多项式多点求值

给定一个多项式和m个x，求相应的y

我们把需要求值的点均分成两个集合S1,S2，构造两个多项式P1,P2，使得这两个多项式分别为这两个集合的零点。则多项式A%P1对于S1满足A%P1对S1内元素求值和A相同，A%P2对于S2内求值和A相同，而它们次数都是n/2,分治递归下去继续求值即可。

由于多项式取模的数组版还不会写，这里借用的以前的vector版，有非常大的优化空间(vector真香)

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

const int p = 998244353;

int qpow(int x, int y)
{
	int res = 1;
	while (y > 0)
	{
		if (y & 1)
			res = 1LL * res * x % p;
		x = 1LL * x * x % p;
		y >>= 1;
	}
	return res;
}

void FNTT(vector<int> &A, int len, int flag)
{
	A.resize(len);
	int *r = new int[len];
	r[0] = 0;
	for (int i = 0; i < len; i++)
		r[i] = (r[i >> 1] >> 1) | ((i & 1) * (len >> 1));
	for (int i = 0; i < len; i++)
		if (i < r[i])
			swap(A[i], A[r[i]]);
	int gn, g, t, A0, A1;
	for (int i = 1; i < len; i <<= 1)
	{
		gn = qpow(3, (p - 1) / (i * 2));
		for (int j = 0; j < len; j += (i << 1))
		{
			g = 1;
			A0 = j;
			A1 = A0 + i;
			for (int k = 0; k < i; k++, A0++, A1++, g = (1LL * g * gn) % p)
			{
				t = (1LL * A[A1] * g) % p;
				A[A1] = ((A[A0] - t) % p + p) % p;
				A[A0] = (A[A0] + t) % p;
			}
		}
	}
	if (flag == -1)
	{
		reverse(A.begin() + 1, A.end());
		int inv = qpow(len, p - 2);
		for (int i = 0; i < len; i++)
			A[i] = 1LL * A[i] * inv % p;
	}
	delete []r;
}

vector<int> operator*(vector<int> a, vector<int> b)
{
	int len = 1;
	int sz = a.size() + b.size() - 1;
	while (len <= sz) len <<= 1;
	FNTT(a, len, 1);
	FNTT(b, len, 1);
	vector<int> res;
	res.resize(len);
	for (int i = 0; i < len; i++)
		res[i] = 1LL * a[i] * b[i] % p;
	FNTT(res, len, -1);
	res.resize(sz);
	return res;
}

vector<int> poly_inv(vector<int> a)
{
	if (a.size() == 1)
	{
		a[0] = qpow(a[0], p - 2);
		return a;
	}
	int n = a.size(), newsz = (n + 1) >> 1;
	vector<int> b(a);
	b.resize(newsz);
	b = poly_inv(b);
	int len = 1;
	while (len <= (n << 1)) len <<= 1;
	vector<int> c(a);
	FNTT(a, len, 1);
	FNTT(b, len, 1);
	for (int i = 0; i < len; i++)
		a[i] = ((1LL * b[i] * (2 - 1LL * a[i] * b[i] % p)) % p + p) % p;
	FNTT(a, len, -1);
	a.resize(n);
	return a;
}

vector<int> poly_r(vector<int> a)
{
	reverse(a.begin(), a.end());
	return a;
}

void div(vector<int> f, vector<int> g, vector<int> &q, vector<int> &r)
{
	int n = f.size() - 1, m = g.size() - 1;
	vector<int> gr = poly_r(g);
	gr.resize(n - m + 1);
	q = poly_r(f) * poly_inv(gr);
	q.resize(n - m + 1);
	q = poly_r(q);
	vector<int> gq = g * q;
	r.resize(m);
	gq.resize(m);
	f.resize(m);
	for (int i = 0; i < m; i++)
		r[i] = ((f[i] - gq[i]) % p + p) % p;
}

int n, m;
vector<int> f, tmp[1000010];
int a[100010], res[100010], le[1000010], re[1000010], tot;

vector<int> prework(int l, int r)
{
	int id = ++tot;
	if (l == r)
	{
		vector<int> res;
		res.push_back(p - a[l]);
		res.push_back(1);
		tmp[id] = res;
		return res;
	}
	int mid = (l + r) / 2;
	le[id] = tot + 1;
	vector<int> res = prework(l, mid);
	re[id] = tot + 1;
	res = res * prework(mid + 1, r);
	return tmp[id] = res;
}

void work(int l, int r, vector<int> sb)
{
	int id = ++tot;
	if (l == r)
	{
		int tmp = 1;
		for (int i = 0; i < (int)sb.size(); i++)
			res[l] = (res[l] + tmp * sb[i]) % p, tmp = tmp * (long long)a[l] % p;
		return;
	}
	vector<int> fl = tmp[le[id]], fr = tmp[re[id]];
	vector<int> tmp1, rel, rer;
	div(sb, fl, tmp1, rel);
	div(sb, fr, tmp1, rer);
	int mid = (l + r) / 2;
	work(l, mid, rel);
	work(mid + 1, r, rer);
}

int main()
{
	scanf("%d%d", &n, &m); f.resize(n + 1);
	for (int i = 0; i <= n; i++) scanf("%d", &f[i]);
	for (int i = 1; i <= m; i++) scanf("%d", &a[i]);
	prework(1, max(n, m));
	tot = 0;
	work(1, max(n, m), f);
	for (int i = 1; i <= m; i++) printf("%d\n", res[i]);
	return 0;
}