[题解] LuoguP5050 【模板】多项式多点求值

https://www.luogu.com.cn/problem/P5050

Description：

字面意思

给一个(n)次多项式(f(x))，让你求(f(x_1),f(x_2),...f(x_m))的值。

对(998244353)取模。

(1le n,m le 64000)

Solution:

考虑分治，咋分治呢？

构造两个多项式

[L(x)=prodlimits_{i=1}^{m/2} (x-x_i) ]

[R(x)=prodlimits_{i=m/2+1}^{m} (x-x_i) ]

那么对于(i in [1,m/2])，有(L(x_i)=0)，考虑将(f)对(L)取模，即

[f(x_i)=Q(x_i)L(x_i)+R(x_i)=R(x_i) ]

即对于左半边的(x_i)，有(f(x_i)=(f mod L)(x_i))，右半边同理。

于是这样递归往下取模就行了。

那个(L,R)珂以分治NTT。于是分治NTT的时候顺便建一棵线段树，然后从根开始往下走并不断取模，到叶子节点的时候(f)已经膜的只剩一个常数，就得到了一个答案。

同时注意要先对根节点上的多项式取膜。

复杂度(O(n log^2 n))，但窝写出来常数贼大......

Code:

这个菜鸡的超大常数代码......最后加了快读才卡过去......

#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
typedef long long ll;
typedef double db;
typedef pair<int,int> PII;
typedef vector<int> VI;

const int N=3e5+10,mod=998244353;
inline int add(int x,int y,int m=mod) {return (x+=y)>=m?x-m:x;}
inline int sub(int x,int y,int m=mod) {return (x-=y)<0?x+m:x;}
inline int powmod(int x,int y=mod-2,int m=mod) {
	int ans=1; for (;y;y>>=1,x=1ll*x*x%m)
		if (y&1) ans=1ll*ans*x%m;
	return ans;
}
const int gn=3,ign=powmod(gn);

namespace Polynomial {
	int rev[N];
	void init(int n) { rep(i,0,n) rev[i]=rev[i>>1]>>1|((i&1)?n>>1:0); }
	int glim(int n) { int lim=1; while (lim<=n) lim<<=1; return lim; }

	void NTT(int *f,int n,int flg) {
		rep(i,0,n) if (rev[i]<i) swap(f[i],f[rev[i]]);
		for (int len=2,k=1;len<=n;len<<=1,k<<=1) {
			int wn=powmod(flg==1?gn:ign,(mod+1)/len);
			for (int i=0;i<n;i+=len)
				for (int j=i,w=1;j<i+k;j++,w=1ll*w*wn%mod) {
					int tmp=1ll*f[j+k]*w%mod;
					f[j+k]=sub(f[j],tmp),f[j]=add(f[j],tmp);
				}
		}
		if (flg==-1) {
			int inv=powmod(n);
			rep(i,0,n) f[i]=1ll*f[i]*inv%mod;
		}
	}

	void Inv(int *f,int n,int *g) {
		if (n==1) { g[0]=powmod(f[0]); return; }
		Inv(f,(n+1)>>1,g);
		static int tf[N]; 
		int lim=glim((n-1)<<1); init(lim);
		rep(i,0,lim) tf[i]=i<n?f[i]:0,g[i]=i<n?g[i]:0;
		NTT(tf,lim,1),NTT(g,lim,1);
		rep(i,0,lim) g[i]=1ll*g[i]*sub(2,1ll*tf[i]*g[i]%mod)%mod;
		NTT(g,lim,-1);
		rep(i,n,lim) g[i]=0;
	}

	struct poly {
		VI a;
		int size()const {return a.size();}
		int deg()const {return size()-1;}
		int &operator[](int i) { assert(i<size()); return a[i]; }
		int operator[](int i)const {return i<size()?a[i]:0;}
		void reverse() {::reverse(all(a));}
		void resize(int n) {a.resize(n);}
		poly(int n=0) {a=VI(n,0);}

		void print() {
			printf("#debug: "); puts("");
			for (auto v:a) printf("%d ",v);
			puts("
-----------------");
		}

		poly inv(int n)const {
			static int f[N],invf[N];
			rep(i,0,n) f[i]=a[i];
			Inv(f,n,invf);
			poly ans(n);
			rep(i,0,n) ans[i]=invf[i];
			return ans;
		}
	};

	poly operator+(const poly &a,const poly &b) {
		poly c(max(a.size(),b.size()));
		rep(i,0,c.size()) c[i]=add(a[i],b[i]);
		return c;
	}
	poly operator-(const poly &a,const poly &b) {
		poly c(max(a.size(),b.size()));
		rep(i,0,c.size()) c[i]=sub(a[i],b[i]);
		return c;
	}

	poly operator*(const poly &a,const poly &b) {
		int n=a.deg(),m=b.deg(),lim=glim(n+m);
		poly c(n+m+1);
		static int ta[N],tb[N];
		rep(i,0,lim) ta[i]=tb[i]=0;
		rep(i,0,n+1) ta[i]=a[i];
		rep(i,0,m+1) tb[i]=b[i];
		init(lim); NTT(ta,lim,1),NTT(tb,lim,1);
		rep(i,0,lim) ta[i]=1ll*ta[i]*tb[i]%mod;
		NTT(ta,lim,-1);
		rep(i,0,n+m+1) c[i]=ta[i];
		return c;
	}

	pair<poly,poly> divmod(const poly &a,const poly &b) {
		if (a.deg()<b.deg()) return mp(poly(0),a);
		int n=a.deg(),m=b.deg();
		poly ra=a,rb=b; ra.reverse(),rb.reverse();
		poly q=ra*rb.inv(n-m+1),r;
		q.resize(n-m+1),q.reverse();
		r=a-q*b,r.resize(m);
		return mp(q,r);
	}

	poly operator/(const poly &a,const poly &b) {return divmod(a,b).fi;}
	poly operator%(const poly &a,const poly &b) {return divmod(a,b).se;}
}
using Polynomial::poly;

poly p[N],a[N];

void build(int x,int l,int r) {
	if (l==r) {p[x]=a[l];return;}
	int mid=(l+r)>>1;
	build(x<<1,l,mid),build(x<<1|1,mid+1,r);
	p[x]=p[x<<1]*p[x<<1|1];
	// printf("%d: [%d,%d]",x,l,r); p[x].print();
}

int ans[N],n,m;
poly A;

void solve(int x,int l,int r,const poly &f) {
	if (l==r) {ans[l]=f[0];return;}
	int mid=(l+r)>>1,lc=x<<1,rc=x<<1|1;
	solve(lc,l,mid,f%p[lc]);
	solve(rc,mid+1,r,f%p[rc]);
}

namespace FakeIO {
	const int SZ=233333;
	char buf[SZ],*p1=buf,*p2=buf;
	inline char nc() {return p1==p2&&(p2=(p1=buf)+fread(buf,1,SZ,stdin),p1==p2)?EOF:*p1++;}
	inline int gi() {
		int x=0,f=1; char ch; 
		while (!isdigit(ch=nc())) if (ch=='-') f=-f;
		while (isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=nc();}
		return x*f;
	}
}
using FakeIO::gi;

int main() {
#ifdef LOCAL
	freopen("A.in","r",stdin);
#endif
	n=gi(),m=gi();
	A=poly(n+1);
	rep(i,0,n+1) {int x=gi();A[i]=x;}
	rep(i,1,m+1) {
		int x=gi(); a[i]=poly(2);
		a[i][0]=mod-x,a[i][1]=1;
	}
	build(1,1,m);
	solve(1,1,m,A%p[1]);
	rep(i,1,m+1) printf("%d
",ans[i]);
	return 0;
}