[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$B_0^2(x)\equiv A(x)\pmod{x^{\lceil\frac n 2\rceil}}$
$$
B(x)-B_0(x)\equiv0\pmod{x^{\lceil\frac n 2\rceil}}\\
B^2(x)−2B(x)B_0(x)+B_0^2(x)≡0\pmod{x^n}\\
A(x)-2B(x)B_0(x)+B_0^2(x)≡0\pmod{x^n}\\
B(x)\equiv\dfrac{A(x)+B_0^2(x)}{2B_0(x)}\pmod{x^n}\\
$$
update：（2019-2-10）
$$
B(x)\equiv\dfrac{A(x)+B_0^2(x)}{2B_0(x)}\pmod{x^n}\\
B(x)\equiv\dfrac{A(x)}{2B_0(x)}+\dfrac{B_0(x)}2\pmod{x^n}\\
$$
发现$\dfrac{B_0(x)}2$只会影响$B(x)$数组的前半部分（即$\pmod{x^{\lceil\frac n2\rceil}}$的部分），但是$B(x)\equiv B_0(x)\pmod{x^{\lceil\frac n2\rceil}}$，所以可以不做考虑，直接把$B_0(x)$拉过来
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 <cctype>
#include <cstdio>
#define maxn 262144
const int mod = 998244353, __2 = mod + 1 >> 1;
namespace std {
struct istream {
#define M (1 << 21 | 3)
char buf[M], *ch = buf - 1;
inline istream() { fread(buf, 1, M, stdin); }
inline istream& operator >> (int &x) {
while (isspace(*++ch));
for (x = *ch & 15; isdigit(*++ch); ) x = x * 10 + (*ch & 15);
return *this;
}
#undef M
} cin;
struct ostream {
#define M (1 << 21 | 3)
char buf[M], *ch = buf - 1;
inline ostream& operator << (int x) {
if (!x) {*++ch = '0'; return *this;}
static int S[20], *top; top = S;
while (x) {*++top = x % 10 ^ 48; x /= 10;}
for (; top != S; --top) *++ch = *top;
return *this;
}
inline ostream& operator << (const char x) {*++ch = x; return *this;}
inline ~ostream() { fwrite(buf, 1, ch - buf + 1, stdout); }
#undef M
} cout;
}
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];
const int len = n + 1 >> 1;
SQRT(A, B, len);
INV(B, D, n), clear(D + n, D + lim);
std::copy(A, A + n, C); clear(C + n, C + lim);
FFT(C), FFT(D);
for (register int i = 0; i < lim; ++i) D[i] = static_cast<long long> (C[i]) * D[i] % mod * __2 % mod;
FFT(D, 0); std::copy(D + len, D + n, B + len);
}
#undef N
}
int n, A[maxn], B[maxn];
int main() {
std::cin >> n;
for (int i = 0; i < n; ++i) std::cin >> A[i];
Poly::SQRT(A, B, n);
for (int i = 0; i < n; ++i) std::cout << B[i] << ' ';
std::cout << '\n';
return 0;
}
