题解 P4725 【模板】多项式对数函数（多项式 ln）

前置知识：求导与积分。

求导公式$$x^{a^{\prime }}=a{x}^{a-1}$$ 不定积分公式$$\int x^{a}dx={\displaystyle \frac{1}{a+1}}{x}^{a+1}$$ 然后开始推式子。$$G(x)\equiv \ln (F(x))(mod{x}^n)$$ 令$$f(x)=\text{ln}(x)$$$$G(x)\equiv f(F(x))(mod{x}^n)$$

观察发现这是个复合函数，两边求导得$$G(x{)}^{\prime }\equiv f^{\prime }(F(x))F^{\prime }(x)(mod{x}^n)$$ 由 ${\text{ln}}^{\prime }(x)=\frac{1}{x}$ 得$$G^{\prime }(x)\equiv {\displaystyle \frac{F^{\prime }(x)}{F(x)}}(mod{x}^n)$$

两边再积起来$$\therefore \int G^{\prime }(x)\equiv \int {\displaystyle \frac{F^{\prime }(x)}{F(x)}}(mod{x}^n)$$$$\therefore G(x)\equiv \int {\displaystyle \frac{F^{\prime }(x)}{F(x)}}(mod{x}^n)$$ 多项式求逆即可，时间复杂度 $O(n\log n)$。

#include <bits/stdc++.h>
using namespace std;
namespace IO {
    //read and write
} using namespace IO;
const int N = 2.7e5 + 10;
namespace Polynomial {
    const int P = 998244353, G = 3, Gi = 332748118, inv2 = 499122177;
    int lim, rev[N], a[N], b[N], c[N], d[N], e[N], f[N];
    int qpow(int n, int k) {
        int res = 1;
        for(; k; n = 1ll * n * n % P, k >>= 1)
            if(k & 1) res = 1ll * res * n % P;
        return res;
    }
    void NTT(int *f, int T) {
        for(int i = 0; i < lim; i++)
            if(i < rev[i]) 
                swap(f[i], f[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int wn = qpow(T == 1 ? G : Gi, (P - 1) / (mid << 1));
            int len = mid << 1;
            for(int i = 0; i < lim; i += (mid << 1)) {
                int w = 1;
                for(int j = 0; j < mid; j++, w = 1ll * w * wn % P) {
                    int x = f[i + j], y = 1ll * w * f[i + j + mid] % P;
                    f[i + j] = (x + y) % P;
                    f[i + j + mid] = (x - y + P) % P;
                }
            }
        }
        if(T == -1) {
            int inv = qpow(lim, P - 2);
            for(int i = 0; i < lim; i++) 
                f[i] = 1ll * f[i] * inv % P;
        }
    }
    void init(int n) {
        for(lim = 1; lim < n; lim <<= 1);
        for(int i = 0; i < lim; i++)
            rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (lim >> 1));
    } 
    void mul(int *f, int *g, int *h, int n, int m) {
        static int a[N], b[N];
        init(n + m - 1);
        memset(a, 0, lim << 2);
        memcpy(a, f, n << 2);
        memset(b, 0, lim << 2);
        memcpy(b, g, m << 2);
        NTT(a, 1), NTT(b, 1);
        for(int i = 0; i < lim; i++) 
            h[i] = 1ll * a[i] * b[i] % P;
        NTT(h, -1);
    }
    void inv(int *f, int *g, int n) {
        if(n == 1) { g[0] = qpow(f[0], P - 2); return; }
        inv(f, g, n + 1 >> 1);
        init(n << 1);
        copy(f, f + n, a);
        fill(a + n, a + lim, 0);
        NTT(a, 1), NTT(g, 1);
        for(int i = 0; i < lim; i++)
            g[i] = (2 - 1ll * a[i] * g[i] % P + P) % P * g[i] % P;
        NTT(g, -1);
        fill(g + n, g + lim, 0);
    }
    void dev(int *f, int *g, int n) {
        for(int i = 1; i < n; i++)
            g[i - 1] = 1ll * i * f[i] % P;
        g[n - 1] = 0;
    }
    void invdev(int *f, int *g, int n) {
        for(int i = n - 1; i; i--)
            g[i] = 1ll * f[i - 1] * qpow(i, P - 2) % P;
        g[0] = 0;
    }
    void ln(int *f, int *g, int n) {
        static int a[N];
        init(n << 1);
        memset(a, 0, n << 2);
        inv(f, a, n);
        dev(f, g, n);
        mul(g, a, g, n, n);
        invdev(g, g, n);
    }
}
using Polynomial::ln;
int n, f[N], g[N];
int main() {
    n = read();
    for(int i = 0; i < n; i++) f[i] = read();
    ln(f, g, n);
    for(int i = 0; i < n; i++) write(g[i]), putc(' ');
    flush();
    return 0;
}