多项式全家桶
补一个多项式
ln
\ln
ln。
给定一个多项式
A
(
x
)
A(x)
A(x),求一个多项式
F
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x
)
F(x)
F(x),满足
F
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x
)
≡
ln
A
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x
)
(
m
o
d
x
n
)
F(x)\equiv \ln A(x)\;\;\;(mod\;\;x^n)
F(x)≡lnA(x)(modxn)
考虑求导。
F
′
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x
)
=
A
′
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x
)
A
(
x
)
F'(x)=\frac{A'(x)}{A(x)}
F′(x)=A(x)A′(x)
因此只需要先求出 H ( x ) = A ′ ( x ) H(x)=A'(x) H(x)=A′(x),再求出 G ( x ) = A − 1 ( x ) G(x)=A^{-1}(x) G(x)=A−1(x),则 F = G ∗ H F=G*H F=G∗H,多项式乘法即能求出 F ′ ( x ) F'(x) F′(x)。
然后在积分回去即可。
在 A ( 0 ) = 1 A(0)=1 A(0)=1时, F ( 0 ) = 0 F(0)=0 F(0)=0。
其他内容见牛顿迭代法
Code——多项式全家桶(多项式除法 咕咕咕)
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <ctime>
#include <cassert>
#include <string.h>
//#include <unordered_set>
//#include <unordered_map>
//#include <bits/stdc++.h>
#define MP(A,B) make_pair(A,B)
#define PB(A) push_back(A)
#define SIZE(A) ((int)A.size())
#define LEN(A) ((int)A.length())
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define fi first
#define se second
using namespace std;
template<typename T>inline bool upmin(T &x,T y) { return y<x?x=y,1:0; }
template<typename T>inline bool upmax(T &x,T y) { return x<y?x=y,1:0; }
typedef long long ll;
typedef unsigned long long ull;
typedef long double lod;
typedef pair<int,int> PR;
typedef vector<int> VI;
const lod eps=1e-11;
const lod pi=acos(-1);
const int oo=1<<30;
const ll loo=1ll<<62;
const int mods=998244353;
const int inv2=(mods+1)>>1;
const int G=3;
const int Gi=(mods+1)/G;
const int MAXN=600005;
const int INF=0x3f3f3f3f;//1061109567
/*--------------------------------------------------------------------*/
inline int read()
{
int f=1,x=0; char c=getchar();
while (c<'0'||c>'9') { if (c=='-') f=-1; c=getchar(); }
while (c>='0'&&c<='9') { x=(x<<3)+(x<<1)+(c^48); c=getchar(); }
return x*f;
}
int a[MAXN],F[MAXN];
namespace Poly
{
int c[MAXN],f[MAXN],b[MAXN],d[MAXN],rev[MAXN],Limit,l;
inline int upd(int x,int y){ return x+y>=mods?x+y-mods:x+y; }
inline int quick_pow(int x,int y)
{
int ret=1;
for (;y;y>>=1)
{
if (y&1) ret=1ll*ret*x%mods;
x=1ll*x*x%mods;
}
return ret;
}
void Number_Theoretic_Transform(int *A,int n,int opt)
{
for (int i=0;i<Limit;i++) if (i<rev[i]) swap(A[i],A[rev[i]]);
for (int mid=1;mid<Limit;mid<<=1)
{
int Wn=quick_pow((opt==1)?G:Gi,(mods-1)/(mid<<1));
for (int j=0;j<Limit;j+=mid<<1)
for (int k=j,w=1;k<j+mid;k++,w=1ll*w*Wn%mods)
{
int x=A[k],y=1ll*A[k+mid]*w%mods;
A[k]=upd(x,y),A[k+mid]=upd(x,mods-y);
}
}
if (opt==-1)
{
int invLimit=quick_pow(Limit,mods-2);
for (int i=0;i<n;i++) A[i]=1ll*A[i]*invLimit%mods;
for (int i=n;i<Limit;i++) A[i]=0;
}
}
void getinv(int *F,int *G,int n) //F->invG
{
if (n==1) { F[0]=quick_pow(G[0],mods-2); return; }
getinv(F,G,(n+1)>>1);
Limit=1,l=0;
while (Limit<(n<<1)) Limit<<=1,l++;
for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
for (int i=0;i<n;i++) c[i]=G[i];
for (int i=n;i<Limit;i++) c[i]=0;
Number_Theoretic_Transform(c,n,1);
Number_Theoretic_Transform(F,n,1);
for (int i=0;i<Limit;i++) F[i]=1ll*upd(2,mods-1ll*F[i]*c[i]%mods)*F[i]%mods;
Number_Theoretic_Transform(F,n,-1);
for (int i=n;i<Limit;i++) F[i]=0;
}
void getln(int *a,int n) //a->ln a (a[0]=1)
{
memset(f,0,sizeof f);
getinv(f,a,n);
for (int i=0;i<n;i++) a[i]=1ll*a[i+1]*(i+1)%mods;
Limit=1,l=0;
while (Limit<(n<<1)) Limit<<=1,l++;
for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
Number_Theoretic_Transform(a,n,1);
Number_Theoretic_Transform(f,n,1);
for (int i=0;i<Limit;i++) f[i]=1ll*f[i]*a[i]%mods;
Number_Theoretic_Transform(f,n,-1);
for (int i=1;i<n;i++) a[i]=1ll*f[i-1]*quick_pow(i,mods-2)%mods;
for (int i=n;i<=Limit;i++) a[i]=0;
a[0]=0;
}
void getexp(int *F,int *a,int n) //F->exp a (a[0]=0)
{
memset(b,0,sizeof b);
if (n==1) { F[0]=1; return; }
getexp(F,a,(n+1)>>1);
for (int i=0;i<n;i++) b[i]=F[i];
getln(b,n);
Limit=1,l=0;
while (Limit<(n<<1)) Limit<<=1,l++;
for (int i=0;i<Limit;i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(l-1));
for (int i=0;i<n;i++) b[i]=upd(a[i],mods-b[i]);
for (int i=n;i<Limit;i++) b[i]=F[i]=0;
b[0]++;
Number_Theoretic_Transform(F,n,1);
Number_Theoretic_Transform(b,n,1);
for (int i=0;i<Limit;i++) F[i]=1ll*F[i]*b[i]%mods;
Number_Theoretic_Transform(F,n,-1);
for (int i=n;i<Limit;i++) F[i]=0;
}
void getsqrt(int *F,int *a,int n,int k) //F->sqrt^k a(a[0]=1)
{
memset(d,0,sizeof d);
for (int i=0;i<n;i++) d[i]=a[i];
getln(d,n);
int invk=quick_pow(k,mods-2);
for (int i=0;i<n;i++) d[i]=1ll*d[i]*invk%mods;
getexp(F,d,n);
}
void getpow(int *F,int *a,int n,int k) //F->pow^k a(a[0]=1)
{
memset(d,0,sizeof d);
for (int i=0;i<n;i++) d[i]=a[i];
getln(d,n);
for (int i=0;i<n;i++) d[i]=1ll*d[i]*k%mods;
getexp(F,d,n);
}
}