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$H(x)$

$$F(x)H(x) equiv 1(mod x^{left lceil frac{n}{2} ight ceil})$$

两式相减，

$$F(x)(G(x) - H(x)) equiv 0(mod x^{left lceil frac{n}{2} ight ceil})$$

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

$$G(x) - H(x) equiv 0(mod x^{left lceil frac{n}{2} ight ceil})$$

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$in [0, left lceil frac{n}{2} ight ceil]$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 = 1 << 20;

int n;
ll f[N], g[N];

namespace Poly {
    const int L = 1 << 20;
    const ll gn = 3;
    const ll Mod[4] = {0, 998244353LL, 1004535809LL, 469762049LL};
    
    int lim, pos[L];
    
    inline ll fmul(ll x, ll y, ll P) {
        ll res = 0;
        for (x %= P; y; y >>= 1) {
            if (y & 1) 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 > 0; y >>= 1) {
            if (y & 1) res = res * x % P;
            x = x * x % P;
        }
        return res;
    }
    
    inline void prework(int len) {
        int l = 0;
        for (lim = 1; lim < len; lim <<= 1, ++l);
        for (int i = 0; i < lim; i++)
            pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
    }
    
    inline void ntt(ll *c, ll opt, ll P) {
        for (int i = 0; i < lim; i++) 
            if (i < pos[i]) swap(c[i], c[pos[i]]);
        for (int i = 1; i < lim; i <<= 1) {
            ll wn = fpow(gn, (P - 1) / (i << 1), P);
            if (opt == -1) wn = fpow(wn, P - 2, P);
            for (int len = i << 1, j = 0; j < lim; j += len) {
                ll w = 1;
                for (int k = 0; 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 == -1) {
            ll inv = fpow(lim, P - 2, P);
            for (int i = 0; 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 == 1) {
            b[0] = fpow(a[0], P - 2, P);
            return;
        }
        inv(a, b, (len + 1) >> 1, P);
        
        prework(len << 1);
        for (int i = 0; i < lim; i++) f[i] = g[i] = 0;
        for (int i = 0; i < len; i++) f[i] = a[i], g[i] = b[i];
        ntt(f, 1, P), ntt(g, 1, P);
        for (int i = 0; i < lim; i++) g[i] = g[i] * (2LL - f[i] * g[i] % P + P) % P;
        ntt(g, -1, P);
        for (int i = 0; i < len; i++) b[i] = g[i];
    }
    
};

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9'|| ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}

int main() {
    read(n);
    for (int i = 0; i < n; i++) read(f[i]);
    Poly :: inv(f, g, n, Poly :: Mod[1]);
    for (int i = 0; i < n; i++)
        printf("%lld%c", g[i], i == (n - 1) ? '
' : ' ');
    return 0;
}

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 3e5 + 5;
const ll P = 998244353LL;

int n, lim, pos[N];
ll a[N], f[2][N], tmp[N];

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9' || ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    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 > 0; y >>= 1) {
        if (y & 1) res = res * x % P;
        x = x * x % P;
    }
    return res;
}

inline void ntt(ll *c, int opt) {
    for (int i = 0; i < lim; i++)
        if(i < pos[i]) swap(c[i], c[pos[i]]);
    for (int i = 1; i < lim; i <<= 1) {
        ll wn = fpow(3, (P - 1) / (i << 1));
        if(opt == -1) wn = fpow(wn, P - 2);
        for (int len = i << 1, j = 0; j < lim; j += len) {
            ll w = 1;
            for (int k = 0; 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 == -1) {
        ll inv = fpow(lim, P - 2);
        for (int i = 0; i < lim; i++) c[i] = c[i] * inv % P;
    }
}

int main() {
    read(n);
    for (int i = 0; i < n; i++) read(a[i]);
    
    f[0][0] = fpow(a[0], P - 2);
    int dep = 1;
    for (int len = 1; len < n; len <<= 1, ++dep) {
        lim = len << 1;
        for (int i = 0; i < lim; i++) tmp[i] = a[i];
        
        lim <<= 1;
        for (int i = 0; i < lim; i++) pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << dep);
        for (int i = (len << 1); i < lim; i++) tmp[i] = 0;

        int now = dep & 1, pre = (dep - 1) & 1;
        ntt(f[pre], 1), ntt(tmp, 1);
        for (int i = 0; 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], -1);
        
        for (int i = (len << 1); i < lim; i++) f[now][i] = 0;
    }
    
    --dep;
    for (int i = 0; i < n; i++)
        printf("%lld%c", f[dep & 1][i], i == (n - 1) ? '
' : ' ');
    
    return 0;
}