【Luogu4238】多项式求逆模板题

Description

给定一个多项式 $F(x)$，请求出一个多项式 $G(x)$ ， 满足 $F(x) * G(x) \equiv 1 \pmod {x^n}$ 。系数对 $998244353$ 取模。

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Solution

设$f_i$满足$F(x)*f_i(x)\equiv 1\pmod {x^{2^i}}$
那么有：$F(x)*f_i(x)\equiv F(x)*f_{i+1}(x)\pmod {x^{2^i}}$
消去$F(x)$后平方：$[f_i(x) - f_{i+1}(x)]^2\equiv 0 \pmod {x^{2^{i+1}}}$
拆开后乘上$F(x)$：$F(x)*f_i^2(x)+f_{i+1}(x)-2f_i(x)\equiv 0$
然后就可以倍增了～

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Code

/************************************************
 * Au: Hany01
 * Date: Jun 10th, 2018
 * Prob: [Luogu4239] 多项式求逆 加强版
 * Email: [email protected]
************************************************/

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (998244353)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaia

template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }

inline int read()
{
    register int _, __; register char c_;
    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;
    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);
    return _ * __;
}

const int maxn = 400000, g0 = 3;

int n, N, rev[maxn], cnt;
LL a[maxn], b[maxn], A[maxn], B[maxn], powg[maxn], invg[maxn];

inline LL Pow(LL a, LL b)
{
    LL Ans = 1;
    for ( ; b; b >>= 1, (a *= a) %= Mod) if (b & 1) (Ans *= a) %= Mod;
    return Ans;
}

inline void init(int n) {
    int invg0 = Pow(g0, Mod - 2);
    for (int i = 1; i <= n; i <<= 1) powg[i] = Pow(g0, (Mod - 1) / i), invg[i] = Pow(invg0, (Mod - 1) / i);
}

inline void getrev(int n, int cnt) {
    rep(i, n) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (cnt - 1));
}

inline void NTT(LL *a, int n, int type)
{
    rep(i, n) if (rev[i] < i) swap(a[i], a[rev[i]]);
    for (int i = 2; i <= n; i <<= 1) {
        LL wn = type ? powg[i] : invg[i];
        for (int j = 0; j < n; j += i) {
            LL w = 1;
            rep(k, i >> 1) {
                LL x = a[j + k], y = a[j + k + (i >> 1)] * w % Mod;
                a[j + k] = (x + y) % Mod, a[j + k + (i >> 1)] = (x - y + Mod) % Mod;
                (w *= wn) %= Mod;
            }
        }
    }
    if (!type) {
        LL invn = Pow(n, Mod - 2);
        rep(i, n) (a[i] *= invn) %= Mod;
    }
}

inline void inverse(LL *a, LL *b, int cnt)
{
    if (!cnt) { b[0] = Pow(a[0], Mod - 2); return ; }
    int len = 1 << cnt;
    inverse(a, b, cnt - 1);
    int N = len << 1;
    getrev(N, cnt + 1);
    rep(i, len) A[i] = a[i], B[i] = b[i];
    NTT(A, N, 1), NTT(B, N, 1);
    rep(i, N) A[i] = A[i] * B[i] % Mod * B[i] % Mod;
    NTT(A, N, 0);
    rep(i, len) b[i] = (b[i] * 2 + Mod - A[i]) % Mod;
}

int main()
{
#ifdef hany01
    File("polyinv");
#endif

    n = read();
    rep(i, n) a[i] = read();
    for (N = 1; N < n; N <<= 1, ++ cnt);
    init(N << 1), inverse(a, b, cnt);
    rep(i, n) printf("%lld ", b[i]);

    return 0;
}
//劳歌一曲解行舟，红叶青山水急流。
//    -- 许浑《谢亭送别》