2019.01.02 洛谷P4512 【模板】多项式除法

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 解析
代码：
#include<bits/stdc++.h>
#define ri register int
using namespace std;
typedef long long ll;
#define add(a,b) ((a)+(b)>=mod?(a)+(b)-mod:(a)+(b))
#define dec(a,b) ((a)>=(b)?(a)-(b):(a)-(b)+mod)
#define mul(a,b) ((ll)(a)*(b)%mod)
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;
}
const int mod=998244353;
int lim,tim;
vector<int>pos,A,B;
inline int ksm(int a,int p){int ret=1;for(;p;p>>=1,a=mul(a,a))if(p&1)ret=mul(ret,a);return ret;}
inline void init(const int&up){
	lim=1,tim=0;
	while(lim<=up)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,const int&type){
	for(ri i=0;i<lim;++i)if(i<pos[i])swap(a[i],a[pos[i]]);
	for(ri mid=1,wn,typ=type==1?3:(mod+1)/3,mult=(mod-1)/2;mid<lim;mid<<=1,mult>>=1){
		wn=ksm(typ,mult);
		for(ri j=0,len=mid<<1;j<lim;j+=len)for(ri k=0,a0,a1,w=1;k<mid;++k,w=mul(w,wn)){
			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(int k,int x=0){a.resize(k+1),a[k]=x;}
	inline int&operator[](const int&k){return a[k];}
	inline const int&operator[](const int&k)const{return a[k];}
	inline int deg()const{return a.size()-1;}
	inline poly extend(const int&k){poly ret=*this;return ret.a.resize(k+1),ret;}
	friend inline poly operator+(const poly&a,const poly&b){
		poly ret(max(a.deg(),b.deg()));
		for(ri i=0;i<=a.deg();++i)ret[i]=add(ret[i],a[i]);
		for(ri i=0;i<=b.deg();++i)ret[i]=add(ret[i],b[i]);
		return ret;
	}
	friend inline poly operator-(const poly&a,const poly&b){
		poly ret(max(a.deg(),b.deg()));
		for(ri i=0;i<=a.deg();++i)ret[i]=add(ret[i],a[i]);
		for(ri i=0;i<=b.deg();++i)ret[i]=dec(ret[i],b[i]);
		return ret;
	}
	friend inline poly operator*(const int&a,const poly&b){
		poly ret(b.deg());
		for(ri i=0;i<=b.deg();++i)ret[i]=mul(a,b[i]);
		return ret;
	}
	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));
		a=a.extend(k);
		poly f0=poly_inv(a,(k+1)>>1);
		return (2*f0-((f0*f0).extend(k)*a).extend(k)).extend(k);
	}
	friend inline poly operator/(const poly&a,const poly&b){
		poly ta=a,tb=b;
		int len=1,up=a.deg()-b.deg();
		reverse(ta.a.begin(),ta.a.end()),reverse(tb.a.begin(),tb.a.end());
		ta.extend(up),tb.extend(up);
		while(len<=up)len<<=1;
		tb=tb.poly_inv(tb,len).extend(up);
		return ta=(ta*tb).extend(up),reverse(ta.a.begin(),ta.a.end()),ta;
	}
};
int n,m;
int main(){
	n=read(),m=read();
	poly a(n),b(m),ans(n-m);
	for(ri i=0;i<=n;++i)a[i]=read();
	for(ri i=0;i<=m;++i)b[i]=read();
	ans=a/b;
	for(ri i=0;i<=n-m;++i)cout<<ans[i]<<' ';
	puts(""),ans=a-ans*b;
	for(ri i=0;i<m;++i)cout<<ans[i]<<' ';
	return 0;
}
2019.01.02 洛谷P4512 【模板】多项式除法

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