Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

FFT模板

用于求多项式卷积。

查看代码
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#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;
const double PI = acos(-1);
const int N = 3e6 + 10;
struct Complex
{
double x, y;
Complex() {}
Complex(double _x, double _y)
{
x = _x;
y = _y;
}
Complex operator+(Complex k)
{
return Complex(x + k.x, y + k.y);
}
Complex operator-(Complex k)
{
return Complex(x - k.x, y - k.y);
}
Complex operator*(Complex k)
{
return Complex(x * k.x - y * k.y, x * k.y + y * k.x);
}
} A[N], B[N];
int n, m, bit, tot, rev[N];
void fft(Complex *x, int op)
{
for (int i = 0; i < tot; i++)
if (rev[i] < i)
swap(x[rev[i]], x[i]);
for (int mid = 1; mid < tot; mid <<= 1)
{
Complex w1 = Complex(cos(PI / mid), sin(op * PI / mid));
for (int i = 0; i < tot; i += (mid << 1))
{
Complex cur = Complex(1, 0);
for (int j = 0; j < mid; j++, cur = cur * w1)
{
Complex p = x[i + j], q = cur * x[i + j + mid];
x[i + j] = p + q, x[i + j + mid] = p - q;
}
}
}
if (op == -1)
for (int i = 0; i < tot; i++)
x[i].x = x[i].x / tot + 0.5;
}
void init()
{
while ((1 << bit) < n + m - 1)
bit++;
tot = 1 << bit;
for (int i = 1; i < tot; i++)
rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
}
int main()
{
scanf("%d%d", &n, &m);
n++, m++;
for (int i = 0; i < n; i++)
scanf("%lf", &A[i].x);
for (int i = 0; i < m; i++)
scanf("%lf", &B[i].x);
init();
fft(A, 1);
fft(B, 1);
for (int i = 0; i < tot; i++)
A[i] = A[i] * B[i];
fft(A, -1);
for (int i = 0; i < n + m - 1; i++)
printf("%d ", (int)A[i].x);
return 0;
}