Luogu4512 【模板】多项式除法（多项式求逆+NTT）

　　http://blog.miskcoo.com/2015/05/polynomial-division 好神啊！

　　通过翻转多项式消除余数的影响，主要原理是商只与次数不小于m的项有关。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 550000
#define P 998244353
int n,m,t,a[N],b[N],r[N],c[N],d[N],e[N],inv3;
int ksm(int a,int k)
{
    if (k==0) return 1;
    int tmp=ksm(a,k>>1);
    if (k&1) return 1ll*tmp*tmp%P*a%P;
    else return 1ll*tmp*tmp%P;
}
int inv(int a){return ksm(a,P-2);}
void DFT(int n,int *a,int p)
{
    for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=2;i<=n;i<<=1)
    {
        int wn=ksm(p,(P-1)/i);
        for (int j=0;j<n;j+=i)
        {
            int w=1;
            for (int k=j;k<j+(i>>1);k++,w=1ll*w*wn%P)
            {
                int x=a[k],y=1ll*w*a[k+(i>>1)]%P;
                a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P;
            }
        }
    }
}
void mul(int n,int *a,int *b)
{
    DFT(n,a,3),DFT(n,b,3);
    for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
    DFT(n,a,inv3);
    int invn=inv(n);
    for (int i=0;i<n;i++) a[i]=1ll*a[i]*invn%P;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("division.in","r",stdin);
    freopen("division.out","w",stdout);
    const char LL[]="%I64d";
#else
    const char LL[]="%lld";
#endif
    n=read(),m=read();
    for (int i=0;i<=n;i++) a[i]=read();
    for (int i=0;i<=m;i++) b[i]=read();
    reverse(b,b+m+1);
    t=1;c[0]=inv(b[0]);
    inv3=inv(3);
    while (t<n-m+1)
    {
        t<<=1;
        for (int i=0;i<t;i++) d[i]=b[i];
        t<<=1;
        for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
        memcpy(e,c,sizeof(e));
        mul(t,e,d);
        for (int i=0;i<t;i++) e[i]=(P-e[i])%P;
        e[0]=(e[0]+2)%P;
        for (int i=(t>>1);i<t;i++) e[i]=0;
        mul(t,c,e);
        for (int i=(t>>1);i<t;i++) c[i]=0;
        t>>=1;
    }
    memcpy(d,a,sizeof(a));
    reverse(d,d+n+1);
    for (int i=n-m+1;i<=n;i++) c[i]=d[i]=0;
    t=1;while (t<=(n-m+1<<1)) t<<=1;
    for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
    mul(t,c,d);
    for (int i=n-m+1;i<t;i++) c[i]=0;
    reverse(c,c+n-m+1);
    for (int i=0;i<=n-m;i++) printf("%d ",c[i]);cout<<endl;
    reverse(b,b+m+1);
    t=1;while (t<=n) t<<=1;
    for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
    mul(t,c,b);
    for (int i=0;i<m;i++) printf("%d ",(a[i]-c[i]+P)%P);
    return 0;
}