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多项式求逆模板题。

简单讲讲？

多项式求逆

• 定义：
  
  对于一个多项式A(x)A(x)A(x)，如果存在一个多项式B(x)B(x)B(x)，满足B(x)B(x)B(x)的次数小于等于A(x)A(x)A(x)且A(x)B(x)≡1mod  xnA(x)B(x)≡1 \mod x^nA(x)B(x)≡1modxn，那么我们称B(x)为A(x)A(x)A(x)在模xnx^nxn意义下的逆元，简单记作A−1(x)A^{−1}(x)A−1(x)
• 求法：
  
  n=1?n=1?n=1?那不就是ccc的逆元么。
  
  n>1?n>1?n>1?我们令B(x)=A−1(x)B(x)=A^{-1}(x)B(x)=A−1(x)
  
  那么有A(x)B(x)≡1mod  xnA(x)B(x)\equiv 1 \mod x^nA(x)B(x)≡1modxn
  
  然后可以用类似倍增的方法求。
  
  假设我们已经知道C(x)C(x)C(x)满足A(x)C(x)≡1mod  xn2A(x)C(x)\equiv 1\mod x^{\frac n2}A(x)C(x)≡1modx2n​（这里的n2\frac n22n​都是向上取整）
  
  显然A(x)B(x)≡1mod  xn2A(x)B(x)\equiv 1\mod x^{\frac n2}A(x)B(x)≡1modx2n​是成立的。
  
  我们将两式相减：
  
  A(x)(B(x)−C(x))≡0mod  xn2A(x)(B(x)-C(x))\equiv 0\mod x^{\frac n2}A(x)(B(x)−C(x))≡0modx2n​
  
  所以B(x)−C(x)≡0mod  xn2B(x)-C(x)\equiv 0\mod x^{\frac n2}B(x)−C(x)≡0modx2n​
  
  然后将两边平方：
  
  B2(x)−2B(x)C(x)+C2(x)≡0mod  xn2B^2(x)-2B(x)C(x)+C^2(x)\equiv 0\mod x^{\frac n2}B2(x)−2B(x)C(x)+C2(x)≡0modx2n​
  
  =>B2(x)−2B(x)C(x)+C2(x)≡0mod  xnB^2(x)-2B(x)C(x)+C^2(x)\equiv 0\mod x^nB2(x)−2B(x)C(x)+C2(x)≡0modxn
  
  这一步很关键，请神犇们仔细思考原因
  
  然后两边同时乘上A(x)A(x)A(x)
  
  =>B(x)−2C(x)+A(x)C2(x)≡0mod  xn)B(x)-2C(x)+A(x)C^2(x)\equiv 0\mod x^n)B(x)−2C(x)+A(x)C2(x)≡0modxn)
  
  于是B(x)≡2C(x)−A(x)C2(x)mod  xnB(x)\equiv2C(x)-A(x)C^2(x)\mod x^nB(x)≡2C(x)−A(x)C2(x)modxn

乘法可以用fft/nttfft/nttfft/ntt加速，因为每次递归的时候多项式最高次项都减少一半，所以总复杂度仍然是O(nlogn)O(nlogn)O(nlogn)

代码：

#include<bits/stdc++.h>
#define ri register int
using namespace std;
inline int read(){
	int ans=0;
	char ch=getchar();
	while(!isdigit(ch))ch=getchar();
	while(isdigit(ch))ans=(ans<<3)+(ans<<1)+(ch^48),ch=getchar();
	return ans;
}
typedef long long ll;
const int mod=998244353;
int n;
inline int ksm(int a,int p){int ret=1;for(;p;p>>=1,a=(ll)a*a%mod)if(p&1)ret=(ll)ret*a%mod;return ret;}
inline int add(const int&a,const int&b){return a+b>=mod?a+b-mod:a+b;}
inline int dec(const int&a,const int&b){return a>=b?a-b:a-b+mod;}
inline int mul(const int&a,const int&b){return (ll)a*b%mod;}
int lim,tim;
vector<int>pos,A,B;
inline void init(const int&n){lim=1,tim=0;
	while(lim<=n)lim<<=1,++tim;
	pos.resize(lim),pos[0]=0;
	for(ri i=0;i<lim;++i)pos[i]=(pos[i>>1]>>1)|((i&1)<<(tim-1));
}
inline void ntt(vector<int>&a,int type){
	for(ri i=0;i<lim;++i)if(i<pos[i])swap(a[i],a[pos[i]]);
	for(ri mult=(mod-1)/2,mid=1,wn,typ=type==1?3:(mod+1)/3;mid<lim;mid<<=1,mult>>=1){
		wn=ksm(typ,mult);
		for(ri j=0,len=mid<<1,w;j<lim;j+=len){
			w=1;
			for(ri k=0,a0,a1;k<mid;w=mul(w,wn),++k){
				a0=a[j+k],a1=mul(w,a[j+k+mid]);
				a[j+k]=add(a0,a1),a[j+k+mid]=dec(a0,a1);
			}
		}
	}
	if(type==-1)for(ri i=0,inv=ksm(lim,mod-2);i<lim;++i)a[i]=mul(a[i],inv);
}
struct poly{
	vector<int>a;
	poly(const int&n,int x=0){a.resize(n+1),a[n]=x;}
	inline int&operator[](const int&i){return a[i];}
	inline const int&operator[](const int&i)const{return a[i];}
	inline poly extend(const int&x){poly ret=*this;return ret.a.resize(x),ret;}
	inline int deg()const{return a.size()-1;}
	friend inline poly operator+(const poly&a,const poly&b){
		poly c(max(a.deg(),b.deg()));
		for(ri i=0;i<=a.deg();++i)c[i]=add(c[i],a[i]);
		for(ri i=0;i<=b.deg();++i)c[i]=add(c[i],b[i]);
		return c;
	}
	friend inline poly operator-(const poly&a,const poly&b){
		poly c(max(a.deg(),b.deg()));
		for(ri i=0;i<=a.deg();++i)c[i]=add(c[i],a[i]);
		for(ri i=0;i<=b.deg();++i)c[i]=dec(c[i],b[i]);
		return c;
	}
	friend inline poly operator*(const poly&a,const int&b){
			poly c=a;
			for(ri i=0;i<=c.deg();++i)c[i]=mul(b,c[i]);
			return c;
	}
	friend inline poly operator *(const poly&a,const poly&b){
		int n=a.deg(),m=b.deg();
		init(n+m),A.resize(lim),B.resize(lim);
		poly ret(lim-1);
		for(ri i=0;i<=n;++i)A[i]=a[i];
		for(ri i=0;i<=m;++i)B[i]=b[i];
		for(ri i=n+1;i<lim;++i)A[i]=0;
		for(ri i=m+1;i<lim;++i)B[i]=0;
		ntt(A,1),ntt(B,1);
		for(ri i=0;i<lim;++i)A[i]=mul(A[i],B[i]);
		return ntt(A,-1),ret.a=A,ret;
	}
	inline poly poly_inv(poly A,const int k){
		if(k==1)return poly(0,ksm(A[0],mod-2));
		poly f0=poly_inv(A.extend((k+1)>>1),(k+1)>>1);
		return (f0*2-(A*((f0*f0).extend(k))).extend(k)).extend(k);
	}
};
int main(){
	freopen("lx.in","r",stdin);
	n=read()-1;
	poly a(n);
	for(ri i=0;i<=n;++i)a[i]=read();
	init(n*2),a=a.extend(lim),a=a.poly_inv(a,lim);
	for(ri i=0;i<=n;++i)cout<<a[i]<<' ';
	return 0;
}