Luogu 4238 【模板】多项式求逆

疯狂补板中。

考虑倍增实现。

假设多项式只有一个常数项，直接对它逆元就可以了。

现在假如要求$G(x)$

$$F(x)G(x) \equiv 1 (\mod x^n)$$

而我们已经求出了$H(x)$

$$F(x)H(x) \equiv 1(\mod x^{\left \lceil \frac{n}{2} \right \rceil})$$

两式相减，

$$F(x)(G(x) - H(x)) \equiv 0(\mod x^{\left \lceil \frac{n}{2} \right \rceil})$$

$F(x) \mod  x^{\left \lceil \frac{n}{2} \right \rceil}$一定不会是$0$，那么

$$G(x) - H(x) \equiv 0(\mod x^{\left \lceil \frac{n}{2} \right \rceil})$$

两边平方，

$$G(x)^2 + H(x)^2 - 2G(x)H(x) \equiv 0(\mod x^n)$$

注意到后面的模数也平方了。

因为多项式$G(x) - H(x)$次数$\in [0, \left \lceil \frac{n}{2} \right \rceil]$的项的系数全都是$0$，所以平方之后次数在$[0, n]$之间的项的系数也全都是$0$。

两边乘上$F(x)$，

$$F(x)G(x)^2 + F(x)H(x)^2 - 2F(x)G(x)H(x) \equiv G(x) + F(x)H(x)^2 - 2H(x) \equiv 0(\mod x^n)$$

就得到了

$$G(x) \equiv 2H(x) - F(x)H(x)^2(\mod x^n)$$

递归实现比较清爽，非递归的比递归的快挺多的。

时间复杂度为$O(nlogn)$。

实现的时候有两个小细节：

1、$H(x)$的长度是$\frac{n}{2}$的，$F(x)$的长度是$n$，所以$F(x)H(x)^2$的长度是$2n$。

2、递归的时候注意那个上取整符号。

Code：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
 
const int N =  << ;
 
int n;
ll f[N], g[N];
 
namespace Poly {
    const int L =  << ;
    const ll gn = ;
    const ll Mod[] = {, 998244353LL, 1004535809LL, 469762049LL};
 
    int lim, pos[L];
 
    inline ll fmul(ll x, ll y, ll P) {
        ll res = ;
        for (x %= P; y; y >>= ) {
            if (y & ) res = (res + x) % P;
            x = (x + x) % P;
        }
        return res;
    }
 
    inline ll fpow(ll x, ll y, ll P) {
        ll res = 1LL;
        for (x %= P; y > ; y >>= ) {
            if (y & ) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }
 
    inline void prework(int len) {
        int l = ;
        for (lim = ; lim < len; lim <<= , ++l);
        for (int i = ; i < lim; i++)
            pos[i] = (pos[i >> ] >> ) | ((i & ) << (l - ));
    }
 
    inline void ntt(ll *c, ll opt, ll P) {
        for (int i = ; i < lim; i++)
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = ; i < lim; i <<= ) {
            ll wn = fpow(gn, (P - ) / (i << ), P);
            if (opt == -) wn = fpow(wn, P - , P);
            for (int len = i << , j = ; j < lim; j += len) {
                ll w = ;
                for (int k = ; k < i; k++, w = w * wn % P) {
                    ll x = c[j + k], y = c[j + k + i] * w % P;
                    c[j + k] = (x + y) % P, c[j + k + i]  = (x - y + P) % P;
                }
            }
        }
 
        if (opt == -) {
            ll inv = fpow(lim, P - , P);
            for (int i = ; i < lim; i++) c[i] = c[i] * inv % P;
        }
    }
 
/*    inline ll get(int k, ll P) {
        ll M = Mod[1] * Mod[2];
        ll t1 = fmul(Mod[2] * ans[1][k] % M, fpow(Mod[1], Mod[2] - 2, Mod[2]), M);
        ll t2 = fmul(Mod[1] * ans[2][k] % M, fpow(Mod[2], Mod[1] - 2, Mod[1]), M);
        ll t = (t1 + t2) % M;
        ll res = (ans[3][k] - t % Mod[3] + Mod[3]) % Mod[3];
        res = res * fpow(M, Mod[3] - 2, Mod[3]) % Mod[3];
        res = ((res % P) * (M % P) % P + (t % P)) % P;
        return res;
    }   */
 
    ll f[L], g[L];
    void inv(ll *a, ll *b, int len, ll P) {
        if (len == ) {
            b[] = fpow(a[], P - , P);
            return;
        }
        inv(a, b, (len + ) >> , P);
 
        prework(len << );
        for (int i = ; i < lim; i++) f[i] = g[i] = ;
        for (int i = ; i < len; i++) f[i] = a[i], g[i] = b[i];
        ntt(f, , P), ntt(g, , P);
        for (int i = ; i < lim; i++) g[i] = g[i] * (2LL - f[i] * g[i] % P + P) % P;
        ntt(g, -, P);
        for (int i = ; i < len; i++) b[i] = g[i];
    }
 
};
 
template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for (; ch > ''|| ch < ''; ch = getchar())
        if (ch == '-') op = -;
    for (; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}
 
int main() {
    read(n);
    for (int i = ; i < n; i++) read(f[i]);
    Poly :: inv(f, g, n, Poly :: Mod[]);
    for (int i = ; i < n; i++)
        printf("%lld%c", g[i], i == (n - ) ? '\n' : ' ');
    return ;
}

递归版

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;
 
const int N = 3e5 + ;
const ll P = 998244353LL;
 
int n, lim, pos[N];
ll a[N], f[][N], tmp[N];
 
template <typename T>
inline void read(T &X) {
    X = ; char ch = ; T op = ;
    for (; ch > '' || ch < ''; ch = getchar())
        if (ch == '-') op = -;
    for (; ch >= '' && ch <= ''; ch = getchar())
        X = (X << ) + (X << ) + ch - ;
    X *= op;
}
 
template <typename T>
inline void swap(T &x, T &y) {
    T t = x; x = y; y = t;
}
 
inline ll fpow(ll x, ll y) {
    ll res = 1LL;
    for (; y > ; y >>= ) {
        if (y & ) res = res * x % P;
        x = x * x % P;
    }
    return res;
}
 
inline void ntt(ll *c, int opt) {
    for (int i = ; i < lim; i++)
        if(i < pos[i]) swap(c[i], c[pos[i]]);
    for (int i = ; i < lim; i <<= ) {
        ll wn = fpow(, (P - ) / (i << ));
        if(opt == -) wn = fpow(wn, P - );
        for (int len = i << , j = ; j < lim; j += len) {
            ll w = ;
            for (int k = ; k < i; k++, w = w * wn % P) {
                ll x = c[j + k], y = c[j + k + i] * w % P;
                c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
            }
        }
    }
 
    if (opt == -) {
        ll inv = fpow(lim, P - );
        for (int i = ; i < lim; i++) c[i] = c[i] * inv % P;
    }
}
 
int main() {
    read(n);
    for (int i = ; i < n; i++) read(a[i]);
 
    f[][] = fpow(a[], P - );
    int dep = ;
    for (int len = ; len < n; len <<= , ++dep) {
        lim = len << ;
        for (int i = ; i < lim; i++) tmp[i] = a[i];
 
        lim <<= ;
        for (int i = ; i < lim; i++) pos[i] = (pos[i >> ] >> ) | ((i & ) << dep);
        for (int i = (len << ); i < lim; i++) tmp[i] = ;
 
        int now = dep & , pre = (dep - ) & ;
        ntt(f[pre], ), ntt(tmp, );
        for (int i = ; i < lim; i++)
            f[now][i] = (2LL * f[pre][i] % P - tmp[i] * f[pre][i] % P * f[pre][i] % P + P) % P;
        ntt(f[now], -);
 
        for (int i = (len << ); i < lim; i++) f[now][i] = ;
    }
 
    --dep;
    for (int i = ; i < n; i++)
        printf("%lld%c", f[dep & ][i], i == (n - ) ? '\n' : ' ');
 
    return ;
}

非递归版