【洛谷 P5273】【模板】多项式快速幂 (加强版)(多项式Ln+Exp)

考虑当 $a_0=\not 1$ 的时候是没法直接用 $Ln$ 的

一个显然的想法是找到第一个 $a_0=\not 0$ 的地方 $a_p$
把 $a$ 左移 $p$ 位后除以 $a_p$ ，这样就保证常数项为 $1$ 了

最后只需要乘上 $a_p^kx^{pk}$ 就可以了

而且我们只需要把前 $n-pk$ 次项拿来做快速幂就可以了

#include<bits/stdc++.h>
using namespace std;
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define re register
#define pb push_back
#define cs const
#define pii pair<int,int>
#define fi first
#define se second
#define ll long long
#define poly vector<int>
#define bg begin
cs int mod=998244353,G=3;
inline int add(int a,int b){return (a+=b)>=mod?a-mod:a;}
inline void Add(int &a,int b){(a+=b)>=mod?(a-=mod):0;}
inline int dec(int a,int b){return (a-=b)<0?a+mod:a;}
inline void Dec(int &a,int b){(a-=b)<0?(a+=mod):0;}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){
	for(;b;b>>=1,a=mul(a,a))(b&1)&&(res=mul(res,a));return res;
}
inline void chemx(int &a,int b){a<b?a=b:0;}
inline void chemn(int &a,int b){a>b?a=b:0;}
cs int N=(1<<21)|5,C=21;
poly w[C+1];
int rev[N],fac[N],ifac[N],inv[N];
inline void init(cs int len=N-5){
	fac[0]=ifac[0]=inv[0]=inv[1]=1;
	for(int i=1;i<=len;i++)fac[i]=mul(fac[i-1],i);
	ifac[len]=ksm(fac[len],mod-2);
	for(int i=len-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	for(int i=2;i<=len;i++)inv[i]=mul(mod-mod/i,inv[mod%i]);
}
inline void init_w(){
	for(int i=1;i<=C;i++)w[i].resize(1<<(i-1));
	int wn=ksm(G,(mod-1)/(1<<C));
	w[C][0]=1;
	for(int i=1;i<(1<<(C-1));i++)w[C][i]=mul(w[C][i-1],wn);
	for(int i=C-1;i;i--)
	for(int j=0;j<(1<<(i-1));j++)
	w[i][j]=w[i+1][j<<1];
}
inline void init_rev(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline void ntt(poly &f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int a0,a1,l=1,mid=1;mid<lim;mid<<=1,l++)
	for(int i=0;i<lim;i+=(mid<<1))
	for(int j=0;j<mid;j++)
	a0=f[i+j],a1=mul(w[l][j],f[i+j+mid]),f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
	if(kd==-1){
		reverse(f.bg()+1,f.bg()+lim);
		for(int i=0;i<lim;i++)Mul(f[i],inv[lim]);
	}
}
inline poly operator +(poly a,poly b){
	if(a.size()<b.size())a.resize(b.size());
	for(int i=0;i<b.size();i++)Add(a[i],b[i]);
	return a;
}
inline poly operator -(poly a,poly b){
	if(a.size()<b.size())a.resize(b.size());
	for(int i=0;i<b.size();i++)Dec(a[i],b[i]);
	return a;
}
inline poly operator *(poly a,int b){
	for(int i=0;i<a.size();i++)Mul(a[i],b);
	return a;
}
inline poly operator *(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	if(deg<=64){
		poly c(deg,0);
		for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
		Add(c[i+j],mul(a[i],b[j]));
		return c;
	}
	while(lim<deg)lim<<=1;
	init_rev(lim);
	a.resize(lim),ntt(a,lim,1);
	b.resize(lim),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1),a.resize(deg);
	return a;
}
inline poly Inv(poly a,int deg){
	poly b(1,ksm(a[0],mod-2)),c;
	for(int lim=4;lim<(deg<<2);lim<<=1){
		c=a,c.resize(lim>>1);
		init_rev(lim);
		c.resize(lim),ntt(c,lim,1);
		b.resize(lim),ntt(b,lim,1);
		for(int i=0;i<lim;i++)Mul(b[i],dec(2,mul(b[i],c[i])));
		ntt(b,lim,-1),b.resize(lim>>1);
	}b.resize(deg);return b;
}
inline poly deriv(poly a){
	for(int i=0;i<a.size()-1;i++)a[i]=mul(a[i+1],i+1);
	a.pop_back();return a;
}
inline poly integ(poly a){
	a.pb(0);
	for(int i=a.size()-1;i;i--)a[i]=mul(a[i-1],inv[i]);
	a[0]=0;return a;
}
inline poly Ln(poly a,int deg){
	a=integ(Inv(a,deg)*deriv(a)),a.resize(deg);return a;
}
inline poly exp(poly a,int deg){
	poly b(1,1),c;
	for(int lim=2;lim<(deg<<1);lim<<=1){
		c=Ln(b,lim),c=a-c;
		c[0]++,b=b*c,b.resize(lim);
	}b.resize(deg);return b;
}
inline poly ksm(poly a,int b,int deg){
	int last=deg-1;
	for(int i=0;i<a.size();i++)if(a[i]){last=i;break;}
	if(1ll*last*b>deg)
		return poly(deg,0);
	int pos=deg-b*last,val=a[last],iv=ksm(val,mod-2);
	for(int i=0;i<pos;i++)a[i]=mul(a[i+last],iv);
	a.resize(pos),a=exp(Ln(a,pos)*b,pos);
	val=ksm(val,b);a.resize(deg);//先resize
	for(int i=pos-1;~i;i--)a[i+last*b]=mul(a[i],val);//逆序做 
	for(int i=0;i<last*b;i++)a[i]=0;
	return a;
}
int n,m;
poly a;
int main(){
	init_w(),init();
	n=read(),m=read();
	a.resize(n);
	for(int i=0;i<n;i++)a[i]=read();
	a=ksm(a,m,n);
	for(int i=0;i<n;i++)cout<<a[i]<<" ";
}

来源：https://blog.csdn.net/qq_42555009/article/details/99657487

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