【总结】多项式基础操作

多项式乘法

\(A * B=\sum_{k=0}^{n}(\sum_{k=i+j}a_ib_j)x^k\)

快速傅里叶变换（FFT）

https://oi.men.ci/fft-notes/

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define ft first
#define sd second
#define PII std::pair<int,int>
#define PLL std::pair<ll,ll>
#define pb(a) push_back(a)
#define mp(a,b) std::make_pair(a,b)
#define ITR(a,b) for(auto a:b)
#define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
#define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
const int MAXN=1<<21;
const db Pi=acos(-1);
int n1,n2,n,m,cnt,rev[MAXN];
struct Complex{
	db real,imag;
	inline Complex operator + (const Complex &A) const {
		return (Complex){real+A.real,imag+A.imag};
	};
	inline Complex operator - (const Complex &A) const {
		return (Complex){real-A.real,imag-A.imag};
	};
	inline Complex operator * (const Complex &A) const {
		return (Complex){real*A.real-imag*A.imag,real*A.imag+imag*A.real};
	};
};
Complex a[MAXN],b[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline bool chkmin(T &x,T y){return y<x?(x=y,true):false;}
template<typename T> inline bool chkmax(T &x,T y){return y>x?(x=y,true):false;}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void FFT(Complex *A,int tp)
{
	REP(i,0,n-1)if(i<rev[i])std::swap(A[i],A[rev[i]]);
	for(register int l=2;l<=n;l<<=1)
	{
		Complex wn=(Complex){cos(2*Pi/l),sin(tp*2*Pi/l)};
		for(register int i=0;i<n;i+=l)
		{
			Complex w=(Complex){1,0};
			for(register int j=0;j<(l>>1);++j)
			{
				Complex A1=A[i+j],A2=A[i+j+(l>>1)]*w;
				A[i+j]=A1+A2;A[i+j+(l>>1)]=A1-A2;
				w=w*wn;
			}
		}
	}
}
int main()
{
	read(n1);read(n2);n1++;n2++;
	REP(i,0,n1-1)scanf("%lf",&a[i].real);
	REP(i,0,n2-1)scanf("%lf",&b[i].real);
	m=n1+n2-1;
	for(n=1;n<m;n<<=1,++cnt);
	REP(i,0,n-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
	FFT(a,1);FFT(b,1);
	REP(i,0,n-1)a[i]=a[i]*b[i];
	FFT(a,-1);
	REP(i,0,m-1)printf("%d ",(int)(a[i].real/n+0.5));
	return 0;
}

快速数论变换（NTT）

用原根代替单位根

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define ft first
#define sd second
#define PII std::pair<int,int>
#define PLL std::pair<ll,ll>
#define pb(a) push_back(a)
#define mp(a,b) std::make_pair(a,b)
#define ITR(a,b) for(auto a:b)
#define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
#define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
const int MAXN=1<<21,Mod=998244353;
int n1,n2,n,m,rev[MAXN],cnt;
ll pw[MAXN],ipw[MAXN],a[MAXN],b[MAXN];
template<typename T> inline void read(T &x)
{
	T data=0,w=1;
	char ch=0;
	while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
	if(ch=='-')w=-1,ch=getchar();
	while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
	x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
	if(ch!='\0')putchar(ch);
}
template<typename T> inline bool chkmin(T &x,T y){return y<x?(x=y,true):false;}
template<typename T> inline bool chkmax(T &x,T y){return y>x?(x=y,true):false;}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline ll qexp(ll a,ll b)
{
	ll res=1;
	while(b)
	{
		if(b&1)res=res*a%Mod;
		a=a*a%Mod;
		b>>=1;
	}
	return res;
}
inline void NTT(ll *A,int tp)
{
	REP(i,0,n-1)if(i<rev[i])std::swap(A[i],A[rev[i]]);
	for(register int l=2;l<=n;l<<=1)
	{
		ll wn=tp>0?pw[l]:ipw[l];
		for(register int i=0;i<n;i+=l)
		{
			ll w=1;
			for(register int j=0;j<(l>>1);++j)
			{
				ll A1=A[i+j],A2=1ll*A[i+j+(l>>1)]*w%Mod;
				A[i+j]=(A1+A2)%Mod;A[i+j+(l>>1)]=(A1-A2+Mod)%Mod;
				w=1ll*w*wn%Mod;
			}
		}
	}
}
int main()
{
	read(n1);read(n2);n1++;n2++;
	REP(i,0,n1-1)read(a[i]);
	REP(i,0,n2-1)read(b[i]);
	m=n1+n2-1;
	for(n=1;n<m;n<<=1,++cnt);
	ll inv3=qexp(3,Mod-2),invn=qexp(n,Mod-2);
	for(register int l=1;l<=n;l<<=1)pw[l]=qexp(3,(Mod-1)/l),ipw[l]=qexp(inv3,(Mod-1)/l);
	REP(i,0,n-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
	NTT(a,1);NTT(b,1);
	REP(i,0,n-1)a[i]*=b[i];
	NTT(a,-1);
	REP(i,0,m-1)write(1ll*a[i]*invn%Mod,' ');
	return 0;
}

任意模数情况下的NTT（MTT）

将每个系数分解为 \(a_i=Ap_i*t+Ak_i~(b_i=Bp_i*t+Bk_i)\)，\(t\) 自己定，相当于转成 \(t\) 进制数，但要保证 \(t\) 范围FFT不会爆精度。

\(\begin{aligned} A * B & = \sum_{k=0}^n(\sum_{k=i+j}a_ib_j)x^k \\ & = \sum_{k=0}^n(\sum_{k=i+j}(Ap_{i}t+Ak_{i})(Bp_jt+Bk_j))x^k \\ & = \sum_{k=0}^n(\sum_{k=i+j}Ap_iBp_jt^2+Ap_iBk_jt+Ak_iBp_jt+Ak_iBk_j)x^k \end{aligned}\)

然后做四次FFT，最后转回 \(10\) 进制累加起来

（图省事，没写共轭优化，并且还多了一次IDFT）

#include<bits/stdc++.h>
#define ui unsigned int
#define ll long long
#define db double
#define ld long double
#define ull unsigned long long
#define ft first
#define sd second
#define PII std::pair<int,int>
#define PLL std::pair<ll,ll>
#define pb(a) push_back(a)
#define mp(a,b) std::make_pair(a,b)
#define ITR(a,b) for(auto a:b)
#define REP(a,b,c) for(register int a=(b),a##end=(c);a<=a##end;++a)
#define DEP(a,b,c) for(register int a=(b),a##end=(c);a>=a##end;--a)
const int MAXN=1<<21,M=32767;
const ld Pi=acos(-1);
int n1,n2,n,m,cnt,p,rev[MAXN];
ll ans[MAXN];
struct Complex{
    ld real,imag;
    inline Complex operator + (const Complex &A) const {
        return (Complex){real+A.real,imag+A.imag};
    };
    inline Complex operator - (const Complex &A) const {
        return (Complex){real-A.real,imag-A.imag};
    };
    inline Complex operator * (const Complex &A) const {
        return (Complex){real*A.real-imag*A.imag,real*A.imag+imag*A.real};
    };
};
Complex a1[MAXN],a2[MAXN],b1[MAXN],b2[MAXN],c[MAXN],w[MAXN];
template<typename T> inline void read(T &x)
{
    T data=0,w=1;
    char ch=0;
    while(ch!='-'&&(ch<'0'||ch>'9'))ch=getchar();
    if(ch=='-')w=-1,ch=getchar();
    while(ch>='0'&&ch<='9')data=((T)data<<3)+((T)data<<1)+(ch^'0'),ch=getchar();
    x=data*w;
}
template<typename T> inline void write(T x,char ch='\0')
{
    if(x<0)putchar('-'),x=-x;
    if(x>9)write(x/10);
    putchar(x%10+'0');
    if(ch!='\0')putchar(ch);
}
template<typename T> inline bool chkmin(T &x,T y){return y<x?(x=y,true):false;}
template<typename T> inline bool chkmax(T &x,T y){return y>x?(x=y,true):false;}
template<typename T> inline T min(T x,T y){return x<y?x:y;}
template<typename T> inline T max(T x,T y){return x>y?x:y;}
inline void FFT(Complex *A,int tp)
{
    REP(i,0,n-1)if(i<rev[i])std::swap(A[i],A[rev[i]]);
    for(register int l=2;l<=n;l<<=1)
    {
		for(register int i=0;i<(l>>1);++i)w[i]=(Complex){cos(2*Pi*i/l),sin(tp*2*Pi*i/l)};
        for(register int i=0;i<n;i+=l)
            for(register int j=0;j<(l>>1);++j)
            {
                Complex A1=A[i+j],A2=A[i+j+(l>>1)]*w[j];
                A[i+j]=A1+A2;A[i+j+(l>>1)]=A1-A2;
            }
    }
}
inline void MTT(Complex *a,Complex *b,ll Bas)
{
    REP(i,0,n-1)c[i]=a[i]*b[i];FFT(c,-1);
    REP(i,0,m-1)(ans[i]+=1ll*(ll)(c[i].real/n+0.5)%p*Bas%p)%=p;
}
int main()
{
    read(n1);read(n2);read(p);n1++;n2++;
    int x;
    REP(i,0,n1-1)read(x),a1[i].real=(ld)((int)x/M),a2[i].real=(ld)((int)x%M);
    REP(i,0,n2-1)read(x),b1[i].real=(ld)((int)x/M),b2[i].real=(ld)((int)x%M);
    m=n1+n2-1;
    for(n=1;n<m;n<<=1,++cnt);
    REP(i,0,n-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
    FFT(a1,1);FFT(a2,1);FFT(b1,1);FFT(b2,1);
    MTT(a1,b1,1ll*M*M%p);
    MTT(a1,b2,M%p);
    MTT(a2,b1,M%p);
    MTT(a2,b2,1);
    REP(i,0,m-1)write(ans[i],' ');
    return 0;
}

多项式求导

\[F=\sum_{i=0}^na_ix^i \]

\[F'=\sum_{i=0}^{n-1}(i+1)a_{i+1}x^i \]

多项式积分

\[F=\sum_{i=0}^na_ix^i \]

\[\int{F}=\sum_{i=1}^{n+1}\frac{a_{i-1}}{i}x^i \]

多项式开根

已知 \(B(x)\) ，求 \(A(x)\) 满足\(A(x)^2 \equiv B(x) \pmod{x^n}\)

假设已知 \(F(x)'\) ，使得 \(F(x)'^2\equiv B(x) \pmod{x^{\lceil\frac{p}{2}\rceil}}\)，想要得到 \(F(x)\) ，使得 \(F(x)^2 \equiv B(x) \pmod{x^p}\)

\[F(x)'^2\equiv B(x) \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)^2 \equiv B(x) \pmod{x^p} \rightarrow F(x)^2 \equiv B(x) \pmod{x^{\lceil\frac{p}{2}\rceil}}\]

\[F(x)^2-F(x)'^2 \equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[(F(x)+F(x)')(F(x)-F(x)')\equiv0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

是有两个解的，这里只推其中一个解的式子

\[F(x)-F(x)'\equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)^2-2F(x)F(x)'+F(x)'^2\equiv 0\pmod{x^p} \]

\[B(x)-2F(x)F(x)'+F(x)'^2\equiv0 \pmod{x^p} \]

\[F(x)\equiv \frac{B(x)+F(x)'^2}{2F(x)'}\pmod{x^p} \]

递归、倍增均可求解

多项式逆元

已知 \(B(x)\) ，求 \(A(x)\) 满足\(A(x)B(x) \equiv 1 \pmod{x^n}\)

假设已知 \(F(x)'\) ，使得 \(F(x)'B(x)\equiv 1 \pmod{x^{\lceil\frac{p}{2}\rceil}}\)，想要得到 \(F(x)\) ，使得 \(F(x)B(x) \equiv 1 \pmod{x^p}\)

\[F(x)'B(x)\equiv 1 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)B(x) \equiv 1 \pmod{x^p} \rightarrow F(x)B(x) \equiv 1 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)-F(x)'\equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)^2-2F(x)F(x)'+F(x)'^2\equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[B(x)F(x)^2-2B(x)F(x)F(x)'+B(x)F(x)'^2\equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)-2F(x)'+B(x)F(x)'^2\equiv 0 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

\[F(x)=2F(x)'-B(x)F(x)'^2 \pmod{x^{\lceil\frac{p}{2}\rceil}} \]

递归、倍增均可求解

多项式带余除法

已知 \(F(x)\) 和 \(G(x)\) ，求 \(Q(x)\) 和 \(R(x)\) 满足\(F(x)=G(x)Q(x)+R(x)\)

前置：对于 \(n\) 次多项式 \(F(x)\)，\(F_R(x)=x^nF(\frac{1}{x})\) ，发现即是 \(reverse\) ；\(F(x)\) 是 \(n\) 次多项式，\(G(x)\) 是 \(m\) 次多项式，那么 \(Q(x)\) 次数为 \(n-m\) ，\(R(x)\) 次数小于 \(m\)

\[F(x)=G(x)Q(x)+R(x) \]

\[F(\frac{1}{x})=G(\frac{1}{x})Q(\frac{1}{x})+R(\frac{1}{x}) \]

\[x^nF(\frac{1}{x})=x^mG(\frac{1}{x})x^{n-m}Q(\frac{1}{x})+x^{m-1}R(\frac{1}{x})x^{n-m+1} \]

\[F_R(x)=G_R(x)Q_R(x)+x^{n-m+1}R_R(x) \]

\[F_R(x) \equiv G_R(x)Q_R(x) \pmod{x^{n-m+1}} \]

\[Q_R(x) \equiv F_R(x)G_R^{-1}(x) \pmod{x^{n-m+1}} \]

因为 \(Q(x)\) 的次数为 \(n-m\) ，所以可以求出。求出 \(Q(x)\) 后，\(R(x)=F(x)-G(x)Q(x)\)

多项式对数

已知 \(B(x)\) ，求 \(A(x)\) 满足\(A(x) \equiv \ln B(x) \pmod{x^n}\)

同时求导

\[A'(x) \equiv (\ln B(x))' \pmod{x^n} \]

\[\frac{\mathrm{d}}{\mathrm{d}x}A'(x) \equiv \frac{\mathrm{d}}{\mathrm{d}x}ln B(x) \pmod{x^n} \]

\[\frac{\mathrm{d}}{\mathrm{d}x}A'(x) \equiv \frac{\mathrm{d}}{\mathrm{d}B(x)}ln B(x)\frac{\mathrm{d}}{\mathrm{d}x}B(x) \pmod{x^n} \]

\[A'\equiv \frac{1}{B}B' \pmod {x^n} \]

\[A'\equiv B^{-1}B' \pmod {x^n} \]

\[A \equiv \int B^{-1}B' \pmod {x^n} \]

多项式指数

已知 \(B(x)\) ，求 \(A(x)\) 满足\(A(x) \equiv \exp B(x) \pmod{x^n}\)

\[A(x) \equiv e^{B(x)} \pmod{x^n} \]

\[\ln A(x) \equiv B(x) \pmod {x^n} \]

\[\ln A(x)-B(x) \equiv 0 \pmod{x^n} \]

带入牛顿迭代公式 \(F(x)=F_0(x)-\frac{G(F_0(x))}{G'(F_0(x))} \pmod {x^n}\) ，\(G(F(x))=\ln F(x)-B(x) \pmod{x^n}\) 可以得到：

\[F(x)=F_0(x)-\frac{\ln F_0(x)-B(x)}{F_0^{-1}(x)} \pmod{x^n} \]

\[F(x)=F_0(x)(1-\ln F_0(x)+B(x)) \pmod{x^n} \]

递归、倍增均可求解

封装

板子要满足的几个条件：

模数是 \(998244353\)
开根，对数时输入多项式的 \(0\) 次项是 \(1\)
指数时输入多项式的 \(0\) 次项是 \(0\)

namespace poly{
	const int Mod=998244353;
	int rev[MAXN];
	ll b[MAXN],c[MAXN],tmp[MAXN],inv2,inv3,invm,pw[MAXN],ipw[MAXN],iv[MAXN];
	inline ll qexp(ll a,ll b)
	{
		ll res=1;
		while(b)
		{
			if(b&1)res=res*a%Mod;
			a=a*a%Mod;
			b>>=1;
		}
		return res;
	}
	inline void init(int n)
	{
		inv2=qexp(2,Mod-2);inv3=qexp(3,Mod-2);
		for(register int l=1;l<=(n<<2);l<<=1)
			pw[l]=qexp(3,(Mod-1)/l),ipw[l]=qexp(inv3,(Mod-1)/l);
		REP(i,1,n+n+1)iv[i]=qexp(i,Mod-2);
	}
	inline void NTT(ll *A,int n,int tp)
	{
		REP(i,0,n-1)if(i<rev[i])std::swap(A[i],A[rev[i]]);
		for(register int l=2;l<=n;l<<=1)
		{
			ll wn=tp>0?pw[l]:ipw[l];
			for(register int i=0;i<n;i+=l)
			{
				ll w=1;
				for(register int j=0;j<(l>>1);++j)
				{
					ll A1=A[i+j],A2=1ll*A[i+j+(l>>1)]*w%Mod;
					A[i+j]=(A1+A2)%Mod,A[i+j+(l>>1)]=(A1-A2+Mod)%Mod;
					w=1ll*w*wn%Mod;
				}
			}
		}
	}
	inline void Get_Mul(int n,ll *A,ll *B,ll *C)
	{
		int m,cnt;
		for(m=1,cnt=0;m<n;m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		NTT(A,m,1);NTT(B,m,1);
		REP(i,0,m-1)A[i]=1ll*A[i]*B[i]%Mod;
		NTT(A,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,n-1)C[i]=1ll*A[i]*invm%Mod;
	}
	inline void Get_Inv(int len,ll *A,ll *B)
	{
		if(len==1)
		{
			B[0]=qexp(A[0],Mod-2);
			return ;
		}
		Get_Inv((len+1)>>1,A,B);
		int m,cnt;
		for(m=1,cnt=0;m<(len<<1);m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		REP(i,0,len-1)tmp[i]=A[i];
		REP(i,len,m-1)tmp[i]=0;
		NTT(tmp,m,1);NTT(B,m,1);
		REP(i,0,m-1)B[i]=(2ll*B[i]%Mod-1ll*B[i]*B[i]%Mod*tmp[i]%Mod+Mod)%Mod;
		NTT(B,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,len-1)B[i]=1ll*B[i]*invm%Mod;
		REP(i,len,m-1)B[i]=0;
	}
	inline void Get_Sqr(int len,ll *A,ll *B)
	{
		if(len==1)
		{
			B[0]=1;
			return ;
		}
		Get_Sqr((len+1)>>1,A,B);
		memset(b,0,sizeof(b));Get_Inv(len,B,b);
		int m,cnt;
		for(m=1,cnt=0;m<(len<<1);m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		REP(i,0,len-1)tmp[i]=A[i];
		REP(i,len,m-1)tmp[i]=0;
		NTT(B,m,1);NTT(tmp,m,1);NTT(b,m,1);
		REP(i,0,m-1)B[i]=1ll*(tmp[i]+1ll*B[i]*B[i]%Mod)%Mod*b[i]%Mod*inv2%Mod;
		NTT(B,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,len-1)B[i]=1ll*B[i]*invm%Mod;
		REP(i,len,m-1)B[i]=0;
	}
	inline void Get_Div(int n1,ll *A,int n2,ll *B,ll *C,ll *D)
	{
		memset(c,0,sizeof(c));
		REP(i,0,n2-1)c[i]=B[n2-1-i];
		Get_Inv(n1-n2+1,c,b);
		int m,cnt;
		for(m=1,cnt=0;m<((n1-n2+1)<<1);m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		REP(i,0,n1-n2)tmp[i]=A[n1-1-i];
		REP(i,n1-n2+1,m-1)tmp[i]=0;
		NTT(tmp,m,1);NTT(b,m,1);
		REP(i,0,m-1)tmp[i]=1ll*tmp[i]*b[i]%Mod;
		NTT(tmp,m,-1);
		invm=qexp(m,Mod-2);
		memset(c,0,sizeof(c));
		REP(i,0,n1-n2)C[n1-n2-i]=c[n1-n2-i]=1ll*tmp[i]*invm%Mod;
		for(m=1,cnt=0;m<n1;m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		NTT(B,m,1);NTT(c,m,1);
		REP(i,0,m-1)B[i]=1ll*B[i]*c[i]%Mod;
		NTT(B,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,n2-2)D[i]=(A[i]-1ll*B[i]*invm%Mod+Mod)%Mod;
	}
	inline void Get_Der(int n,ll *A,ll *B)
	{
		REP(i,0,n-2)B[i]=1ll*A[i+1]*(i+1)%Mod;B[n-1]=0;
	}
	inline void Get_Int(int n,ll *A,ll *B)
	{
		REP(i,1,n-1)B[i]=1ll*A[i-1]*iv[i]%Mod;B[0]=0;
	}
	inline void Get_Ln(int n,ll *A,ll *B)
	{
		memset(b,0,sizeof(b));
		Get_Inv(n,A,b);Get_Der(n,A,c);
		int m,cnt;
		for(m=1,cnt=0;m<(n<<1);m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		NTT(b,m,1);NTT(c,m,1);
		REP(i,0,m-1)b[i]=1ll*b[i]*c[i]%Mod;
		NTT(b,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,n-1)b[i]=1ll*b[i]*invm%Mod;
		Get_Int(n,b,B);
	}
	inline void Get_Exp(int len,ll *A,ll *B)
	{
		if(len==1)
		{
			B[0]=1;
			return ;
		}
		Get_Exp((len+1)>>1,A,B);Get_Ln(len,B,c);
		int m,cnt;
		for(m=1,cnt=0;m<(len<<1);m<<=1,++cnt);
		REP(i,0,m-1)rev[i]=(rev[i>>1]>>1)|((i&1)<<(cnt-1));
		REP(i,0,len-1)tmp[i]=(A[i]-c[i]+(i==0)+Mod)%Mod;
		REP(i,len,m-1)tmp[i]=0;
		NTT(tmp,m,1);NTT(B,m,1);
		REP(i,0,m-1)B[i]=1ll*tmp[i]*B[i]%Mod;
		NTT(B,m,-1);
		invm=qexp(m,Mod-2);
		REP(i,0,len-1)B[i]=1ll*B[i]*invm%Mod;
		REP(i,len,m-1)B[i]=0;
	}
}

posted @ 2019-02-10 11:33 HYJ_cnyali 阅读(283) 评论(0) 编辑收藏举报

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【总结】多项式基础操作

多项式乘法

快速傅里叶变换（FFT）

快速数论变换（NTT）

任意模数情况下的NTT（MTT）

多项式求导

多项式积分

多项式开根

多项式逆元

多项式带余除法

多项式对数

多项式指数

封装

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