多项式对数函数$\ln f(x)$

如果存在解必然有$[x^0]f(x)=1$，

对$\ln f(x)$求导，有$\frac{d\ln f(x)}{dx}\equiv \frac{f^{\prime }(x)}{f(x)}(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n)$，

$dx$乘到右边，再求积分有：
$$\begin{gathered} \int d\ln f(x)\equiv \int \frac{f^{\prime }(x))}{f(x)}dx(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n) \\ \ln f(x)\equiv \int \frac{f^{\prime }(x)}{f(x)}(\mathrm{m}\mathrm{o}\mathrm{d}{x}^n) \end{gathered}$$
然后只要对先对$f(x)$求个导，求个逆，最后求一次积分即可，整体复杂度$O(n\log n)$。

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N = 5e6 + 10, mod = 998244353, inv2 = mod + 1 >> 1;

int a[N], b[N], c[N], d[N], r[N], inv[N];

int quick_pow(int a, int n) {
   
  int ans = 1;
  while (n) {
   
    if (n & 1) {
   
      ans = 1ll * ans * a % mod;
    }
    a = 1ll *  a * a % mod;
    n >>= 1;
  }
  return ans;
}

void get_r(int lim) {
   
  for (int i = 0; i < lim; i++) {
   
    r[i] = (i & 1) * (lim >> 1) + (r[i >> 1] >> 1);
  }
}

void get_inv(int n) {
   
  inv[1] = 1;
  for (int i = 2; i < n; i++) {
   
    inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
  }
}

void NTT(int *f, int lim, int rev) {
   
  for (int i = 0; i < lim; i++) {
   
    if (i < r[i]) {
   
      swap(f[i], f[r[i]]);
    }
  }
  for (int mid = 1; mid < lim; mid <<= 1) {
   
    int wn = quick_pow(3, (mod - 1) / (mid << 1));
    for (int len = mid << 1, cur = 0; cur < lim; cur += len) {
   
      int w = 1;
      for (int k = 0; k < mid; k++, w = 1ll * w * wn % mod) {
   
        int x = f[cur + k], y = 1ll * w * f[cur + mid + k] % mod;
        f[cur + k] = (x + y) % mod, f[cur + mid + k] = (x - y + mod) % mod;
      }
    }
  }
  if (rev == -1) {
   
    int inv = quick_pow(lim, mod - 2);
    reverse(f + 1, f + lim);
    for (int i = 0; i < lim; i++) {
   
      f[i] = 1ll * f[i] * inv % mod;
    }
  }
}

void polyinv(int *a, int *b, int n) {
   
  if (n == 1) {
   
    b[0] = quick_pow(a[0], mod - 2);
    return ;
  }
  polyinv(a, b, n + 1 >> 1);
  int lim = 1;
  while (lim < 2 * n) {
   
    lim <<= 1;
  }
  get_r(lim);
  for (int i = 0; i < n; i++) {
   
    c[i] = a[i];
  }
  for (int i = n; i < lim; i++) {
   
    c[i] = 0;
  }
  NTT(b, lim, 1);
  NTT(c, lim, 1);
  for (int i = 0; i < lim; i++) {
   
    int cur = (2 - 1ll * c[i] * b[i] % mod + mod) % mod;
    b[i] = 1ll * b[i] * cur % mod;
  }
  NTT(b, lim, -1);
  for (int i = n; i < lim; i++) {
   
    b[i] = 0;
  }
}

void derivative(int *a, int *b, int n) {
   
  for (int i = 0; i < n; i++) {
   
    b[i] = 1ll * a[i + 1] * (i + 1) % mod;
  }
}

void integrate(int *a, int n) {
   
  get_inv(n);
  for (int i = n - 1; i >= 1; i--) {
   
    a[i] = 1ll * a[i - 1] * inv[i] % mod;
  }
  a[0] = 0;
}

void polyln(int *a, int *b, int n) {
   
  derivative(a, b, n);
  polyinv(a, d, n);
  int lim = 1;
  while (lim < 2 * n) {
   
    lim <<= 1;
  }
  get_r(lim);
  NTT(b, lim, 1);
  NTT(d, lim, 1);
  for (int i = 0; i < lim; i++) {
   
    b[i] = 1ll * b[i] * d[i] % mod;
  }
  NTT(b, lim, -1);
  integrate(b, n);
}

int main() {
   
  // freopen("in.txt", "r", stdin);
  // freopen("out.txt", "w", stdout);
  // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
   
    scanf("%d", &a[i]);
  }
  polyln(a, b, n);
  for (int i = 0; i < n; i++) {
   
    printf("%d%c", b[i], i + 1 == n ? '\n' : ' ');
  }
  return 0;
}