「多项式乘法逆」
前置芝士
基本问题
给定一个 \(n\) 次多项式 \(F(x)\)，求 \(G(x)\) 满足：
\[F(x)\times G(x)\equiv 1\mod x^n \]
假设有一个 \(0\) 次多项式 \(F(x)\)，易得 \(F(x)\) 为 \(G(x)\) 的逆元， 这给我们提供了一个分治的思路。
假设已有：
\[G'(x)\equiv F(x)\mod x^{\left \lceil \frac {n}{2} \right \rceil}
\]
- 考虑为什么要向上取整？
我们分治的思路是得到两个区间来合并到一个区间，我们要保证合并后的区间长度要大于 \(n\) 。
则有：
\[G(x)-G'(x)\equiv 0\mod x^{\left \lceil \frac {n}{2} \right \rceil}
\]
两边同时平方：
\[(G(x)-G'(x))^2\equiv 0\mod x^n
\]
\[G^2(x)-2G(x)G'(x)+G'^2(x)\equiv 0\mod x^n
\]
两边同乘 \(F(x)\)：
\[F(x)G^2(x)-2F(x)G(x)G'(x)+F(x)G'^2(x)\equiv 0\mod x^n
\]
\[\because F(x)G(x)\equiv 1\mod x^n
\]
\[G(x)-2G'(x)+F(x)G'^2(x)\equiv 0\mod x^n
\]
\[G(x)\equiv 2G'(x)-F(x)G'^2(x)\mod x^n
\]
最后递归求解即可
代码
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
const int maxn = 3e5 + 50, INF = 0x3f3f3f3f, mod = 998244353, inv3 = 332748118;
inline int read () {
register int x = 0, w = 1;
register char ch = getchar ();
for (; ch < '0' || ch > '9'; ch = getchar ()) if (ch == '-') w = -1;
for (; ch >= '0' && ch <= '9'; ch = getchar ()) x = x * 10 + ch - '0';
return x * w;
}
inline void write (register int x) {
if (x / 10) write (x / 10);
putchar (x % 10 + '0');
}
int n, f[maxn], g[maxn], rev[maxn], res[maxn], a[maxn], b[maxn];
inline int qpow (register int a, register int b, register int ans = 1) {
for (; b; b >>= 1, a = 1ll * a * a % mod)
if (b & 1) ans = 1ll * ans * a % mod;
return ans;
}
inline void NTT (register int len, register int * a, register int opt) {
for (register int i = 1; i < len; i ++) if (i < rev[i]) swap (a[i], a[rev[i]]);
for (register int d = 1; d < len; d <<= 1) {
register int w1 = qpow (opt, (mod - 1) / (d << 1));
for (register int i = 0; i < len; i += d << 1) {
register int w = 1;
for (register int j = 0; j < d; j ++, w = 1ll * w * w1 % mod) {
register int x = a[i + j], y = 1ll * w * a[i + j + d] % mod;
a[i + j] = (x + y) % mod, a[i + j + d] = (x - y + mod) % mod;
}
}
}
}
inline void Poly_Inv (register int d, register int * a, register int * b) {
if (d == 1) return b[0] = qpow (a[0], mod - 2), void (); // 长度为1,只有常数项
Poly_Inv ((d + 1) >> 1, a, b); // 向下递归
register int len = 1, bit = 0;
while (len <= d << 1) len <<= 1, bit ++;
for (register int i = 0; i < len; i ++) res[i] = 0, rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << bit - 1);
for (register int i = 0; i < d; i ++) res[i] = a[i];
NTT (len, res, 3), NTT (len, b, 3);
for (register int i = 0; i < len; i ++) b[i] = ((2ll * b[i] % mod - 1ll * res[i] * b[i] % mod * b[i] % mod) % mod + mod) % mod; // 套公式
NTT (len, b, inv3);
register int inv = qpow (len, mod - 2);
for (register int i = 0; i < d; i ++) b[i] = 1ll * b[i] * inv % mod;
for (register int i = d; i < len; i ++) b[i] = 0; // 记得清零，下次NTT要用
}
inline void P4238 () {
n = read() - 1;
for (register int i = 0; i <= n; i ++) f[i] = read(); Poly_Inv (n + 1, f, g);
for (register int i = 0; i <= n; i ++) printf ("%d ", g[i]); putchar ('\n');
}
int main () {
return P4238 (), 0;
}
