[洛谷3803] 【模板】多项式乘法（FFT）

题目大意：FFT，给你两个多项式，请输出乘起来后的多项式。

题解：FFT

卡点：无

C++ Code：

#include <cstdio>
#include <cmath>
using namespace std;
const double Pi = acos(-1);
int n, m;
struct complex {
	double r, i;
	complex (double a = 0, double b = 0) {r = a, i = b;}
	complex operator + (complex a) {return (complex) {r + a.r, i + a.i};}
	complex operator - (complex a) {return (complex) {r - a.r, i - a.i};}
	complex operator * (complex a) {return (complex) {r * a.r - i * a.i, r * a.i + i * a.r};}
} a[500000], b[500000];
int rev[500000], dig, l;
void swap(complex &a, complex &b) {complex t = a; a = b; b = t;}
void FFT(complex *a, int op) {
	for (int i = 0; i < l; i++) if (i < rev[i]) swap(a[i], a[rev[i]]);
	for (int mid = 1; mid < l; mid <<= 1) {
		complex Wn(cos(Pi / mid), op * sin(Pi / mid));
		for (int i = 0; i < l; i += (mid << 1)) {
			complex W(1, 0);
			for (int j = 0; j < mid; j++, W = W * Wn) {
				complex X = a[i + j], Y = W * a[i + j + mid];
				a[i + j] = X + Y;
				a[i + j + mid] = X - Y;
			}
		}
	}
	if (op == -1) for (int i = 0; i <= l; i++) a[i].r /= l;
}
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);
	l = 1; while (l <= (n + m)) l <<= 1, dig++;
	for (int i = 0; i < l; i++) rev[i] =  (rev[i >> 1] >> 1) | ((i & 1) << (dig - 1));
//	for (int i = 0; i < l; i++) printf("::%d\n ", rev[i]);
	FFT(a, 1); FFT(b, 1);
//	for (int i = 0; i < l; i++) printf("::%lf %lf\n", b[i].r, b[i].i);
	for (int i = 0; i < l; i++) a[i] = a[i] * b[i];
	FFT(a, -1);
	for (int i = 0; i <= n + m; i++) printf("%d ", (int)(a[i].r + 0.5));
	puts("");
	return 0;
}