多项式 ln

多项式 ln

定义

\(给一多项式F(x),求G(x)\equiv lnF(x)\pmod x^n\)

前置知识

推式子:

\[\because G(x)\equiv lnF(x)\pmod x^n \]

\[又\because lnF(x)=\int dlnF(x) \]

\[=\int (lnF(x))'dx \]

\[=\int \frac{F'(x)}{F(x)}dx \]

步骤:

\(1.求F(x)的导数\)

\(2.求F(x)的逆F_r(x)\)

\(3.求\int F(x)F_r(x)dx\)

\[\\ \]

【模板】多项式对数函数(多项式 ln)

\[\\ \]

\(\mathfrak{Talk\ is\ cheap,show\ you\ the\ code.}\)

#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;
# define read read1<int>()
# define Type template<typename T>
Type inline T read1(){
    T n=0;
    char k;
    bool fl=0;
    do (k=getchar())=='-'&&(fl=1);while('9'<k||k<'0');
    while(47<k&&k<58)n=(n<<3)+(n<<1)+(k^48),k=getchar();
    return fl?-n:n;
}
# define f(i,l,r) for(int i=(l);i<=(r);++i)
# define fre(k) freopen(k".in","r",stdin);freopen(k".ans","w",stdout)
# define ll int64_t
class Array{
    private:
        vector<int>a;
    public:
        Array(const int size,const int f):a(size,f){}
        void push(int n){a.push_back(n);}
        Array(int* l=NULL,int* r=NULL){while(l!=r)push(*l),++l;}
        inline int size(){return a.size();}
        inline int& operator [] (const int x){return a[x];}
        void resize(int n){a.resize(n);}
        void clear(){a.clear();}
        void swap(){reverse(a.begin(),a.end());}
        int& top(){return a[a.size()-1];}
        void pop(){a.pop_back();}
};
const int mod=998244353,g=3,inv2=499122177;
Array operator -(Array a,Array b){
    int N=a.size(),M=b.size();
    Array t;
    for(int i=0;i<N&&i<M;++i){
        t.push((i<N?a[i]:0)-(i<M?b[i]:0));
        if(t.top()<0)t.top()+=mod;
    }
    return t;
}
Array operator +(Array a,Array b){
    int N=a.size(),M=b.size();
    Array t;
    for(int i=0;i<N&&i<M;++i){
        t.push((i<N?a[i]:0)+(i<M?b[i]:0));
        if(t.top()>=mod)t.top()-=mod;
    }
    return t;
}
Array operator *(Array a,int n){
    int N=a.size();
    for(int i=0;i<N;a[i]=(ll)a[i]*n%mod,++i);
    return a;
}
Array operator *(int n,Array a){return a*n;}
int qkpow(int b,int m,int mod){
    int tem=b,ans=1;
    for(;m;m>>=1,tem=(ll)tem*tem%mod)
        if(m&1)ans=(ll)ans*tem%mod;
    return ans;
}
int* NTT(const int len,Array& a,const bool Ty,int* r=NULL){
    if(!r){
        r=new int[len];
        r[0]=0;int L=log2(len);
        f(i,0,len-1)
            r[i]=(r[i>>1]>>1)|((i&1)<<L-1);
    }
    f(i,0,len-1)
        if(i<r[i])swap(a[i],a[r[i]]);
    for(int i=1;i<len;i<<=1){
        int T=qkpow(Ty?g:332748118,(mod-1)/(i<<1),mod);
        for(int W=i<<1,j=0;j<len;j+=W){
            ll omega=1;
            for(int k=0;k<i;++k,omega=omega*T%mod){
                ll x(a[j+k]),y(omega*a[i+j+k]%mod);
                a[j+k]=x+y;(a[j+k]>mod)&&(a[j+k]-=mod);
                a[i+j+k]=x-y+mod;(a[i+j+k]>mod)&&(a[i+j+k]-=mod);
            }
        }
    }
    return r;
}
Array operator * (Array x,Array y){
    int n=x.size()-1,m=y.size()-1;
    int limit=1;
    while(limit<=n+m)limit<<=1;
    Array ans;
    x.resize(limit+1);
    y.resize(limit+1);
    int *r;
    r=NTT(limit,x,1);
    NTT(limit,y,1,r);
    f(i,0,limit)x[i]=(ll)x[i]*y[i]%mod;
    NTT(limit,x,0,r);
    int tem=qkpow(limit,mod-2,mod);
    f(i,0,n+m)ans.push((ll)x[i]*tem%mod);
    return ans;
}
Array& operator *= (Array& x,Array y){
    return x=x*y;
}
void Rev(Array &x,Array y){
    int n=x.size()-1,m=y.size()-1;
    int limit=1;
    while(limit<=n+m)limit<<=1;
    Array ans;
    x.resize(limit+1);
    y.resize(limit+1);
    int *r;
    r=NTT(limit,x,1);
    NTT(limit,y,1,r);
    f(i,0,limit)x[i]=(ll)(2ll-(ll)x[i]*y[i]%mod+mod)%mod*y[i]%mod;
    NTT(limit,x,0,r);
    int tem=qkpow(limit,mod-2,mod);
    f(i,0,n+m)x[i]=(ll)x[i]*tem%mod;
    x.resize(n+m+1);
}
Array Inv(Array a){
    int N=a.size();
    // printf("%d\n",N);
    if(N==1)return Array(1,qkpow(a[0],mod-2,mod));
    Array b=a;b.resize(N+1>>1);
    b=Inv(b);b.resize(N);
    Rev(a,b);
    a.resize(N);
    return a;
}
Array operator / (Array x,Array y){
    int N=x.size()-1,M=y.size()-1;
    if(N<M)return Array(1,0);
    x.swap();y.swap();
    y.resize(N-M+1);
    x*=Inv(y);
    x.resize(N-M+1);
    x.swap();
    return x;
}
Array sqrt(Array x){
    int N=x.size();
    if(N==1)return Array(1,1);
    Array y=x;
    y.resize(N+1>>1);
    y=sqrt(y);
    y.resize(N);
    Array z=Inv(y);
    return inv2*(y+x*z);
}
Array diff(Array x){
    for(int i=0;i+1<x.size();++i)
        x[i]=(ll)x[i+1]*(i+1)%mod;
    x.pop();
    return x;
}
Array integral(Array x){
    for(int i=x.size();--i;)
        x[i]=(ll)x[i-1]*qkpow(i,mod-2,mod)%mod;
    x[0]=0;
    return x;
}
Array ln(Array x){
    int N=x.size();
    x=integral(diff(x)*Inv(x));
    x.resize(N);
    return x;
}
int n;
Array x;
int main(){
    n=read;
    for(int i=0;i<n;++i)x.push(read);
    x=ln(x);
    for(int i=0;i<n;++i)printf("%d ",x[i]);
    return 0;
}
posted @ 2019-12-21 16:43  ファイナル  阅读(334)  评论(0)    收藏  举报