[Luogu P4512] 【模板】多项式除法

洛谷传送门

题目描述

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 $Q(x)$ , $R(x)$ ，满足以下条件：

$Q(x)$ 次数为 $n-m$ ， $R(x)$ 次数小于 $m$
$F(x) = Q(x) * G(x) + R(x)$

所有的运算在模 $998244353$ 意义下进行。

输入输出格式

输入格式：

第一行两个整数 $n$ ， $m$ ，意义如上。
第二行 $n+1$ 个整数，从低到高表示 $F(x)$ 的各个系数。
第三行 $m+1$ 个整数，从低到高表示 $G(x)$ 的各个系数。

输出格式：

第一行 $n-m+1$ 个整数，从低到高表示 $Q(x)$ 的各个系数。
第二行 $m$ 个整数，从低到高表示 $R(x)$ 的各个系数。
如果 $R(x)$ 不足 $m-1$ 次，多余的项系数补 $0$ 。

输入输出样例

输入样例#1：

5 1
1 9 2 6 0 8
1 7

输出样例#1：

237340659 335104102 649004347 448191342 855638018
760903695

说明

对于所有数据， $1 \le m < n \le 10^5$ ，给出的系数均属于 $[0, 998244353) \cap \mathbb{Z}$ 。

解题分析

设 $F_R(x)=x^nF(\frac{1}{x})$ ，可以发现 $F_R(x)$ 和 $F(x)$ 的系数正好从大到小正好相反。

那么就有：
$F(\frac{1}{x})=Q(\frac{1}{x})G(\frac{1}{x})+R(\frac{1}{x}) \\ F_R(x)=Q_R(x)G_R(x)+x^{n-m+1}R_R(x) \\ F_R(x)\equiv Q_R(x)G_R(x)(mod\ x^{n-m+1}) \\ Q_R(x)\equiv \frac{F_R(x)}{G_R(x)} (mod\ x^{n-m+1})$
然后就是一波求逆，然后带入原式即可得到 $R(x)$ 。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>
#include <cctype>
#define R register
#define IN inline
#define W while
#define gc getchar()
#define ll long long
#define MOD 998244353
#define g 3
#define ginv 332748118
#define MX 400500
template <class T>
IN void in(T &x)
{
	x = 0; R char c = gc;
	for (; !isdigit(c); c = gc);
	for (;  isdigit(c); c = gc)
	x = (x << 1) + (x << 3) + c - 48;
}
IN int fpow(R int base, R int tim)
{
	int ret = 1;
	W (tim)
	{
		if (tim & 1) ret = 1ll * ret * base % MOD;
		base = 1ll * base * base % MOD, tim >>= 1;
	}
	return ret;
}
int n, m;
int F[MX], G[MX], FR[MX], GR[MX], Q[MX], rev[MX], a[MX], b[MX], GRinv[MX], sur[MX];
namespace Poly
{
	IN void NTT(int *dat, R int len, R int typ)
	{
		for (R int i = 1; i < len; ++i) if (rev[i] > i) std::swap(dat[rev[i]], dat[i]);
		R int seg, step, bd, now, cur, buf1, buf2, base, deal;
		for (seg = 1; seg < len; seg <<= 1)
		{
			base = fpow(typ ? g : ginv, (MOD - 1) / (seg << 1)), step = seg << 1;
			for (now = 0; now < len; now += step)
			{
				bd = now + seg, deal = 1;
				for (cur = now; cur < bd; ++cur, deal = 1ll * deal * base % MOD)
				{
					buf1 = dat[cur], buf2 = 1ll * dat[cur + seg] * deal % MOD;
					dat[cur] = (buf1 + buf2) % MOD, dat[cur + seg] = (buf1 - buf2 + MOD) % MOD;
				}
			}
		}
		if (typ) return; int inv = fpow(len, MOD - 2);
		for (R int i = 0; i < len; ++i) dat[i] = 1ll * dat[i] * inv % MOD;
	}
	void Getinv(R int up, R int lg, int *ans, int *ind)
	{
		if (up == 1) return ans[0] = fpow(ind[0], MOD - 2), void();
		Getinv(up >> 1, lg - 1, ans, ind); R int len = up << 1;
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < len; ++i) a[i] = b[i] = 0;
		for (R int i = 0; i < up; ++i) a[i] = ans[i], b[i] = ind[i];
		NTT(a, len, 1), NTT(b, len, 1);
		for (R int i = 0; i < len; ++i) a[i] = (2 * a[i] % MOD - 1ll * b[i] * a[i] % MOD * a[i] % MOD + MOD) % MOD;
		NTT(a, len, 0);
		for (R int i = 0; i < up; ++i) ans[i] = a[i];
	}
	IN void Div(int n, int m)
	{
		int bd = n - m + 1, lg = 0, len = 1, up = n + m;
		for (; len < bd; len <<= 1, lg++);
		Getinv(len, lg + 1, GRinv, GR);
		for (len = 1, lg = 0; len <= (bd << 1); len <<= 1, ++lg);
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < len; ++i) a[i] = b[i] = 0;
		for (R int i = 0; i < bd; ++i) a[i] = FR[i];
		for (R int i = 0; i < bd; ++i) b[i] = GRinv[i];
		NTT(a, len, 1), NTT(b, len, 1);
		for (R int i = 0; i < len; ++i) a[i] = 1ll * a[i] * b[i] % MOD;
		NTT(a, len, 0);
		for (R int i = 0; i < bd; ++i) Q[i] = a[bd - 1 - i];
		up = n; for (len = 1, lg = 0; len <= up; len <<= 1, ++lg);
		for (R int i = 1; i < len; ++i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << lg - 1);
		for (R int i = 0; i < len; ++i) a[i] = b[i] = 0;
		for (R int i = 0; i < bd; ++i) a[i] = Q[i];
		for (R int i = 0; i <= m; ++i) b[i] = G[i];
		NTT(a, len, 1), NTT(b, len, 1);
		for (R int i = 0; i < len; ++i) a[i] = 1ll * a[i] * b[i] % MOD;
		NTT(a, len, 0);
		for (R int i = 0; i < m; ++i) sur[i] = (F[i] - a[i] + MOD) % MOD;
		for (R int i = 0; i < bd; ++i) printf("%d ", Q[i]); puts("");
		for (R int i = 0; i < m; ++i) printf("%d ", sur[i]);
	}
}
int main(void)
{
	in(n), in(m);
	for (R int i = 0; i <= n; ++i) in(F[i]), FR[n - i] = F[i];
	for (R int i = 0; i <= m; ++i) in(G[i]), GR[m - i] = G[i];
	Poly::Div(n, m);
}