[洛谷P4512]【模板】多项式除法

题目大意：给定一个$n$次多项式$F(x)$和一个$m$次多项式$G(x)$，请求出多项式$Q(x),R(x)$，满足：

1. $Q(x)$次数为$n-m$，$R(x)$次数小于$m$
2. $F(x)=Q(x)\times G(x)+R(x)$

题解：多项式除法。
$$
F(x)\equiv Q(x)G(x)+R(x)(\bmod{x^n})\\
F(\dfrac 1 x)\equiv Q(\dfrac 1 x)G(\dfrac 1 x)+R(\dfrac 1 x)(\bmod{x^n})\\
x^nF(\dfrac 1 x)\equiv x^{n-m}Q(\dfrac 1 x)\cdot x^mG(\dfrac 1 x)+x^nR(\dfrac 1 x)(\bmod{x^n})\\
F_R(x)\equiv Q_R(x)G_R(x)+x^{n-m+1}R_R(x)(\bmod{x^n})\\
F_R(x)\equiv Q_R(x)G_R(x)(\bmod{x^{n-m+1}})\\
Q_R(x)\equiv F_R(x)G_R^{-1}(x)(\bmod{x^{n-m+1}})\\
R_R(x)=F_R(x)-G_R(x)Q_R(x)
$$
卡点：无

C++ Code：

#include <algorithm>
#include <cstdio>
#include <cctype>
namespace __IO {
	namespace R {
		int x, ch;
		inline int read() {
			ch = getchar();
			while (isspace(ch)) ch = getchar();
			for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x = x * 10 + (ch & 15);
			return x;
		}
	}
}
using __IO::R::read;
 
const int mod = 998244353, G = 3;
 
namespace Math {
	int x, y;
	inline int pw(int base, int p) {
		int res = 1;
		for (; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	void exgcd(int a, int b, int &x, int &y) {
		if (!b) x = 1, y = 0;
		else exgcd(b, a % b, y, x), y -= a / b * x;
	}
	inline int inv(int a) {
		exgcd(a, mod, x, y);
		return x + (x >> 31 & mod);
	}
}
 
#define N 262144
inline void reduce(int &a) {a += a >> 31 & mod;}
inline void clear(int *l, const int *r) {
	if (l >= r) return ;
	while (l != r) *l++ = 0;
}
 
namespace Poly {
	int lim, ilim, rev[N], s;
	int Wn[N + 1];
	inline void init(int n) {
		s = -1, lim = 1; while (lim < n) lim <<= 1, s++; ilim = Math::inv(lim);
		for (register int i = 0; i < lim; i++) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
		const int t = Math::pw(G, (mod - 1) / lim);
		*Wn = 1; for (register int i = 1; i <= lim; i++) Wn[i] = static_cast<long long> (Wn[i - 1]) * t % mod;
	}
	void NTT(int *A, int op = 1) {
		for (register int i = 0; i < lim; i++) if (i < rev[i]) std::iter_swap(A + i, A + rev[i]);
		for (register int mid = 1; mid < lim; mid <<= 1) {
			const int t = lim / mid >> 1;
			for (register int i = 0; i < lim; i += mid << 1) {
				for (register int j = 0; j < mid; j++) {
					const int W = op ? Wn[t * j] : Wn[lim - t * j];
					const int X = A[i + j], Y = static_cast<long long> (A[i + j + mid]) * W % mod;
					reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
				}
			}
		}
		if (!op) for (register int i = 0; i < lim; i++) A[i] = static_cast<long long> (A[i]) * ilim % mod;
	}
	int C[N];
	void INV(int *A, int *B, int n) {
		if (n == 1) {
			*B = Math::inv(*A);
			return ;
		}
		INV(A, B, n + 1 >> 1);
		std::copy(A, A + n, C);
		init(n << 1), clear(C + n, C + lim);
		NTT(B), NTT(C);
		for (int i = 0; i < lim; i++) B[i] = (2 + mod - static_cast<long long> (C[i]) * B[i] % mod) * B[i] % mod;
		NTT(B, 0);
		clear(B + n, B + lim);
	}
	int D[N], E[N], F[N];
	void DIV(int *A, int *B, int *Q, int n, int m) {
		std::reverse_copy(A, A + n, D), std::reverse_copy(B, B + m, E);
		clear(D + n - m + 1, D + n);
		clear(E + n - m + 1, E + m);
		INV(E, F, n - m + 1), init(n - m + 1 << 1);
		NTT(D), NTT(F);
		for (int i = 0; i < lim; i++) Q[i] = static_cast<long long> (D[i]) * F[i] % mod;
		NTT(Q, 0);
		std::reverse(Q, Q + n - m + 1);
		for (int i = n - m + 1; i < lim; i++) Q[i] = 0;
	}
	int G[N];
	void DIV_MOD(int *A, int *B, int *Q, int *R, int n, int m) {
		DIV(A, B, Q, n, m);
		std::copy(Q, Q + n - m + 1, G);
		init(n << 1);
		NTT(A), NTT(B), NTT(G);
		for (int i = 0; i < lim; i++) R[i] = (A[i] + mod - static_cast<long long> (B[i]) * G[i] % mod) % mod;
		NTT(R, 0);
	}
}
 
int A[N], B[N], Q[N], R[N];
int n, m;
int main() {
	n = read() + 1, m = read() + 1;
	for (int i = 0; i < n; i++) A[i] = read();
	for (int i = 0; i < m; i++) B[i] = read();
	Poly::DIV_MOD(A, B, Q, R, n, m);
	for (int i = 0; i < n - m + 1; i++) printf("%d ", Q[i]); puts("");
	for (int i = 0; i < m - 1; i++) printf("%d ", R[i]); puts("");
	return 0;
}