并不对劲的多项式求ln,exp
ln
解释
设\(g(x)=ln(f(x))\)，两边同时求导，则有：\(g'(x)=ln'(f(x))*f'(x)=f^{-1}(x)*f'(x)\)(1)
因为\(f(x)\)是个多项式，所以设\(f(x)=\sum_{i=0}^{n}a_i*x^i\)，则有\(f'(x)=\sum_{i=0}^{n-1}a_{i+1}*(i+1)*x^i\)
设\(h(x)=f^{-1}(x)*f'(x)\)，对(1)式两边同时求积分，得：\(g(x)=\int h(x)\space dx\)
因为\(h(x)\)是个多项式，设\(h(x)=\sum_{i=0}^{n}b_i*x^i\)就有：\(\int h(x)\space dx=\sum_{i=1}^{n}\frac{b_{i-1}}{i}x^i\)
代码
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define maxn 800010
#define LL long long
#define mo(x) (x>=mod?x-mod:(x<0?x+mod:x))
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)&&ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
void write(int x)
{
if(x==0){putchar('0');return;}
int f=0;char ch[20];
if(x<0)putchar('-'),x=-x;
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
return;
}
const LL mod=998244353;
int f[maxn],g[maxn],h[maxn],nowlen,nown,r[maxn],tmp[maxn],n;
int mul(int x,int y){int res=1;while(y){if(y&1)res=(LL)res*(LL)x%mod;x=(LL)x*(LL)x%mod,y>>=1;}return res;}
void dnt(int * u,int fh)
{
rep(i,0,nown-1)r[i]=(r[i>>1]>>1)|((i&1)<<(nowlen-1));
rep(i,0,nown-1)if(i<r[i])swap(u[i],u[r[i]]);
for(int i=1;i<nown;i<<=1)
{
int wn=mul(3,(mod-1)/(i<<1));
if(fh==-1)wn=mul(wn,mod-2);
for(int j=0;j<nown;j+=(i<<1))
{
int w=1;
rep(k,0,i-1){int x=u[j+k],y=(LL)w*(LL)u[i+j+k]%mod;u[j+k]=mo(x+y),u[i+j+k]=mo(x-y),w=(LL)w*(LL)wn%mod;}
}
}
if(fh==-1)
{
int inv=mul(nown,mod-2);
rep(i,0,nown-1)u[i]=(LL)u[i]*(LL)inv%mod;
}
}
void getny(int * u,int * v,int nlen)
{
v[0]=mul(u[0],mod-2);
for(int len=0,tmpn=1;tmpn<nlen;len++,tmpn<<=1)
{
nown=tmpn<<1,nowlen=len+1;
rep(i,0,nown-1)tmp[i]=u[i];
nown<<=1,nowlen++;
rep(i,(tmpn<<1),nown-1)tmp[i]=0;
dnt(tmp,1),dnt(v,1);
rep(i,0,nown-1)v[i]=mo(2ll-(LL)tmp[i]*(LL)v[i]%mod)*(LL)v[i]%mod;
dnt(v,-1);
rep(i,(tmpn<<1),nown-1)v[i]=0;
}
rep(i,nlen,nown)v[i]=0;
}
void getup(int * u,int * v,int nlen)
{
rep(i,1,nlen-1)v[i-1]=(LL)i*(LL)u[i]%mod;v[nlen-1]=0;
}
void getdx(int * u,int * v,int nlen)
{
rep(i,1,nlen-1)v[i]=(LL)u[i-1]*(LL)mul(i,mod-2)%mod;v[0]=0;
}
void getln(int * u,int nlen)
{
getup(u,h,nlen),getny(u,g,nlen);
for(nowlen=0,nown=1;nown<(nlen+nlen);nowlen++,nown<<=1);
dnt(h,1),dnt(g,1);
rep(i,0,nown-1)h[i]=(LL)h[i]*(LL)g[i]%mod;
rep(i,0,nown-1)g[i]=0;
dnt(h,-1);getdx(h,g,nlen);
rep(i,nlen,nown)g[i]=0;
}
int main()
{
n=read();
rep(i,0,n-1)f[i]=read();
getln(f,n);
rep(i,0,n-1)write(g[i]),putchar(' ');
return 0;
}
exp
解释
这个人（点这里）讲得很清楚
代码
#include<algorithm>
#include<cmath>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<iomanip>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define rep(i,x,y) for(register int i=(x);i<=(y);++i)
#define dwn(i,x,y) for(register int i=(x);i>=(y);--i)
#define maxn 800010
#define LL long long
#define mo(x) (x>=mod?x-mod:(x<0?x+mod:x))
using namespace std;
int read()
{
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)&&ch!='-')ch=getchar();
if(ch=='-')f=-1,ch=getchar();
while(isdigit(ch))x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
void write(int x)
{
if(x==0){putchar('0');return;}
int f=0;char ch[20];
if(x<0)putchar('-'),x=-x;
while(x)ch[++f]=x%10+'0',x/=10;
while(f)putchar(ch[f--]);
return;
}
const LL mod=998244353;
int f[maxn],g[maxn],h[maxn],ans[maxn],nowlen,nown,r[maxn],tmp[maxn],tmp2[maxn],n;
int mul(int x,int y){int res=1;while(y){if(y&1)res=(LL)res*(LL)x%mod;x=(LL)x*(LL)x%mod,y>>=1;}return res;}
void dnt(int * u,int fh)
{
rep(i,0,nown-1)r[i]=(r[i>>1]>>1)|((i&1)<<(nowlen-1));
rep(i,0,nown-1)if(i<r[i])swap(u[i],u[r[i]]);
for(int i=1;i<nown;i<<=1)
{
int wn=mul(3,(mod-1)/(i<<1));
if(fh==-1)wn=mul(wn,mod-2);
for(int j=0;j<nown;j+=(i<<1))
{
int w=1;
rep(k,0,i-1){int x=u[j+k],y=(LL)w*(LL)u[i+j+k]%mod;u[j+k]=mo(x+y),u[i+j+k]=mo(x-y),w=(LL)w*(LL)wn%mod;}
}
}
if(fh==-1)
{
int inv=mul(nown,mod-2);
rep(i,0,nown-1)u[i]=(LL)u[i]*(LL)inv%mod;
}
}
void getny(int * u,int * v,int nlen)
{
v[0]=mul(u[0],mod-2);
for(int len=0,tmpn=1;tmpn<nlen;len++,tmpn<<=1)
{
nown=tmpn<<1,nowlen=len+1;
rep(i,0,nown-1)tmp[i]=u[i];
nown<<=1,nowlen++;
rep(i,(tmpn<<1),nown-1)tmp[i]=0;
dnt(tmp,1),dnt(v,1);
rep(i,0,nown-1)v[i]=mo(2ll-(LL)tmp[i]*(LL)v[i]%mod)*(LL)v[i]%mod;
dnt(v,-1);
rep(i,(tmpn<<1),nown-1)v[i]=0;
}
rep(i,0,nown)tmp[i]=0;
rep(i,nlen,nown)v[i]=0;
}
void getup(int * u,int * v,int nlen)
{
rep(i,1,nlen-1)v[i-1]=(LL)i*(LL)u[i]%mod;v[nlen-1]=0;
}
void getdx(int * u,int * v,int nlen)
{
rep(i,1,nlen-1)v[i]=(LL)u[i-1]*(LL)mul(i,mod-2)%mod;v[0]=0;
}
void getln(int * u,int * v,int nlen)
{
getup(u,h,nlen),getny(u,g,nlen);
for(nowlen=0,nown=1;nown<(nlen+nlen);nowlen++,nown<<=1);
dnt(h,1),dnt(g,1);
rep(i,0,nown-1)h[i]=(LL)h[i]*(LL)g[i]%mod;
rep(i,0,nown-1)g[i]=0;
dnt(h,-1);getdx(h,v,nlen);
rep(i,nlen,nown)v[i]=0;
}
void getexp(int * u,int * v,int nlen)
{
rep(i,0,(nlen<<2))v[i]=0;
v[0]=1;
for(int len=0,tmpn=1;tmpn<nlen;len++,tmpn<<=1)
{
rep(i,0,(tmpn<<1))tmp2[i]=0;
getln(v,tmp2,(tmpn<<1));
nown=(tmpn<<2),nowlen=len+2;
rep(i,(tmpn<<1),nown)tmp2[i]=0;
rep(i,0,(tmpn<<1)-1)tmp[i]=u[i];
rep(i,(tmpn<<1),nown)tmp[i]=0;
dnt(tmp2,1),dnt(v,1),dnt(tmp,1);
rep(i,0,nown-1)v[i]=(LL)v[i]*(LL)mo(mo(1ll-tmp2[i])+tmp[i])%mod;
dnt(v,-1);
rep(i,(tmpn<<1),nown)v[i]=0;
}
rep(i,nlen,nown)v[i]=0;
}
int main()
{
n=read();
rep(i,0,n-1)f[i]=read();
getexp(f,ans,n);
rep(i,0,n-1)write(ans[i]),putchar(' ');
return 0;
}
