[洛谷P5205]【模板】多项式开根

题目大意：给你$n$项多项式$A(x)$，求出$B(x)$满足$B^2(x)\equiv A(x)\pmod{x^n}$

题解：考虑已经求出$B_0(x)$满足$B_0^2(x)\equiv A(x)\pmod{x^{\lceil\frac n 2\rceil}}$
$$
B(x)-B_0(x)\equiv0\pmod{x^{\lceil\frac n 2\rceil}}\\
B^2(x)−2B(x)B_0(x)+B_0^2(x)≡0\pmod{x^n}\\
A(x)-2B(x)B_0(x)+B_0^2(x)≡0\pmod{x^n}\\
B(x)\equiv\dfrac{A(x)+B_0^2(x)}{2B_0(x)}\pmod{x^n}\\
$$
卡点：求$INV$时注意清空数组，防止因为$B$数组不干净导致出锅

C++ Code：

#include <algorithm>
#include <cstdio>
#define maxn 262144
const int mod = 998244353, __2 = mod + 1 >> 1;

namespace Math {
	inline int pw(int base, int p) {
		static int res;
		for (res = 1; p; p >>= 1, base = static_cast<long long> (base) * base % mod) if (p & 1) res = static_cast<long long> (res) * base % mod;
		return res;
	}
	inline int inv(int x) { return pw(x, mod - 2); }
}

inline void reduce(int &x) { x += x >> 31 & mod; }
inline void clear(register int *l, const int *r) {
	if (l >= r) return ;
	while (l != r) *l++ = 0;
}

namespace Poly {
#define N maxn
	int lim, s, rev[N];
	int Wn[N + 1];

	inline void init(const int n) {
		lim = 1, s = -1; while (lim < n) lim <<= 1, ++s;
		for (register int i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
		const int t = Math::pw(3, (mod - 1) / lim);
		*Wn = 1; for (register int *i = Wn; i != Wn + lim; ++i) *(i + 1) = static_cast<long long> (*i) * t % mod;
	}
	inline void FFT(int *A, const int op = 1) {
		for (register int i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
		for (register int mid = 1; mid < lim; mid <<= 1) {
			const int t = lim / mid >> 1;
			for (register int i = 0; i < lim; i += mid << 1)
				for (register int j = 0; j < mid; ++j) {
					const int X = A[i + j], Y = static_cast<long long> (A[i + j + mid]) * Wn[t * j] % mod;
					reduce(A[i + j] += Y - mod), reduce(A[i + j + mid] = X - Y);
				}
		}
		if (!op) {
			const int ilim = Math::inv(lim);
			for (register int *i = A; i != A + lim; ++i) *i = static_cast<long long> (*i) * ilim % mod;
			std::reverse(A + 1, A + lim);
		}
	}

	void INV(int *A, int *B, int n) {
		if (n == 1) { *B = Math::inv(*A); return ; }
		const int len = n + 1 >> 1;
		INV(A, B, len); init(len * 3);
		static int C[N], D[N];
		std::copy(A, A + n, C); clear(C + n, C + lim);
		std::copy(B, B + len, D); clear(D + len, D + lim);
		FFT(D), FFT(C);
		for (register int i = 0; i < lim; ++i) D[i] = (2 - static_cast<long long> (D[i]) * C[i] % mod + mod) * D[i] % mod;
		FFT(D, 0); std::copy(D + len, D + n, B + len);
	}
	void SQRT(int *A, int *B, int n) {
		if (n == 1) { *B = 1; return ; }
		static int C[N], D[N];
		SQRT(A, B, n + 1 >> 1);
		INV(B, D, n);
		init(n + n);
		std::copy(A, A + n, C); clear(C + n, C + lim);
		clear(D + n, D + lim);
		FFT(C), FFT(D), FFT(B);
		for (int i = 0; i < lim; ++i) B[i] = (static_cast<long long> (B[i]) * B[i] % mod + C[i]) * D[i] % mod * __2 % mod;
		FFT(B, 0), clear(B + n, B + lim);
	}
#undef N
}

int n, A[maxn], B[maxn];
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n; ++i) scanf("%d", A + i);
	Poly::SQRT(A, B, n);
	for (int i = 0; i < n; ++i) printf("%d ", B[i]);
	puts("");
	return 0;
}

标签：洛谷,P5205,int,lim,FFT,len,static,开根,mod
来源： https://www.cnblogs.com/Memory-of-winter/p/10335437.html