg b)
k = min ( deg a , deg b )
a
,
b
a, b
a , b
(
x
1
y
1
x
2
y
2
)
\begin{pmatrix} x_1 & y_1 \\ x_2 & y_2 \end{pmatrix}
( x 1 x 2 y 1 y 2 )
deg
x
≤
k
2
+
o
(
k
)
\deg x \le \frac k 2 + o(k)
deg x ≤ 2 k + o ( k )
deg
(
a
x
1
+
b
y
1
)
,
deg
(
a
x
2
+
b
y
2
)
≤
k
2
+
o
(
k
)
\deg (a x_1 + b y_1), \deg (a x_2 + b y_2) \le \frac k 2 + o(k)
deg ( a x 1 + b y 1 ) , deg ( a x 2 + b y 2 ) ≤ 2 k + o ( k )
gcd
\gcd
g cd
deg
\deg
deg
睿智的你想必已经意识到了这个东西如何得出真正的欧几里得的解。我们只需要调用一次这个算法，然后就规约到了一个
n
2
\frac n2
2 n
O
(
M
(
n
)
)
O(M(n))
O ( M ( n ) )
Θ
(
M
(
n
)
)
+
T
(
n
)
+
T
(
n
/
2
)
+
T
(
n
/
4
)
+
⋯
\Theta(M(n)) + T(n) + T(n/2) + T(n/4) + \cdots
Θ ( M ( n ) ) + T ( n ) + T ( n / 2 ) + T ( n / 4 ) + ⋯
那么接下来考虑通过分治来解决 HALF-GCD：设
k
=
min
(
deg
a
,
deg
b
)
k=\min(\deg a, \deg b)
k = min ( deg a , deg b )
a
/
z
k
/
2
,
b
/
z
k
/
2
a/z^{k/2}, b/z^{k/2}
a / z k / 2 , b / z k / 2
a
=
a
0
+
a
1
z
k
/
2
,
b
=
b
0
+
b
1
z
k
/
2
a=a_0 + a_1 z^{k/2}, b=b_0 + b_1 z^{k/2}
a = a 0 + a 1 z k / 2 , b = b 0 + b 1 z k / 2
a
x
+
b
y
ax + by
a x + b y
deg
(
(
a
1
x
+
b
1
y
)
z
k
/
2
)
=
3
4
k
+
o
(
k
)
\deg ((a_1 x + b_1 y)z^{k/2}) = \frac 34 k + o(k)
deg ( ( a 1 x + b 1 y ) z k / 2 ) = 4 3 k + o ( k )
deg
(
a
0
x
+
b
0
y
)
=
3
4
k
+
o
(
k
)
\deg (a_0 x + b_0 y) = \frac 34 k + o(k)
deg ( a 0 x + b 0 y ) = 4 3 k + o ( k )
3
4
k
\frac 34 k
4 3 k
z
k
/
4
z^{k/4}
z k / 4
1
4
k
+
o
(
k
)
\frac 14 k + o(k)
4 1 k + o ( k )
k
2
+
o
(
k
)
\frac k 2 + o(k)
2 k + o ( k )
T
(
n
)
=
2
T
(
n
/
2
)
+
M
(
n
)
T(n) = 2T(n/2) + M(n)
T ( n ) = 2 T ( n / 2 ) + M ( n )
M
(
n
)
log
n
M(n)\log n
M ( n ) log n
M
(
n
)
M(n)
M ( n )
综上所述，我们在
M
(
n
)
log
n
M(n)\log n
M ( n ) log n
这里 是一份代码，这里处理的时候是令
k
=
max
(
deg
a
,
deg
b
)
k = \max (\deg a, \deg b)
k = max ( deg a , deg b )