多项式除法

多项式除法解决这样一个问题：
有一个\(n\)次多项式\(A(x)\)和一个\(m\)次多项式\(B(x)\)，你希望求得多项式\(Q(x)\)和\(R(x)\)，使得

\[A(x) = Q(x)B(x) + R(x) \]

其中\(deg_Q \le n - m\)，\(deg_R < m\)

那个余数特别难处理，考虑如何去掉
我们记\(A^{R}(x) = x^nA(\frac{1}{x})\)表示系数翻转后的多项式
那么对于

\[A(x) = Q(x)B(x) + R(x) \]

我们写成

\[A(\frac{1}{x}) = Q(\frac{1}{x})B(\frac{1}{x}) + R(\frac{1}{x}) \]

两边同乘\(x^{n}\)

\[x^{n}A(\frac{1}{x}) = x^{n - m}Q(\frac{1}{x})x^{n}B(\frac{1}{x}) + x^{n}R(\frac{1}{x}) \]

即

\[A^{R}(x) = Q^{R}(x)B^{R}(x) + x^{n - m + 1}R^{R}(x) \]

发现此时在模\(x^{n - m + 1}\)意义下\(R^{R}(x)\)就不存在了，而\(Q(x)\)次的上界恰好就是\(n - m + 1\)，所以在此模意义下我们求出的\(Q^{R}(x)\)就是实际我们所需的\(Q(x)\)的系数翻转后的多项式

化为

\[Q^{R}(x) \equiv A^{R}(x)(B^{R}(x))^{-1} \pmod {x^{n - m + 1}} \]

多项式求逆即可
\(Q^{R}(x)\)翻转之后得到\(Q(x)\)，回代求出\(R(x)\)
复杂度\(O(nlogn)\)

模板题

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
const int G = 3,P = 998244353;
inline int qpow(int a,int b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
int R[maxn];
inline void NTT(int* a,int n,int f){
	for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
	for (int i = 1; i < n; i <<= 1){
		int gn = qpow(G,(P - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1)){
			int g = 1,x,y;
			for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
				x = a[j + k],y = 1ll * g * a[j + k + i] % P;
				a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
			}
		}
	}
	if (f == 1) return;
	int nv = qpow(n,P - 2); reverse(a + 1,a + n);
	for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int c[maxn];
void Inv(int* a,int* b,int deg){
    if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
    Inv(a,b,(deg + 1) >> 1);
    int L = 0,n = 1;
    while (n < (deg << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i < deg; i++) c[i] = a[i];
    for (int i = deg; i < n; i++) c[i] = 0;
    NTT(c,n,1); NTT(b,n,1);
    for (int i = 0; i < n; i++)
        b[i] = 1ll * ((2ll - 1ll * c[i] * b[i] % P) + P) % P * b[i] % P;
    NTT(b,n,-1);
    for (int i = deg; i < n; i++) b[i] = 0;
}
int f[maxn],g[maxn],q[maxn],r[maxn],gv[maxn],N,M;
void work(){
	reverse(f,f + N + 1);
	reverse(g,g + M + 1);
	Inv(g,gv,N - M + 1);
	
	int n = 1,L = 0,E = N - M;
	while (n <= (E << 1)) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	for (int i = 0; i <= E; i++) c[i] = f[i];
	for (int i = E + 1; i < n; i++) c[i] = 0;
	for (int i = E + 1; i < n; i++) gv[i] = 0;
	NTT(c,n,1); NTT(gv,n,1);
	for (int i = 0; i < n; i++) q[i] = 1ll * c[i] * gv[i] % P;
	NTT(q,n,-1);
	reverse(q,q + E + 1);
	for (int i = E + 1; i < n; i++) q[i] = 0;
	reverse(g,g + M + 1);
	reverse(f,f + N + 1);
	
	n = 1,L = 0;
	while (n <= N) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	for (int i = 0; i <= E; i++) c[i] = q[i];
	for (int i = E + 1; i < n; i++) c[i] = 0;
	for (int i = M + 1; i < n; i++) g[i] = 0;
	NTT(g,n,1); NTT(c,n,1);
	for (int i = 0; i < n; i++) c[i] = 1ll * c[i] * g[i] % P;
	NTT(c,n,-1);
	for (int i = 0; i < M; i++) r[i] = ((f[i] - c[i]) % P + P) % P;
	
	for (int i = 0; i <= E; i++) printf("%d ",q[i]); puts("");
	for (int i = 0; i < M; i++) printf("%d ",r[i]);
}
int main(){
	N = read(); M = read();
	for (int i = 0; i <= N; i++) f[i] = read();
	for (int i = 0; i <= M; i++) g[i] = read();
	work();
	return 0;
}