Luogu4238 模板多项式求逆（NTT）

Posted 2020-12-28 gloid

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

　　B(x)=B‘(x)·(2-A(x)B‘(x))

　　注意ntt的时候防止项数溢出，即将多项式补零成n位后，相乘时次数最高的非零项不超过n次。

#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,t,r[N],a[N],b[N],c[N],d[N];
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 x){return ksm(x,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;
            }
        }
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("inverse.in","r",stdin);
    freopen("inverse.out","w",stdout);
    const char LL[]="%I64d";
#else
    const char LL[]="%lld";
#endif
    n=read();
    for (int i=0;i<n;i++) a[i]=read();
    t=1;b[0]=inv(a[0]);
    int inv3=inv(3);
    while (t<n)
    {
        t<<=1;
        for (int i=0;i<t;i++) d[i]=a[i];
        t<<=1;
        memcpy(c,b,sizeof(b));
        for (int i=0;i<t;i++) r[i]=(r[i>>1]>>1)|(i&1)*(t>>1);
        DFT(t,c,3);DFT(t,d,3);
        for (int i=0;i<t;i++) c[i]=1ll*c[i]*d[i]%P;
        DFT(t,c,inv3);
        int invt=inv(t);
        for (int i=0;i<t;i++) c[i]=1ll*c[i]*invt%P;
        for (int i=0;i<t;i++) c[i]=(P-c[i])%P;
        c[0]=(c[0]+2)%P;
        for (int i=(t>>1);i<t;i++) c[i]=0;
        DFT(t,b,3);DFT(t,c,3);
        for (int i=0;i<t;i++) b[i]=1ll*b[i]*c[i]%P;
        DFT(t,b,inv3);
        for (int i=0;i<t;i++) b[i]=1ll*b[i]*invt%P;
        for (int i=(t>>1);i<t;i++) b[i]=0;
        t>>=1;
    }
    for (int i=0;i<n;i++) printf("%d ",b[i]);
    return 0;
}