Luogu4726 【模板】多项式指数函数(NTT+多项式求逆)

  https://www.cnblogs.com/HocRiser/p/8207295.html 安利!

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define P 998244353
#define N 550000
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
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;
}
int n,a[N],r[N],b[N],c[N],d[N],A[N],B[N],t;
int ksm(int a,int k)
{
    int s=1;
    for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P;
    return s;
}
int inv(int a){return ksm(a,P-2);}
void DFT(int n,int *a,int g)
{
    for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1);
    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(g,(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 IDFT(int *a,int n)
{
    DFT(n,a,inv(3));
    int u=inv(n);
    for (int i=0;i<n;i++) a[i]=1ll*a[i]*u%P;
}
void mul(int *a,int *b,int n)
{
    DFT(n,a,3),DFT(n,b,3);
    for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
    IDFT(a,n);
}
void Inv(int *a,int *b,int n) 
{
    if (n==1) {for (int i=0;i<t;i++) b[i]=0;b[0]=inv(a[0]);return;}
    Inv(a,b,n>>1);
    for (int i=0;i<n;i++) A[i]=a[i];
    for (int i=n;i<(n<<1);i++) A[i]=0;
    n<<=1;
    DFT(n,A,3),DFT(n,b,3);
    for (int i=0;i<n;i++) b[i]=1ll*b[i]*(P+2-1ll*A[i]*b[i]%P)%P;
    IDFT(b,n);
    n>>=1;
    for (int i=n;i<(n<<1);i++) b[i]=0;
}
void trans(int *a,int *b,int n){for (int i=0;i<n-1;i++) b[i]=1ll*a[i+1]*(i+1)%P;}
void dx(int *a,int *b,int n){b[0]=0;for (int i=1;i<n;i++) b[i]=1ll*a[i-1]*inv(i)%P;}
void Ln(int *a,int t)
{
    for (int i=0;i<t;i++) b[i]=c[i]=0;
    trans(a,c,t>>1);
    Inv(a,b,t>>1);
    mul(c,b,t);
    dx(c,a,t);
}
void Exp(int *a,int *b,int n) 
{
    if (n==1) {b[0]=1;return;}
    Exp(a,b,n>>1);
    for (int i=0;i<(n>>1);i++) B[i]=b[i];
    for (int i=(n>>1);i<n;i++) B[i]=0;
    Ln(B,n);
    for (int i=0;i<n;i++) B[i]=(a[i]-B[i]+P)%P;
    B[0]=(B[0]+1)%P;
    for (int i=n;i<(n<<1);i++) B[i]=0;
    mul(b,B,n<<1);
    for (int i=n;i<(n<<1);i++) b[i]=0;
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("ln.in","r",stdin);
    freopen("ln.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=0;i<n;i++) a[i]=read();
    t=1;while (t<=(n<<1)) t<<=1;
    Exp(a,d,t);
    for (int i=0;i<n;i++) printf("%d ",d[i]);
    return 0;
}
//ln(F(x))=G(x)
//ln(F(x))'=G(x)'
//F(x)'/F(x)=G(x)'
//G(x)=dx(F(x)'/F(x))

//exp(F(x))=G(x)
//F(x)=ln(G(x))
//ln(G(x))-F(x)=0
//H(G(x))=ln(G(x))-F(x)
//G1(x)=G0(x)-H(G0(x))/H(G0(x))'
//G1(x)=[F(x)+1-ln(G0(x))]*G0(x)

 

  

 

posted @ 2019-02-16 00:51  Gloid  阅读(186)  评论(0编辑  收藏  举报