[洛谷P3803] 【模板】多项式乘法（FFT, NTT）

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：$FFT$，由于给的$n$不是很大，也可以用$NTT$做

卡点：无

C++ Code：

 FFT：

#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 /= (int a) {r /= a, i /= a;}
	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] /= 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));
	FFT(a, 1), FFT(b, 1);
	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));
	return 0;
}

NTT：

#include <cstdio>
#define int long long
using namespace std;
const int maxn = 2100010;
const int mod = 998244353;
const int P = 3, invP = 332748118;
int n, m;
int a[maxn], b[maxn], rev[maxn], l, dig;
int Inv[2040826], invl;
inline void swap(int &a, int &b) {a ^= b ^= a ^= b;}
int inv(int i) {
	if (i < 2040826) {
		if (Inv[i]) return Inv[i];
		return (Inv[i] = inv(mod % i) * (mod - mod / i) % mod);
	}else return inv(mod % i) * (mod - mod / i) % mod;
}
inline int pw(int base, int p) {
	int ans = 1;
	for (p <<= 1; p >>= 1; (base *= base) %= mod) if (p & 1) (ans *= base) %= mod;
	return ans;
}
void NTT(int *a, int op) {
	int Yx;
	if (op == 1) Yx = P; else Yx = invP;
	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) {
		int Wn = pw(Yx, (mod - 1) / (mid << 1));
		for (int i = 0; i < l; i += (mid << 1)) {
			int W = 1;
			for (int j = 0; j < mid; j++, W = W * Wn % mod) {
				int X = a[i + j], Y = W * a[i + j + mid] % mod;
				a[i + j] = (X + Y) % mod;
				a[i + j + mid] = (X - Y + mod) % mod;
			}
		}
	}
	if (op == -1) for (int i = 0; i < l; i++) a[i] = (a[i] * invl) % mod;
}
signed main() {
	Inv[0] = Inv[1] = 1;
	scanf("%lld%lld", &n, &m);
	for (int i = 0; i <= n; i++) scanf("%lld", &a[i]);
	for (int i = 0; i <= m; i++) scanf("%lld", &b[i]);
	l = 1; while (l <= n + m) l <<= 1, dig++; invl = inv(l);
	for (int i = 1; i < l; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (dig - 1));
	NTT(a, 1), NTT(b, 1);
	for (int i = 0; i < l; i++) (a[i] *= b[i]) %= mod;
	NTT(a, -1);
	for (int i = 0; i <= n + m; i++) printf("%lld ", a[i]);
	return 0;
}