多項式開方

\[\begin{align} A(x) &\equiv B'(x)^2 &\mod x^{n/2}\\ A(x) &\equiv B(x)^2 &\mod x^{n}\\ A(x) &\equiv B(x)^2 &\mod x^{n/2}\\ B(x)^2 - B'(x)^2 &\equiv 0 &\mod x^{n/2}\\ B^4(x)+B'(x)^4-2B(x)^2B'(x)^2 &\equiv 0 &\mod x^n\\ B^4(x)+B'(x)^4 + 2B(x)^2B'(x)^2 &\equiv 4B(x)^2B'(x)^2 &\mod x^n\\ B^2(x)+B'(x)^2 &\equiv 2B(x)B'(x) &\mod x^n\\ A(x)+B'(x)^2 &\equiv 2B(x)B'(x) &\mod x^n\\ B(x) &\equiv \dfrac{A(x)+B'(x)^2}{2B'(x)} &\mod x^n \end{align} \]

跟求逆用的那種倍增方法極度相似。

就是要用 O(n log² n) 的時間來求了。真的要學牛頓迭代了嗎（

但是這個好像可以在倍增的過程中維護 B 的逆， 有點可以， 可以單 log， 以後再補。

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

const int N = 4e5 + 233, mo = 998244353;
LL ksm(LL a, LL b) {
	LL res = 1ll;
	for(; b; b >>= 1, a = a * a % mo)
		if(b & 1) res = res * a % mo;
	return res;
}
const LL g = 3, ig = ksm(g, mo - 2), inv_2 = ksm(2, mo - 2);

int rv[N];
void NTT(LL *a, int n, int type) {
	for(int i = 0; i < n; ++i) if(i < rv[i]) swap(a[i], a[rv[i]]);
	for(int m = 2; m <= n; m = m << 1) {
		LL w = ksm(type == 1 ? g : ig, (mo - 1) / m);
		for(int i = 0; i < n; i += m) {
			LL tmp = 1ll;
			for(int j = 0; j < (m >> 1); ++j) {
				LL p = a[i + j], q = tmp * a[i + j + (m >> 1)] % mo;
				a[i + j] = (p + q) % mo, a[i + j + (m >> 1)] = (p - q + mo) % mo;
				tmp = tmp * w % mo;
			}
		}
	}
	if(type == -1) {
		LL Inv = ksm(n, mo - 2);
		for(int i = 0; i < n; ++i) a[i] = a[i] * Inv % mo;
	}
}

LL t[N];
void poly_inv(int deg, LL *a, LL *b) {
	if(deg == 1) { b[0] = ksm(a[0], mo - 2); return; }
	poly_inv((deg + 1) >> 1, a, b);
	int len = 1; while(len < (deg << 1)) len = len << 1;
	for(int i = 0; i < deg; ++i) t[i] = a[i];
	for(int i = deg; i < len; ++i) b[i] = t[i] = 0ll;
	for(int i = 1; i < len; ++i) rv[i] = (rv[i>>1]>>1) | (i&1?len>>1:0);
	NTT(b, len, 1), NTT(t, len, 1);
	for(int i = 0; i < len; ++i) b[i] = b[i] * (2ll - t[i] * b[i] % mo) % mo;
	NTT(b, len, -1);
	for(int i = deg; i < len; ++i) b[i] = 0ll;
}

LL binv[N], c[N];
void poly_sqrt(int deg, LL *a, LL *b) {
	if(deg == 1) { b[0] = 1ll; return; }
	poly_sqrt((deg + 1) >> 1, a, b);
	for(int i = 0; i < deg; ++i) c[i] = 2ll * b[i] % mo;
	poly_inv(deg, c, binv);
	int len = 1; while(len < (deg << 1)) len = len << 1;
	for(int i = 1; i < len; ++i) rv[i] = (rv[i>>1]>>1) | (i&1?len>>1:0);
	for(int i = 0; i < deg; ++i) c[i] = a[i];
	for(int i = deg; i < len; ++i) b[i] = binv[i] = c[i] = 0ll;
	NTT(b, len, 1), NTT(binv, len, 1), NTT(c, len, 1);
	for(int i = 0; i < len; ++i) b[i] = (c[i] + b[i] * b[i] % mo ) % mo * binv[i] % mo;
	NTT(b, len, -1);
	for(int i = deg; i < len; ++i) b[i] = 0ll;
}

int n;
LL a[N], b[N];
int main() {
	scanf("%d", &n);
	for(int i = 0; i < n; ++i) scanf("%lld", &a[i]);
	poly_sqrt(n, a, b);
	for(int i = 0; i < n; ++i) cout << (b[i]%mo + mo) % mo << ' ';
	return 0;
}