如果给你一个多项式
A
(
x
)
=
∑
a
n
x
n
A(x) = ∑a_nx^n
A ( x ) = ∑ a n x n
B
(
x
)
=
∑
b
n
x
n
B(x) = ∑b_nx^n
B ( x ) = ∑ b n x n
A
(
x
)
⋅
B
(
x
)
A(x) · B(x)
A ( x ) ⋅ B ( x )
O
(
n
2
)
O(n^2)
O ( n 2 )
∑
i
=
1
n
∑
j
=
1
m
a
i
∗
b
j
\sum_{i=1}^n\sum_{j=1}^mai*bj
∑ i = 1 n ∑ j = 1 m a i ∗ b j
O
(
n
2
)
O(n^2)
O ( n 2 )
F
F
T
FFT
F F T
O
(
n
l
o
g
n
)
O(nlogn)
O ( n l o g n )
对于求一个
n
−
1
n-1
n − 1
f
(
x
)
f(x)
f ( x )
f
(
x
)
=
∑
i
=
0
n
−
1
c
i
∗
x
i
f(x)=\sum_{i=0}^{n-1}c_i*x^i
f ( x ) = i = 0 ∑ n − 1 c i ∗ x i
f
(
x
)
=
(
a
0
∗
x
0
+
a
1
∗
x
1
+
a
2
∗
x
2
)
∗
(
b
0
∗
x
0
+
b
1
∗
x
1
+
b
2
∗
x
2
)
f(x)=(a_0*x^0+a_1*x^1+a_2*x^2)*(b_0*x^0+b_1*x^1+b_2*x^2)
f ( x ) = ( a 0 ∗ x 0 + a 1 ∗ x 1 + a 2 ∗ x 2 ) ∗ ( b 0 ∗ x 0 + b 1 ∗ x 1 + b 2 ∗ x 2 )
f
(
x
)
=
a
0
b
0
∗
x
0
+
a
0
b
1
∗
x
1
+
a
0
b
2
∗
x
2
+
a
1
b
0
∗
x
1
+
a
1
b
1
∗
x
2
+
a
1
b
2
∗
x
3
+
a
2
b
0
∗
x
2
+
a
2
b
1
∗
x
3
+
a
2
b
2
∗
x
4
f(x)=a_0b_0*x^0+a_0b_1*x^1+a_0b_2*x^2+a_1b_0*x^1+a_1b_1*x^2+a_1b_2*x^3+a_2b_0*x^2+a_2b_1*x^3+a_2b_2*x^4
f ( x ) = a 0 b 0 ∗ x 0 + a 0 b 1 ∗ x 1 + a 0 b 2 ∗ x 2 + a 1 b 0 ∗ x 1 + a 1 b 1 ∗ x 2 + a 1 b 2 ∗ x 3 + a 2 b 0 ∗ x 2 + a 2 b 1 ∗ x 3 + a 2 b 2 ∗ x 4
f
(
x
)
=
a
0
b
0
∗
x
0
+
(
a
0
b
1
+
a
1
b
0
)
∗
x
1
+
(
a
0
b
2
+
a
1
b
1
+
a
2
b
0
)
∗
x
2
+
(
a
1
b
2
+
a
2
b
1
)
∗
x
3
+
a
2
b
2
∗
x
4
f(x)=a_0b_0*x^0+(a_0b_1+a_1b_0)*x_1+(a_0b_2+a_1b_1+a_2b_0)*x^2+(a_1b_2+a_2b_1)*x^3+a_2b_2*x^4
f ( x ) = a 0 b 0 ∗ x 0 + ( a 0 b 1 + a 1 b 0 ) ∗ x 1 + ( a 0 b 2 + a 1 b 1 + a 2 b 0 ) ∗ x 2 + ( a 1 b 2 + a 2 b 1 ) ∗ x 3 + a 2 b 2 ∗ x 4
c
0
=
a
0
b
0
,
c
1
=
(
a
0
b
1
+
a
1
b
0
)
,
c
2
=
(
a
0
b
2
+
a
1
b
1
+
a
2
b
0
)
,
c
3
=
(
a
1
b
2
+
a
2
b
1
)
,
c
4
=
a
2
b
2
c_0=a_0b_0,c_1=(a_0b_1+a_1b_0),c2_=(a_0b_2+a_1b_1+a_2b_0),c_3=(a_1b_2+a_2b_1),c_4=a_2b_2
c 0 = a 0 b 0 , c 1 = ( a 0 b 1 + a 1 b 0 ) , c 2 = ( a 0 b 2 + a 1 b 1 + a 2 b 0 ) , c 3 = ( a 1 b 2 + a 2 b 1 ) , c 4 = a 2 b 2
c
i
c_i
c i
x
i
x^i
x i
f
(
x
)
f(x)
f ( x )
就是把
i
i
i
i
i
i
y
i
=
∑
j
=
0
n
−
1
c
j
∗
x
i
j
y_i=\sum_{j=0}^{n-1}c_j*x_i^j
y i = j = 0 ∑ n − 1 c j ∗ x i j
优
点
优点
优 点
A
(
x
)
=
a
0
+
a
1
∗
x
1
+
a
2
∗
x
2
.
.
.
a
n
∗
x
n
A(x)=a_0+a_1*x^1+a_2*x^2...a_n*x^n
A ( x ) = a 0 + a 1 ∗ x 1 + a 2 ∗ x 2 . . . a n ∗ x n
x
x
x
A
(
x
)
=
a
0
+
x
(
a
1
+
a
2
∗
x
+
.
.
.
a
n
∗
x
n
−
1
)
A(x)=a_0+x(a_1+a_2*x+...a_n*x^{n-1})
A ( x ) = a 0 + x ( a 1 + a 2 ∗ x + . . . a n ∗ x n − 1 )
x
x
x
a
0
a_0
a 0 霍纳法则
O
(
n
)
O(n)
O ( n )
多项式的加减计算效率也高，对于
A
(
x
)
=
a
0
+
a
1
∗
x
1
+
a
2
∗
x
2
+
.
.
.
+
a
n
−
1
∗
x
n
−
1
,
B
(
x
)
=
b
0
+
b
1
∗
x
1
+
.
.
.
+
b
n
−
1
∗
x
n
−
1
A(x)=a_0+a_1*x^1+a_2*x^2+...+a_{n-1}*x^{n-1},B(x)=b_0+b_1*x^1+...+b_{n-1}*x^{n-1}
A ( x ) = a 0 + a 1 ∗ x 1 + a 2 ∗ x 2 + . . . + a n − 1 ∗ x n − 1 , B ( x ) = b 0 + b 1 ∗ x 1 + . . . + b n − 1 ∗ x n − 1
O
(
n
)
O(n)
O ( n )
C
(
x
)
=
c
0
+
c
1
∗
x
1
+
.
.
.
+
c
n
−
1
∗
x
n
−
1
C(x)=c_0+c_1*x^1+...+c_{n-1}*x^{n-1}
C ( x ) = c 0 + c 1 ∗ x 1 + . . . + c n − 1 ∗ x n − 1
i
∈
[
0
,
n
)
i∈[0,n)
i ∈ [ 0 , n )
c
i
=
a
i
+
b
i
c_i=a_i+b_i
c i = a i + b i
缺
点
缺点
缺 点
O
(
n
2
)
O(n^2)
O ( n 2 )
A
A
A
B
B
B
2
n
−
2
2n-2
2 n − 2
A
,
B
A,B
A , B
x
n
−
1
x^{n-1}
x n − 1
C
C
C
x
2
n
−
2
x^{2n-2}
x 2 n − 2
C
(
x
)
=
∑
i
=
0
2
n
−
2
c
i
∗
x
i
,
c
i
=
∑
j
=
0
j
a
j
b
i
−
j
C(x)=\sum_{i=0}^{2n-2}c_i*x^i,c_i=\sum_{j=0}^ja_jb_{i-j}
C ( x ) = i = 0 ∑ 2 n − 2 c i ∗ x i , c i = j = 0 ∑ j a j b i − j
给一堆点对
(
x
1
,
x
2
,
x
3
.
.
.
x
n
)
,
(
y
1
,
y
2
,
y
3
.
.
.
y
n
)
(x_1,x_2,x_3...x_n),(y_1,y_2,y_3...y_n)
( x 1 , x 2 , x 3 . . . x n ) , ( y 1 , y 2 , y 3 . . . y n )
f
(
x
i
)
=
y
i
f(x_i)=y_i
f ( x i ) = y i
(
x
i
,
y
i
)
(x_i,y_i)
( x i , y i )
y
=
f
(
x
)
y=f(x)
y = f ( x )
n
n
n
拓展一下：举个例子
f
(
y
)
=
x
3
+
1
f(y)=x^3+1
f ( y ) = x 3 + 1
f
(
x
)
=
a
∗
x
3
+
b
∗
x
2
+
c
∗
x
1
+
d
f(x)=a*x^3+b*x^2+c*x^1+d
f ( x ) = a ∗ x 3 + b ∗ x 2 + c ∗ x 1 + d 感觉好像是一样的证明，别管这么多了，反正都是简单证明，口胡口胡
f
(
x
)
=
∑
i
=
0
n
−
1
y
i
∏
j
≠
i
(
x
−
x
j
)
∏
j
≠
i
(
x
i
−
x
j
)
f(x)=\sum_{i=0}^{n-1}y_i\frac{\prod_{j\neq i}(x-x_j)}{\prod_{j\ne i}(x_i-x_j)}
f ( x ) = i = 0 ∑ n − 1 y i ∏ j = i ( x i − x j ) ∏ j = i ( x − x j ) 拉格朗日插值法 ，但是因为我们一般不用这玩意儿，老子就不搞了，太现实了
优
点
优点
优 点
n
n
n
O
(
n
)
O(n)
O ( n )
C
(
x
)
=
A
(
x
)
+
b
(
x
)
C(x)=A(x)+b(x)
C ( x ) = A ( x ) + b ( x )
C
(
x
k
)
=
A
(
x
k
)
+
B
(
x
k
)
C(x_k)=A(x_k)+B(x_k)
C ( x k ) = A ( x k ) + B ( x k )
A
(
x
0
,
y
0
)
,
(
x
1
,
y
1
)
.
.
.
(
x
n
−
1
,
y
n
−
1
)
)
,
B
(
x
0
,
y
0
′
)
,
(
x
1
,
y
1
′
)
.
.
.
(
x
n
−
1
,
y
n
−
1
′
)
)
A{(x_0,y_0),(x_1,y_1)...(x_{n-1},y_{n-1}))},B{(x_0,y_0'),(x_1,y_1')...(x_{n-1},y_{n-1}'))}
A ( x 0 , y 0 ) , ( x 1 , y 1 ) . . . ( x n − 1 , y n − 1 ) ) , B ( x 0 , y 0 ′ ) , ( x 1 , y 1 ′ ) . . . ( x n − 1 , y n − 1 ′ ) )
A
A
A
B
B
B
n
n
n
C
C
C
C
(
x
0
,
y
0
+
y
0
′
)
,
(
x
1
,
y
1
+
y
1
′
)
.
.
.
(
x
n
−
1
,
y
n
−
1
+
y
n
−
1
′
)
)
C{(x_0,y_0+y_0'),(x_1,y_1+y_1')...(x_{n-1},y_{n-1}+y_{n-1}'))}
C ( x 0 , y 0 + y 0 ′ ) , ( x 1 , y 1 + y 1 ′ ) . . . ( x n − 1 , y n − 1 + y n − 1 ′ ) )
乘法计算效率也高：对两个点值表达的次数界为
n
n
n
O
(
n
)
O(n)
O ( n )
C
(
x
)
=
A
(
x
)
∗
b
(
x
)
C(x)=A(x)*b(x)
C ( x ) = A ( x ) ∗ b ( x )
C
(
x
k
)
=
A
(
x
k
)
∗
B
(
x
k
)
C(x_k)=A(x_k)*B(x_k)
C ( x k ) = A ( x k ) ∗ B ( x k )
A
,
B
A,B
A , B
C
C
C
C
C
C
2
n
2n
2 n
A
,
B
A,B
A , B
n
n
n
A
,
B
A,B
A , B
C
C
C
A
(
x
0
,
y
0
)
,
(
x
1
,
y
1
)
.
.
.
(
x
2
n
−
1
,
y
2
n
−
1
)
)
,
B
(
x
0
,
y
0
′
)
,
(
x
1
,
y
1
′
)
.
.
.
(
x
2
n
−
1
,
y
2
n
−
1
′
)
)
A{(x_0,y_0),(x_1,y_1)...(x_{2n-1},y_{2n-1}))},B{(x_0,y_0'),(x_1,y_1')...(x_{2n-1},y_{2n-1}'))}
A ( x 0 , y 0 ) , ( x 1 , y 1 ) . . . ( x 2 n − 1 , y 2 n − 1 ) ) , B ( x 0 , y 0 ′ ) , ( x 1 , y 1 ′ ) . . . ( x 2 n − 1 , y 2 n − 1 ′ ) )
C
C
C
C
(
x
0
,
y
0
+
y
0
′
)
,
(
x
1
,
y
1
+
y
1
′
)
.
.
.
(
x
n
−
1
,
y
2
n
−
1
+
y
2
n
−
1
′
)
)
C{(x_0,y_0+y_0'),(x_1,y_1+y_1')...(x_{n-1},y_{2n-1}+y_{2n-1}'))}
C ( x 0 , y 0 + y 0 ′ ) , ( x 1 , y 1 + y 1 ′ ) . . . ( x n − 1 , y 2 n − 1 + y 2 n − 1 ′ ) )
缺
点
缺点
缺 点
O
(
n
)
O(n)
O ( n )
O
(
n
2
)
O(n^2)
O ( n 2 )
换言之：对于多项式
A
(
x
)
A(x)
A ( x )
B
(
x
)
B(x)
B ( x )
d
e
g
A
+
d
e
g
B
<
n
degA + degB < n
d e g A + d e g B < n
A
A
A
B
B
B
x
0
,
x
1
,
.
.
.
,
x
n
−
1
{x_0, x_1, . . . , x_{n-1}}
x 0 , x 1 , . . . , x n − 1
(
A
⋅
B
)
(A · B)
( A ⋅ B )
(
A
⋅
B
)
(
x
i
)
=
A
(
x
i
)
⋅
B
(
x
i
)
(A · B)(x_i) = A(x_i) · B(x_i)
( A ⋅ B ) ( x i ) = A ( x i ) ⋅ B ( x i )
O
(
N
)
O(N)
O ( N )
(
A
⋅
B
)
(A · B)
( A ⋅ B )
O
(
n
2
)
O(n^2)
O ( n 2 )
举个栗子：所以如果有一道题给我们系数表达式，最后又让我们输出结果的系数表达式，我们用以上的方法虽然计算成点值表达式只用了
O
(
n
)
O(n)
O ( n )
O
(
n
2
)
O(n^2)
O ( n 2 )
i
,
j
i,j
i , j
但是我们的
F
F
T
FFT
F F T
O
(
n
l
o
g
n
)
O(nlogn)
O ( n l o g n )
1.把已知的一个多项式转化成对应的点值表示
说了这么多，还是没有告诉我怎么做啊！！！不急慢慢往下看
我们把形如
z
=
a
+
b
i
z=a+bi
z = a + b i 复数 ，其中a称为实部 ，b称为虚部 ，i称为虚数单位 ，
i
=
−
1
i=\sqrt{-1}
i = − 1
i
2
=
−
1
i^2=-1
i 2 = − 1
我们可以把复数当做一个向量丢在二维平面，即平面直角坐标系
百度百科说：复数之间的加减乘（除）是可以直接算的 ，除法因为不怎么用就不说了
z
1
=
a
+
b
i
，
z
2
=
c
+
d
i
z1=a+bi，z2=c+di
z 1 = a + b i ， z 2 = c + d i
(
a
+
b
i
)
+
(
c
+
d
i
)
=
(
a
+
c
)
+
(
b
+
d
)
i
(a+bi)+(c+di)=(a+c)+(b+d)i
( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i 两个复数的和依然是复数，它的实部是原来两个复数实部的和，它的虚部是原来两个虚部的和 ，当然复数的加法满足交换律和结合律
2.减法法则
z
1
=
a
+
b
i
，
z
2
=
c
+
d
i
z1=a+bi，z2=c+di
z 1 = a + b i ， z 2 = c + d i
(
a
+
b
i
)
−
(
c
+
d
i
)
=
(
a
−
c
)
+
(
b
−
d
)
i
(a+bi)-(c+di)=(a-c)+(b-d)i
( a + b i ) − ( c + d i ) = ( a − c ) + ( b − d ) i 两个复数的差依然是复数，它的实部是原来两个复数实部的差，它的虚部是原来两个虚部的差
3.乘法法则
z
1
=
a
+
b
i
，
z
2
=
c
+
d
i
(
a
、
b
、
c
、
d
∈
R
)
z1=a+bi，z2=c+di(a、b、c、d∈R)
z 1 = a + b i ， z 2 = c + d i ( a 、 b 、 c 、 d ∈ R )
(
a
+
b
i
)
(
c
+
d
i
)
=
(
a
c
−
b
d
)
+
(
b
c
+
a
d
)
i
(a+bi)(c+di)=(ac-bd)+(bc+ad)i
( a + b i ) ( c + d i ) = ( a c − b d ) + ( b c + a d ) i 类似两个多项式相乘 ，展开得:
a
c
+
a
d
i
+
b
c
i
+
b
d
i
2
ac+adi+bci+bdi^2
a c + a d i + b c i + b d i 2
i
2
=
−
1
i^2=-1
i 2 = − 1
(
a
c
－
b
d
)
+
(
b
c
+
a
d
)
i
(ac－bd)+(bc+ad)i
( a c － b d ) + ( b c + a d ) i 两个复数的积仍然是一个复数 幅角相加，模长相乘
定义
n
n
n
ω
n
i
\omega_n^i
ω n i
x
n
=
1
x^n=1
x n = 1
x
x
x
单位复根满足的性质 如下：可以想象成在单位圆上的旋转
ω
n
a
∗
ω
n
b
=
ω
n
a
+
b
\omega_n^a*\omega_n^b=\omega_n^{a+b}
ω n a ∗ ω n b = ω n a + b
ω
n
i
=
ω
n
i
+
n
\omega_n^i=\omega_n^{i+n}
ω n i = ω n i + n
ω
k
n
k
i
=
ω
n
i
\omega_{kn}^{ki}=\omega_n^i
ω k n k i = ω n i
ω
n
i
=
−
ω
n
i
+
n
/
2
\omega_n^i=-\omega_n^{i+n/2}
ω n i = − ω n i + n / 2
−
-
−
单位复根的作用：因为单位复根刚好有
n
n
n
n
−
1
n-1
n − 1
{
(
ω
n
0
,
f
(
ω
n
0
)
)
,
(
ω
n
1
,
f
(
ω
n
1
)
)
,
(
ω
n
2
,
f
(
ω
n
2
)
)
.
.
.
,
(
ω
n
n
−
1
,
f
(
ω
n
n
−
1
)
)
}
\{(\omega_n^0,f(\omega_n^0)),(\omega_n^1,f(\omega_n^1)),(\omega_n^2,f(\omega_n^2))...,(\omega_n^{n-1},f(\omega_n^{n-1}))\}
{ ( ω n 0 , f ( ω n 0 ) ) , ( ω n 1 , f ( ω n 1 ) ) , ( ω n 2 , f ( ω n 2 ) ) . . . , ( ω n n − 1 , f ( ω n n − 1 ) ) }
提前剧透一下：
F
F
T
FFT
F F T
n
n
n
FFT 的正变换实现，是基于对多项式进行奇偶项分开递归再合并的分治进行的
n
−
1
n-1
n − 1
n
n
n 这就跟我们引入单位复根的原因相结合了
设
f
(
x
)
=
a
0
+
a
1
∗
x
1
+
a
2
∗
x
2
+
.
.
.
+
a
n
−
1
∗
x
n
−
1
f(x)=a_0+a_1*x^1+a_2*x^2+...+a_{n-1}*x^{n-1}
f ( x ) = a 0 + a 1 ∗ x 1 + a 2 ∗ x 2 + . . . + a n − 1 ∗ x n − 1
f
0
(
x
)
=
a
0
+
a
2
∗
x
+
a
4
∗
x
2
+
a
6
∗
x
3
+
.
.
.
f_0(x)=a_0+a_2*x+a_4*x^2+a_6*x^3+...
f 0 ( x ) = a 0 + a 2 ∗ x + a 4 ∗ x 2 + a 6 ∗ x 3 + . . .
f
1
(
x
)
=
a
1
+
a
3
∗
x
+
a
5
∗
x
2
+
.
.
.
f_1(x)=a_1+a_3*x+a_5*x^2+...
f 1 ( x ) = a 1 + a 3 ∗ x + a 5 ∗ x 2 + . . .
f
(
x
)
=
f
0
(
x
2
)
+
x
∗
f
1
(
x
2
)
f(x)=f_0(x^2)+x*f_1(x^2)
f ( x ) = f 0 ( x 2 ) + x ∗ f 1 ( x 2 ) 证明的话把这个式子展开就行了，跳过
f
(
x
)
f(x)
f ( x )
接着，令
n
=
2
∗
p
n=2*p
n = 2 ∗ p 如果你看不懂证明推到请回到单位复根性质一处重新来过谢谢配合
f
(
ω
n
i
)
=
f
0
(
(
ω
n
/
2
i
/
2
)
2
)
+
ω
n
i
∗
f
1
(
(
w
n
/
2
i
/
2
)
2
)
=
f
0
(
ω
p
i
)
+
ω
n
i
∗
f
1
(
ω
p
i
)
f(\omega_n^i)=f_0((\omega_{n/2}^{i/2})^2)+\omega_n^i*f1((w_{n/2}^{i/2})^2)=f_0(\omega_p^i)+\omega_n^i*f_1(\omega_p^i)
f ( ω n i ) = f 0 ( ( ω n / 2 i / 2 ) 2 ) + ω n i ∗ f 1 ( ( w n / 2 i / 2 ) 2 ) = f 0 ( ω p i ) + ω n i ∗ f 1 ( ω p i )
f
(
ω
n
i
+
p
)
=
f
0
(
(
ω
n
/
2
(
i
+
p
)
/
2
)
2
)
+
ω
n
i
+
p
∗
f
1
(
(
w
n
/
2
(
i
+
p
)
/
2
)
2
)
f(\omega_n^{i+p})=f_0((\omega_{n/2}^{(i+p)/2})^2)+\omega_n^{i+p}*f1((w_{n/2}^{(i+p)/2})^2)
f ( ω n i + p ) = f 0 ( ( ω n / 2 ( i + p ) / 2 ) 2 ) + ω n i + p ∗ f 1 ( ( w n / 2 ( i + p ) / 2 ) 2 )
=
f
0
(
ω
p
i
+
p
)
+
ω
n
i
+
p
∗
f
1
(
ω
p
i
+
p
)
=
f
0
(
ω
p
i
)
−
ω
n
i
∗
f
1
(
ω
p
i
)
=f_0(\omega_p^{i+p})+\omega_n^{i+p}*f_1(\omega_p^{i+p})=f_0(\omega_p^i)-\omega_n^i*f_1(\omega_p^i)
= f 0 ( ω p i + p ) + ω n i + p ∗ f 1 ( ω p i + p ) = f 0 ( ω p i ) − ω n i ∗ f 1 ( ω p i )
f
0
(
ω
p
i
)
,
f
1
(
ω
p
i
)
f_0(\omega_p^i),f_1(\omega_p^i)
f 0 ( ω p i ) , f 1 ( ω p i )
O
(
1
)
O(1)
O ( 1 )
f
(
ω
n
i
)
,
f
(
ω
n
i
+
n
/
2
)
f(\omega_n^i),f(\omega_n^{i+n/2})
f ( ω n i ) , f ( ω n i + n / 2 )
f
0
(
x
)
,
f
1
(
x
)
f_0(x),f_1(x)
f 0 ( x ) , f 1 ( x )
n
/
2
n/2
n / 2 插值 ，我们就能
O
(
n
)
O(n)
O ( n )
f
(
x
)
f(x)
f ( x )
n
n
n
O
(
n
l
o
g
n
)
O(nlogn)
O ( n l o g n )
x
x
x
n
n
n
x
2
x^2
x 2
(
n
/
2
)
(n/2)
( n / 2 )
struct complex {
double real, i;
complex ( ) { }
complex ( double xx, double yy ) {
real = xx;
i = yy;
}
} a[ MAXN] , b[ MAXN] ;
complex operator + ( complex s, complex t ) {
return complex ( s. real + t. real, s. i + t. i ) ;
}
complex operator - ( complex s, complex t ) {
return complex ( s. real - t. real, s. i - t. i ) ;
}
complex operator * ( complex s, complex t ) {
return complex ( s. real * t. real - s. i * t. i, s. real * t. i + s. i * t. real ) ;
}
const double pi = acos ( - 1.0 ) ;
void FFT ( int limit, complex * a, int inv ) {
if ( limit == 1 )
return ;
complex a1[ limit >> 1 ] , a2[ limit >> 1 ] ;
for ( int i = 0 ; i < limit; i + = 2 ) {
a1[ i >> 1 ] = a[ i] ;
a2[ i >> 1 ] = a[ i + 1 ] ;
}
FFT ( limit >> 1 , a1, inv ) ;
FFT ( limit >> 1 , a2, inv ) ;
complex w = complex ( cos ( 2 * pi / limit ) , inv * sin ( 2 * pi / limit ) ) , p = complex ( 1 , 0 ) ;
for ( int i = 0 ; i < ( limit >> 1 ) ; i ++ , p = p * w ) {
a[ i] = a1[ i] + p * a2[ i] ;
a[ i + ( limit >> 1 ) ] = a1[ i] - p * a2[ i] ;
}
}
其实正变换的实现就是下列的矩阵相乘，反正我是没看出来
[
(
ω
n
0
)
0
(
ω
n
0
)
1
.
.
.
(
ω
n
0
)
n
−
1
(
ω
n
1
)
0
(
ω
n
1
)
1
.
.
.
(
ω
n
1
)
n
−
1
.
.
.
.
.
.
.
.
.
.
.
.
(
ω
n
n
−
1
)
0
(
ω
n
n
−
1
)
1
.
.
.
(
ω
n
n
−
1
)
n
−
1
]
×
[
a
0
a
1
a
2
.
.
.
a
n
−
1
]
=
[
f
(
w
n
0
)
f
(
w
n
1
)
f
(
w
n
2
)
.
.
.
f
(
w
n
n
−
1
)
]
\begin{bmatrix} \ (\omega_n^0)^0 & (\omega_n^0)^1 & ... &(\omega_n^0)^{n-1} \\ \\ (\omega_n^1)^0 & (\omega_n^1)^1 & ... & (\omega_n^1)^{n-1} \\ \\.&.&...&. \\.&.&...&. \\ (\omega_n^{n-1})^0 & (\omega_n^{n-1})^1 & ... & (\omega_n^{n-1})^{n-1} \\ \end{bmatrix} \times \begin{bmatrix} \ a_0 \\ \ a_1\\ \ a_2 \\ \ .\\ \ .\\ \ .\\ \ a_{n-1} \end{bmatrix} = \begin{bmatrix} \ f(w_n^0) \\ \ f(w_n^1) \\ \ f(w_n^2)\\ \ . \\ \ . \\ \ . \\ \ f(w_n^{n-1}) \\ \end{bmatrix}
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ ( ω n 0 ) 0 ( ω n 1 ) 0 . . ( ω n n − 1 ) 0 ( ω n 0 ) 1 ( ω n 1 ) 1 . . ( ω n n − 1 ) 1 . . . . . . . . . . . . . . . ( ω n 0 ) n − 1 ( ω n 1 ) n − 1 . . ( ω n n − 1 ) n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ × ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ a 0 a 1 a 2 . . . a n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ f ( w n 0 ) f ( w n 1 ) f ( w n 2 ) . . . f ( w n n − 1 ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
V
V
V
D
D
D
D
=
[
(
ω
n
−
0
)
0
(
ω
n
−
0
)
1
.
.
.
(
ω
n
−
0
)
n
−
1
(
ω
n
−
1
)
0
(
ω
n
−
1
)
1
.
.
.
(
ω
n
−
1
)
n
−
1
.
.
.
.
.
.
.
.
.
.
.
.
(
ω
n
−
n
+
1
)
0
(
ω
n
−
n
+
1
)
1
.
.
.
(
ω
n
−
n
+
1
)
n
−
1
]
D= \begin{bmatrix} \ (\omega_n^{-0})^0 & (\omega_n^{-0})^1 & ... &(\omega_n^{-0})^{n-1} \\ \\ (\omega_n^{-1})^0 & (\omega_n^{-1})^1 & ... & (\omega_n^{-1})^{n-1} \\ \\.&.&...&. \\.&.&...&. \\ (\omega_n^{-n+1})^0 & (\omega_n^{-n+1})^1 & ... & (\omega_n^{-n+1})^{n-1} \\ \end{bmatrix}
D = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ ( ω n − 0 ) 0 ( ω n − 1 ) 0 . . ( ω n − n + 1 ) 0 ( ω n − 0 ) 1 ( ω n − 1 ) 1 . . ( ω n − n + 1 ) 1 . . . . . . . . . . . . . . . ( ω n − 0 ) n − 1 ( ω n − 1 ) n − 1 . . ( ω n − n + 1 ) n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
那么计算矩阵
D
∗
V
D*V
D ∗ V
i
,
j
i,j
i , j
i
=
j
i=j
i = j
(
D
∗
V
)
i
,
j
=
∑
k
=
0
n
−
1
D
i
,
k
∗
V
k
,
j
=
∑
k
=
0
n
−
1
ω
n
−
i
k
∗
ω
n
j
k
=
∑
k
=
0
n
−
1
ω
n
(
j
−
i
)
k
=
∑
k
=
0
n
−
1
ω
n
0
=
n
(D*V)_{i,j}=\sum_{k=0}^{n-1}D_{i,k}*V_{k,j}=\sum_{k=0}^{n-1}\omega_n^{-ik}*\omega_n^{jk}=\sum_{k=0}^{n-1}\omega_n^{(j-i)k}=\sum_{k=0}^{n-1}\omega_n^0=n
( D ∗ V ) i , j = k = 0 ∑ n − 1 D i , k ∗ V k , j = k = 0 ∑ n − 1 ω n − i k ∗ ω n j k = k = 0 ∑ n − 1 ω n ( j − i ) k = k = 0 ∑ n − 1 ω n 0 = n
i
≠
j
i\ne j
i = j
(
D
∗
V
)
i
,
j
=
∑
k
=
0
n
−
1
D
i
,
k
∗
V
k
,
j
=
∑
k
=
0
n
−
1
ω
n
−
i
k
∗
ω
n
j
k
=
∑
k
=
0
n
−
1
ω
n
(
j
−
i
)
k
(D*V)_{i,j}=\sum_{k=0}^{n-1}D_{i,k}*V_{k,j}=\sum_{k=0}^{n-1}\omega_n^{-ik}*\omega_n^{jk}=\sum_{k=0}^{n-1}\omega_n^{(j-i)k}
( D ∗ V ) i , j = k = 0 ∑ n − 1 D i , k ∗ V k , j = k = 0 ∑ n − 1 ω n − i k ∗ ω n j k = k = 0 ∑ n − 1 ω n ( j − i ) k
=
(
ω
n
j
−
i
)
0
+
(
ω
n
j
−
i
)
1
+
(
ω
n
j
−
i
)
2
+
.
.
.
+
(
ω
n
j
−
i
)
n
−
1
=
n
=(\omega_n^{j-i})^0+(\omega_n^{j-i})^1+(\omega_n^{j-i})^2+...+(\omega_n^{j-i})^{n-1}=n
= ( ω n j − i ) 0 + ( ω n j − i ) 1 + ( ω n j − i ) 2 + . . . + ( ω n j − i ) n − 1 = n
ω
n
j
−
1
\omega_n^{j-1}
ω n j − 1 摘自百度百科 ，那么就可以转换为
(
ω
n
j
−
i
)
0
−
(
ω
n
j
−
i
)
1
∗
(
ω
n
j
−
i
)
n
−
1
1
−
(
ω
n
j
−
i
)
1
=
1
−
(
ω
n
j
−
i
)
n
1
−
ω
n
j
−
i
=
0
1
−
ω
n
j
−
i
=
0
\frac{(\omega_n^{j-i})^0-(\omega_n^{j-i})^1*(\omega_n^{j-i})^{n-1}}{1-(\omega_n^{j-i})^1}=\frac{1-(\omega_n^{j-i})^n}{1-\omega_n^{j-i}}=\frac{0}{1-\omega_n^{j-i}}=0
1 − ( ω n j − i ) 1 ( ω n j − i ) 0 − ( ω n j − i ) 1 ∗ ( ω n j − i ) n − 1 = 1 − ω n j − i 1 − ( ω n j − i ) n = 1 − ω n j − i 0 = 0
n
n
n
=
0
=0
= 0
j
≠
i
j\ne i
j = i
0
0
0
j
≠
i
j\ne i
j = i
(
D
∗
V
)
i
,
j
=
0
(D*V)_{i,j}=0
( D ∗ V ) i , j = 0
用
D
∗
V
∗
f
D*V*f
D ∗ V ∗ f
[
(
ω
n
−
0
)
0
(
ω
n
−
0
)
1
.
.
.
(
ω
n
−
0
)
n
−
1
(
ω
n
−
1
)
0
(
ω
n
−
1
)
1
.
.
.
(
ω
n
−
1
)
n
−
1
.
.
.
.
.
.
.
.
.
.
.
.
(
ω
n
−
n
+
1
)
0
(
ω
n
−
n
+
1
)
1
.
.
.
(
ω
n
−
n
+
1
)
n
−
1
]
×
[
(
ω
n
0
)
0
(
ω
n
0
)
1
.
.
.
(
ω
n
0
)
n
−
1
(
ω
n
1
)
0
(
ω
n
1
)
1
.
.
.
(
ω
n
1
)
n
−
1
.
.
.
.
.
.
.
.
.
.
.
.
(
ω
n
n
−
1
)
0
(
ω
n
n
−
1
)
1
.
.
.
(
ω
n
n
−
1
)
n
−
1
]
×
[
f
(
w
n
0
)
f
(
w
n
1
)
f
(
w
n
2
)
.
.
.
f
(
w
n
n
−
1
)
]
=
\begin{bmatrix} \ (\omega_n^{-0})^0 & (\omega_n^{-0})^1 & ... &(\omega_n^{-0})^{n-1} \\ \\ (\omega_n^{-1})^0 & (\omega_n^{-1})^1 & ... & (\omega_n^{-1})^{n-1} \\ \\.&.&...&. \\.&.&...&. \\ (\omega_n^{-n+1})^0 & (\omega_n^{-n+1})^1 & ... & (\omega_n^{-n+1})^{n-1} \\ \end{bmatrix} \times \begin{bmatrix} \ (\omega_n^0)^0 & (\omega_n^0)^1 & ... &(\omega_n^0)^{n-1} \\ \\ (\omega_n^1)^0 & (\omega_n^1)^1 & ... & (\omega_n^1)^{n-1} \\ \\.&.&...&. \\.&.&...&. \\ (\omega_n^{n-1})^0 & (\omega_n^{n-1})^1 & ... & (\omega_n^{n-1})^{n-1} \\ \end{bmatrix} \times \begin{bmatrix} \ f(w_n^0) \\ \ f(w_n^1) \\ \ f(w_n^2)\\ \ . \\ \ . \\ \ . \\ \ f(w_n^{n-1}) \\ \end{bmatrix} =
⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ ( ω n − 0 ) 0 ( ω n − 1 ) 0 . . ( ω n − n + 1 ) 0 ( ω n − 0 ) 1 ( ω n − 1 ) 1 . . ( ω n − n + 1 ) 1 . . . . . . . . . . . . . . . ( ω n − 0 ) n − 1 ( ω n − 1 ) n − 1 . . ( ω n − n + 1 ) n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ × ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ ( ω n 0 ) 0 ( ω n 1 ) 0 . . ( ω n n − 1 ) 0 ( ω n 0 ) 1 ( ω n 1 ) 1 . . ( ω n n − 1 ) 1 . . . . . . . . . . . . . . . ( ω n 0 ) n − 1 ( ω n 1 ) n − 1 . . ( ω n n − 1 ) n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ × ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ f ( w n 0 ) f ( w n 1 ) f ( w n 2 ) . . . f ( w n n − 1 ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ =
[
n
0
0
.
.
.
0
0
n
0
.
.
.
0
0
0
n
.
.
0
.
.
.
.
.
.
.
.
.
.
.
.
0
0
0
0
.
.
.
n
]
×
[
f
(
w
n
0
)
f
(
w
n
1
)
f
(
w
n
2
)
.
.
.
f
(
w
n
n
−
1
)
]
=
[
n
∗
a
0
n
∗
a
1
n
∗
a
2
.
.
.
n
∗
a
n
−
1
]
\begin{bmatrix} \ n&0&0&...&0 \\ \ 0&n&0&...&0 \\ \ 0 &0&n&..&0\\ \ ...&...&...&... &0\\ \ 0&0&0&...&n \\ \end{bmatrix} \times \begin{bmatrix} \ f(w_n^0) \\ \ f(w_n^1) \\ \ f(w_n^2)\\ \ . \\ \ . \\ \ . \\ \ f(w_n^{n-1}) \\ \end{bmatrix} = \begin{bmatrix} \ n*a_0 \\ \ n*a_1\\ \ n*a_2 \\ \ .\\ \ .\\ \ .\\ \ n*a_{n-1} \end{bmatrix}
⎣ ⎢ ⎢ ⎢ ⎢ ⎡ n 0 0 . . . 0 0 n 0 . . . 0 0 0 n . . . 0 . . . . . . . . . . . . . . 0 0 0 0 n ⎦ ⎥ ⎥ ⎥ ⎥ ⎤ × ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ f ( w n 0 ) f ( w n 1 ) f ( w n 2 ) . . . f ( w n n − 1 ) ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎡ n ∗ a 0 n ∗ a 1 n ∗ a 2 . . . n ∗ a n − 1 ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎤
/
n
/n
/ n
D
,
V
D,V
D , V
逆变换就相当于把正变换过程中的
ω
n
k
\omega^k_n
ω n k
w
n
−
k
w^{-k}_n
w n − k
之前的思路全都是递归思想，实现出来后发现吓死个人，所以我们考虑转成迭代
可以发现终序列是原序列每个元素的翻转。
所以我们才会把这个
F
F
T
FFT
F F T 盗一波百度的图片
void FFT ( complex * c, int f ) {
for ( int i = 0 ; i < len; i ++ )
if ( i < r[ i] )
swap ( c[ i] , c[ r[ i] ] ) ;
for ( int i = 1 ; i < len; i <<= 1 ) {
complex omega ( cos ( pi / i ) , f * sin ( pi / i ) ) ;
for ( int j = 0 ; j < len; j + = ( i << 1 ) ) {
complex w ( 1 , 0 ) ;
for ( int k = 0 ; k < i; k ++ , w = w * omega ) {
complex x = c[ j + k] , y = w * c[ j + k + i] ;
c[ j + k] = x + y;
c[ i + j + k] = x - y;
}
}
}
}
题目
#include <cmath>
#include <cstdio>
using namespace std;
#define MAXN 3000005
struct complex {
double real, i;
complex ( ) { }
complex ( double xx, double yy ) {
real = xx;
i = yy;
}
} a[ MAXN] , b[ MAXN] ;
complex operator + ( complex s, complex t ) {
return complex ( s. real + t. real, s. i + t. i ) ;
}
complex operator - ( complex s, complex t ) {
return complex ( s. real - t. real, s. i - t. i ) ;
}
complex operator * ( complex s, complex t ) {
return complex ( s. real * t. real - s. i * t. i, s. real * t. i + s. i * t. real ) ;
}
const double pi = acos ( - 1.0 ) ;
void FFT ( int limit, complex * a, int inv ) {
if ( limit == 1 )
return ;
complex a1[ limit >> 1 ] , a2[ limit >> 1 ] ;
for ( int i = 0 ; i < limit; i + = 2 ) {
a1[ i >> 1 ] = a[ i] ;
a2[ i >> 1 ] = a[ i + 1 ] ;
}
FFT ( limit >> 1 , a1, inv ) ;
FFT ( limit >> 1 , a2, inv ) ;
complex w = complex ( cos ( 2 * pi / limit ) , inv * sin ( 2 * pi / limit ) ) , p = complex ( 1 , 0 ) ;
for ( int i = 0 ; i < ( limit >> 1 ) ; i ++ , p = p * w ) {
a[ i] = a1[ i] + p * a2[ i] ;
a[ i + ( limit >> 1 ) ] = a1[ i] - p * a2[ i] ;
}
}
int main ( ) {
int n, m;
scanf ( "%d %d" , & n, & m ) ;
for ( int i = 0 ; i <= n; i ++ )
scanf ( "%lf" , & a[ i] . real ) ;
for ( int i = 0 ; i <= m; i ++ )
scanf ( "%lf" , & b[ i] . real ) ;
int limit = 1 ;
while ( limit <= n + m )
limit <<= 1 ;
FFT ( limit, a, 1 ) ;
FFT ( limit, b, 1 ) ;
for ( int i = 0 ; i <= limit; i ++ )
a[ i] = a[ i] * b[ i] ;
FFT ( limit, a, - 1 ) ;
for ( int i = 0 ; i <= n + m; i ++ )
printf ( "%d " , ( int ) ( a[ i] . real / limit + 0.5 ) ) ;
return 0 ;
}
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
#define MAXN 3000005
struct complex {
double real, i;
complex ( ) { }
complex ( double x, double y ) {
real = x;
i = y;
}
} a[ MAXN] , b[ MAXN] ;
int len, l;
int r[ MAXN] ;
const double pi = acos ( - 1.0 ) ;
complex operator + ( complex a, complex b ) {
return complex ( a. real + b. real, a. i + b. i ) ;
}
complex operator - ( complex a, complex b ) {
return complex ( a. real - b. real, a. i - b. i ) ;
}
complex operator * ( complex a, complex b ) {
return complex ( a. real * b. real - a. i * b. i, a. real * b. i + a. i * b. real ) ;
}
void FFT ( complex * c, int f ) {
for ( int i = 0 ; i < len; i ++ )
if ( i < r[ i] )
swap ( c[ i] , c[ r[ i] ] ) ;
for ( int i = 1 ; i < len; i <<= 1 ) {
complex omega ( cos ( pi / i ) , f * sin ( pi / i ) ) ;
for ( int j = 0 ; j < len; j + = ( i << 1 ) ) {
complex w ( 1 , 0 ) ;
for ( int k = 0 ; k < i; k ++ , w = w * omega ) {
complex x = c[ j + k] , y = w * c[ j + k + i] ;
c[ j + k] = x + y;
c[ i + j + k] = x - y;
}
}
}
}
int main ( ) {
int n, m;
scanf ( "%d %d" , & n, & m ) ;
for ( int i = 0 ; i <= n; i ++ )
scanf ( "%lf" , & a[ i] . real ) ;
for ( int i = 0 ; i <= m; i ++ )
scanf ( "%lf" , & b[ i] . real ) ;
len = 1 ;
while ( len <= n + m ) {
len <<= 1 ;
l ++ ;
}
for ( int i = 0 ; i < len; i ++ )
r[ i] = ( r[ i >> 1 ] >> 1 ) | ( ( i & 1 ) << ( l - 1 ) ) ;
FFT ( a, 1 ) ;
FFT ( b, 1 ) ;
for ( int i = 0 ; i < len; i ++ )
a[ i] = a[ i] * b[ i] ;
FFT ( a, - 1 ) ;
for ( int i = 0 ; i <= n + m; i ++ )
printf ( "%d " , ( int ) ( a[ i] . real / len + 0.5 ) ) ;
return 0 ;
}
https://www.cnblogs.com/Tiw-Air-OAO/p/10162034.html
