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题目传送门

题目大意

给出一个\(n\)次多项式\(f\)，有\(m\)个点，分别为\(\{a_1,a_2,...,a_m\}\)，请您求出对于任意\(i\in [1,m]\)，求出\(f(a_i)\)。

\(n,m\le 64000\)

思路

我用的是一种人尽皆知的方法，即多项式取模的\(\Theta(n\log ^2n)\)的方法，常数极大，而且因为我自身封装的问题，导致我的空间消耗也极大。（自闭了）

我们发现，如果我们构造\(P(x)=\prod_{i=1}^{m} (x-a_i)\)，那么我们假设\(F(x)=P(x)D(x)+R(x)\)，那么对于\(i\in [1,m],F(a_i)=R(a_i)\)。于是，我们可以得到一个非常\(\texttt{naive}\)的想法，即一直多项式取模，模到最后只剩下常数项就是答案。

根据主定理，时间复杂度为\(\Theta(n\log^2n)\)，因为多项式取模时间复杂度为\(\Theta(n\log n)\)。

\(\texttt{Code}\)

#include <bits/stdc++.h>
using namespace std;

#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 1000005

int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
	int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
	return res;
}
int inv (int x){return qkpow (x,mod - 2);}

typedef vector <int> poly;

int w[MAXN],rev[MAXN];

void init_ntt (){
	int lim = 1 << 18;
	for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
	int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
	for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
	for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}

void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
	static int d[MAXN];
	for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
	for (Int i = 1;i < lim;i <<= 1)
		for (Int j = 0;j < lim;j += i << 1)
			for (Int k = 0;k < i;++ k){
				int x = mul (w[i + k],d[i + j + k]);
				d[i + j + k] = dec (d[j + k],x),d[j + k] = add (d[j + k],x);
			}
	for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
	if (type == -1){
		reverse (a.begin() + 1,a.begin() + lim);
		for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
	}
#undef G
#undef Gi 
}

poly operator + (poly a,poly b){
	a.resize (max (SZ (a),SZ (b)));
	for (Int i = 0;i < SZ (b);++ i) a[i] = add (a[i],b[i]);
	return a;
}

poly operator - (poly a,poly b){
	a.resize (max (SZ (a),SZ (b)));
	for (Int i = 0;i < SZ (b);++ i) a[i] = dec (a[i],b[i]);
	return a;
}

poly operator * (poly a,int b){
	for (Int i = 0;i < SZ (a);++ i) a[i] = mul (a[i],b);
	return a;
}

poly operator * (poly a,poly b){
	int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
	a.resize (lim),b.resize (lim);
	ntt (a,lim,1),ntt (b,lim,1);
	for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
	ntt (a,lim,-1),a.resize (d);
	return a;
}

poly inv (poly a,int n){
	poly b(1,inv (a[0])),c;
	for (Int l = 4;(l >> 2) < n;l <<= 1){
		c.resize (l >> 1);
		for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
		c.resize (l),b.resize (l);
		ntt (c,l,1),ntt (b,l,1);
		for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
		ntt (b,l,-1),b.resize (l >> 1);
	}
	b.resize (n);
	return b;
}

poly inv (poly a){return inv (a,SZ (a));}

poly Mod (poly F,poly G){
	int n = SZ (F) - 1,m = SZ (G) - 1;poly Q;Q.resize (m + 1);for (Int i = 0;i <= m;++ i) Q[i] = G[i];
	reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),reverse (Q.begin(),Q.end()),Q.resize (n - m + 1),Q = inv (Q) * F,Q.resize (n - m + 1),reverse (Q.begin(),Q.end());
	reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),Q = G * Q,Q.resize (m),Q = F - Q,Q.resize (m);
	return Q;
}

template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < ‘0‘ || c > ‘9‘){if (c == ‘-‘) f = -f;c = getchar();}while (c >= ‘0‘ && c <= ‘9‘){t = (t << 3) + (t << 1) + c - ‘0‘;c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar (‘-‘);}if (x > 9) write (x / 10);putchar (x % 10 + ‘0‘);}

int n,m,a[MAXN];
poly A,DR[MAXN << 2];

void divide1 (int i,int l,int r){
	if (l == r) return DR[i].resize (2),DR[i][0] = mod - a[l],DR[i][1] = 1,void ();
	int mid = (l + r) >> 1;divide1 (i << 1,l,mid),divide1 (i << 1 | 1,mid + 1,r);
	DR[i] = DR[i << 1] * DR[i << 1 | 1];
}

void divide2 (int i,int l,int r,poly AA){
	if (l == r) return write (AA[0]),putchar (‘\n‘),void ();
	poly B = Mod (AA,DR[i << 1]);int mid = (l + r) >> 1;divide2 (i << 1,l,mid,B);B = Mod (AA,DR[i << 1 | 1]);divide2 (i << 1 | 1,mid + 1,r,B);
}

signed main(){
	init_ntt(),read (n,m),A.resize (n + 1);for (Int i = 0;i <= n;++ i) read (A[i]);for (Int i = 1;i <= m;++ i) read (a[i]);
	divide1 (1,1,m),divide2 (1,1,m,A);
	return 0;
}

题解 多项式多点求值

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原文地址：https://www.cnblogs.com/Dark-Romance/p/13340941.html