多项式
\[ \begin{aligned} A(x)& =\sum_{i=0}^na_ix_i \\ &=(x_0,y_0),(x_1,y_1),\cdots,(x_n,y_n) \end{aligned} \]
卷积
\[ A(x)B(x)=\sum_{i=0}^n\sum_{j=0}^na_ib_jx^{i+j} \]
若 \(A\ast B=C\)，可知求 \(C\) 的系数数列时间复杂度 \(O(n^2)\)。
考虑点值表达式的卷积
\[ A(x)=(x_0,y_0),(x_1,y_1),\cdots,(x_n,y_n) \]
\[ B(x)=(x_0,y_0'),(x_1,y_1'),\cdots,(x_n,y_n')) \]
可得出 \(C\) 的一个不完整表达
\[C(x)=(x_0,y_0y_0'),(x_1,y_1y_1'),\cdots,(x_n,y_ny_n')\]
若将点对个数增至 \(2n+1\) 个，可知求出 \(C\) 的点值数列时间复杂度 \(O(n)\)。
DFT & IDFT
DFT(Discrete Fourier Transform，离散傅里叶变换) 是系数数列 \(a_0,a_1,\cdots,a_n\) 转为点值数列的过程，IDFT(Inverse Discrete Fourier Transform，逆离散傅里叶变换) 是 DFT 的逆运算。
FFT
考虑 DFT 和 IDFT 时间复杂度难以接受，可以利用 FFT(Fast Fourier Transform，快速傅里叶变换) 将其加速。
单位复根
\(\omega^n=1\) 在复数域下有 \(n\) 个不同的根。
\[ e^{ix}=\cos x+i\sin x \\ e^{2\pi i}=1=\omega^n \\ \omega=e^{\frac{2\pi i}n} \]
定义主次单位根
\[ \omega_n=e^{\frac{2\pi i}n} \]
对于其它单位根，记作
\[ \omega_n^k=e^{\frac{2\pi ik}n} \]
存在以下性质
\[ \omega_{2n}^{k}+\omega_{2n}^{n+k}=0 \]
\[ (\omega_{2n}^k)^2=\omega_{n}^k \]
考虑用单位复根的性质优化 DFT。
对于多项式 \(A(x)=\sum_{i=0}^{2n-1}a_ix^i\)，其中 \(2n=2^k\)，我们将 \(\omega_{2n}^0,\omega_{2n}^1,\cdots,\omega_{2n}^{{2n}-1}\) 代入公式计算点值。
现在重定义两个多项式
\[ A_0(x)=a_0+a_2x+a_4x^2+\cdots+a_{2n}x^n \]
\[ A_1(x)=a_1+a_3x+a_5x^2+\cdots+a_{2n-1}x^n \]
显然
\[ A(x)=A_0(x^2)+xA_1(x^2) \]
将单位复根代入上式
\[ \begin{aligned} A(\omega_{2n}^k)&=A_0((\omega_{2n}^k)^2)+\omega_{2n}^kA_1((\omega_{2n}^k)^2)\\ &=A_0(\omega_n^k)+\omega_{2n}^kA_1(\omega_n^k)\\ A(\omega_{2n}^{n+k})&=A_0((\omega_{2n}^{n+k})^2)+\omega_{2n}^{n+k}A_1((\omega_{2n}^{n+k})^2)\\ &=A_0(\omega_n^k)-\omega_{2n}^kA_1(\omega_n^k) \end{aligned} \]
发现对于 \(k\in[0,1,\cdots,n-1]\) \(A(\omega_{2n}^k)\) 和 \(A(\omega_{2n}^{n+k})\) 是可以在一起计算的。
于是有以下伪代码
function FFT(complex A[], int Length)
if Length == 1 return
complex A0[Length / 2], A1[Length / 2]
for int i = 0 to Length / 2 - 1 with i += 1
A0[i] = A[i * 2]
A1[i] = A[i * 2 + 1]
FFT(A0, Length / 2)
FFT(A1, Length / 2)
complex wn = (cos(2 * Pi / Length), sin(2 * Pi / Length))
complex w = (1, 0)
for int i = 0 to Length / 2 - 1 with i += 1, w *= wn
A[i] = A0[i] + A1[i] * w
A[i + Length / 2] = A0[i] - A1[i] * w
考虑 IDFT，IDFT 是 DFT 的逆运算，令 \(DFT(\{a_i\})=\{b_i\}\)，已知
\[b_k=\sum_{i=0}^{n-1}a_i\omega_n^{ki}\]
存在结论
\[a_k=\frac 1 n\sum_{i=0}^{n-1}b_i\omega_n^{-ki}\]
证明，将前式代入后式
\[ \begin{aligned} a_k&=\frac 1 n\sum_{i=0}^{n-1}b_i\omega_n^{-ki}\\ &=\frac 1 n\sum_{i=0}^{n-1}\omega_n^{-ki}\sum_{j=0}^{n-1}a_j\omega_n^{ij}\\ &=\frac 1 n\sum_{j=0}^{n-1}a_j\sum_{i=0}^{n-1}\omega_n^{-ki}\omega_{n}^{ij}\\ &=\frac 1 n\sum_{j=0}^{n-1}a_j\sum_{i=0}^{n-1}\omega_{n}^{i(j-k)} \end{aligned} \]
考虑 \(\sum_{i=0}^{n-1}\omega_n^{i(j-k)}\)
当 \(j=k\)，
\[\sum_{i=0}^{n-1}\omega_n^{i(j-k)}=\sum1=n\]
当 \(j\neq k\)，
\[ \begin{aligned} \sum_{i=0}^{n-1}\omega_n^{i(j-k)}&=\sum_{i=0}^{n-1}(\omega_n^{j-k})^i\\ &=\frac{1-(\omega_n^{j-k})^n}{1-\omega_n^{j-k}}\\ &=\frac{1-(\omega_n^n)^{j-k}}{1-\omega_n^{j-k}}=\frac{1-1}{1-\omega_n^{j-k}}=0 \end{aligned} \]
所以，
\[ \begin{aligned} a_k&=\frac 1 n\sum_{j=0}^{n-1}a_j\sum_{i=0}^{n-1}\omega_{n}^{i(j-k)}\\ &=\frac 1 n\cdot na_k=a_k \end{aligned} \]
发现 DFT 和 IDFT 的公式形式一样，对于指数上的 \(-1\) 只需要在代码中加入一个开关即可。
蝴蝶定理
考虑 FFT 的分治过程，以 \(n=16\) 为例
\[\{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15\}\]
\[\{0,2,4,6,8,10,12,14\},\{1,3,5,7,9,11,13,15\}\]
\[\{0,4,8,12\},\{2,6,10,14\},\{1,5,9,13\},\{3,7,11,15\}\]
\[\{0,8\},\{4,12\},\{2,10\},\{6,14\},\{1,9\},\{5,13\},\{3,11\},\{7,15\}\]
其二进制下表示，
\[\{0000,1000\},\{0100,1100\},\{0010,1010\},\{0110,1110\},\]
\[\{0001,1001\},\{0101,1101\},\{0011,1011\},\{0111,1111\}\]
观察发现，若干次蝴蝶操作（奇偶分治）后，其数位二进制翻转后是连续的。
对于二进制翻转，可用递推计算，rev[i] = rev[i >> 1] >> 1 | (i & 1 ? n >> 1 : 0)。
考虑用蝴蝶定理使 \(FFT\) 的过程避免递归，可以先将 \(\{a_i\}\) 按 \(rev\) 序重新排列，在分治树上从下往上，
function FFT(complex A[], int Length, int on)
for int i = 0 to Length - 1 with i += 1
if i < Rev[i]
swap(A[i], A[Rev[i]])
for int k = 1 to n - 1 with k *= 2
complex wn = (cos(Pi / k), sin(on * Pi / k))
for int i = 0 to n with i += k * 2
complex w = (1, 0)
for int j = i to i + k - 1 with j += 1
complex x = a[j]
complex y = a[j + k]
a[j] = x + y
a[j + k] = x - y
if on == -1
for int i = 0 to n - 1 with i += 1
a[i] /= n
然后枚举当前合并的长度 \(2k\)，枚举合并区间开始位置 \(i\)，枚举区间中的元素 \(a_j\)。
Luogu P3803 多项式乘法(FFT)
#include <iostream>
#include <cmath>
using namespace pl;
const int N = 4200000 + 7;
const double PI = acos(-1);
int n, m;
complex a[N], b[N];
int rev[N];
void mul(complex a[], int n, complex b[], int m);
void fft(complex a[], int n, int on);
int main() {
scanf("%d%d", &n, &m);
for (int i = 0; i <= n; ++i) scanf("%lf", &a[i].r);
for (int i = 0; i <= m; ++i) scanf("%lf", &b[i].r);
mul(a, n, b, m);
for (int i = 0; i <= n + m; ++i)
printf("%d", (int)(a[i].r + 0.5));
return 0;
}
void mul(complex a[], int n, complex b[], int m) {
for (n = m = n + m - 1; n ^ (n & -n); n += n & -n);
fft(a, n, 1), fft(b, n, 1);
for (int i = 0; i < n; ++i) a[i] = a[i] * b[i];
fft(a, n, -1);
}
void fft(complex a[], int n, int on) {
static int rev[N], lstn;
if (n != lstn) {
lstn = n;
for (int i = 0; i < n; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1 ? n >> 1 : 0);
}
complex t, w, x;
for (int k = 1; k < n; k <<= 1) {
t = (complex){ cos(PI / k), on * sin(PI / k) };
for (int i = 0; i < n; i += k << 1) {
w = (complex){ 1, 0 }；
for (int j = i; j < i + k; ++j, w = w * t)
x = w * a[j + k], a[j + k] = a[j] - x, a[j] = a[j] + x;
}
}
if (on == -1)
for (int i = 0; i < n; ++i) a[i].r /= n;
}
\[\mathcal{by\ Ryedii}\]
来源：https://www.cnblogs.com/Ryedii-blog/p/12178719.html