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
- \(不定积分\)
- \(微分\)
- \(多项式乘法逆\)
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
#include
#include
#include
using namespace std;
# define read read1()
# define Type template
Type inline T read1(){
T n=0;
char k;
bool fl=0;
do (k=getchar())=='-'&&(fl=1);while('9'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=mod)t.top()-=mod;
}
return t;
}
Array operator *(Array a,int n){
int N=a.size();
for(int i=0;i>=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)<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>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