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| // Problem: P5205 【模板】多项式开根
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P5205
// Memory Limit: 125 MB
// Time Limit: 1000 ms
// Author: Ybw051114
//
// Powered by CP Editor (https://cpeditor.org)
#include "ybwhead/ios.h"
const int maxn = 400000 + 10;
const int mod = 998244353;
const int inv2 = 499122177;
int n, a[maxn], b[maxn], A[maxn], B[maxn], C[maxn], D[maxn], r[maxn];
int fpow(int a, int b)
{
int ret = 1;
for (; b; b >>= 1, a = 1ll * a * a % mod)
if (b & 1)
ret = 1ll * ret * a % mod;
return ret;
}
void NTT(int *f, int n, int op)
{
for (int i = 0; i < n; i++)
if (i < r[i])
swap(f[i], f[r[i]]);
int buf, tmp, x, y;
for (int len = 1; len < n; len <<= 1)
{
tmp = fpow(3, (mod - 1) / (len << 1));
if (op == -1)
tmp = fpow(tmp, mod - 2);
for (int i = 0; i < n; i += len << 1)
{
buf = 1;
for (int j = 0; j < len; j++)
{
x = f[i + j];
y = 1ll * buf * f[i + j + len] % mod;
f[i + j] = (x + y) % mod;
f[i + j + len] = (x - y + mod) % mod;
buf = 1ll * buf * tmp % mod;
}
}
}
if (op == 1)
return;
int inv = fpow(n, mod - 2);
for (int i = 0; i < n; i++)
f[i] = 1ll * f[i] * inv % mod;
}
void Inv(int *a, int *b, int n)
{
b[0] = fpow(a[0], mod - 2);
int len, lim;
for (len = 1; len < (n << 1); len <<= 1)
{
lim = len << 1;
for (int i = 0; i < len; i++)
A[i] = a[i], B[i] = b[i];
for (int i = 0; i < lim; i++)
r[i] = (r[i >> 1] >> 1) | ((i & 1) ? len : 0);
NTT(A, lim, 1);
NTT(B, lim, 1);
for (int i = 0; i < lim; i++)
b[i] = ((2ll - 1ll * A[i] * B[i] % mod) * B[i] % mod + mod) % mod;
NTT(b, lim, -1);
for (int i = len; i < lim; i++)
b[i] = 0;
}
for (int i = 0; i < len; i++)
A[i] = B[i] = 0;
for (int i = n; i < len; i++)
b[i] = 0;
}
void Sqrt(int *a, int *b, int n)
{
b[0] = 1;
int *A = C, *B = D, len, lim;
for (len = 1; len < (n << 1); len <<= 1)
{
lim = len << 1;
for (int i = 0; i < len; i++)
A[i] = a[i];
Inv(b, B, lim >> 1);
for (int i = 0; i < lim; i++)
r[i] = (r[i >> 1] >> 1) | ((i & 1) ? len : 0);
NTT(A, lim, 1);
NTT(B, lim, 1);
for (int i = 0; i < lim; i++)
A[i] = 1ll * A[i] * B[i] % mod;
NTT(A, lim, -1);
for (int i = 0; i < len; i++)
b[i] = 1ll * (b[i] + A[i]) % mod * inv2 % mod;
for (int i = len; i < lim; i++)
b[i] = 0;
}
for (int i = 0; i < len; i++)
A[i] = B[i] = 0;
for (int i = n; i < len; i++)
b[i] = 0;
}
int main()
{
yin >> n;
for (int i = 0; i < n; ++i)
yin >> a[i];
Sqrt(a, b, n);
for (int i = 0; i < n; i++)
yout << b[i] << " ";
yout << endl;
return 0;
}
|