[学习笔记] 多项式 ln & exp
前置知识
-
\[\begin{aligned} (\ln x)' &= \frac{1}{x} \\ (\exp x)' &= x \\ \end{aligned} \]
-
复合函数的求导（链式法则）
\[(g\circ f)' (x) = g(f(x))'f'(x) \] -
多项式求逆，分治FFT。
多项式 ln
设 \(g(x) = \ln{f(x)}\)，则有
\[\begin{aligned}
g'(x) = f'(x) \frac{1}{f(x)}
\end{aligned}
\]
先把 \(g'(x)\) 求出来然后再积分就行了。
多项式求导，积分都是 \(O(n)\) 的，多项式乘法为 \(O(n\log n)\)，所以总复杂度为 \(O(n\log n)\)。
多项式 exp
普通方法一
设 \(g(x) = e^{f(x)}\)，则有
\[\begin{aligned}
g'(x) = g(x)f'(x)
\end{aligned}
\]
写成卷积的形式就是（\(f_i\) 表示 \(f(x)\) 的 \(i\) 次项次数）
\[g'_i = \sum_{j = 0}^i g_j f_{i - j}
\]
然后因为 \(f'_i = (i + 1)f_{i + 1}\)，所以就有
\[\begin{aligned}
g_{i + 1} &= \frac{1}{i + 1} \sum_{j = 0}^i g_{j} (i + 1 - j) f_{i + 1 - j} \\
g_i &= \frac{1}{i} \sum_{j = 0}^{i - 1} g_{j} (i - j) f_{i - j}
\end{aligned}
\]
可以用分治 FFT 解决，时间复杂度为 \(O(n \log^2 n)\)。
牛顿迭代
时间复杂度为 \(O(n \log n)\) ，但由于实现过程中需要求 \(\ln\)，所以实际上快不了多少（至少在洛谷的模板上跑得差不多）。
代码
\(\ln\)
#include <cstdio>
#include <iostream>
using namespace std;
typedef long long ll;
const int _ = (1 << 18) + 7;
const int mod = 998244353;
const int rt = 3;
int n，f[_];
int Pw(int a，int p) {
int res = 1;
while (p) {
if (p & 1) res = (ll)res * a % mod;
a = (ll)a * a % mod;
p >>= 1;
}
return res;
}
namespace POLY {
int tot，num[_]，inv[_]，pwrt[2][_]，tmp[6][_];
void Init() {
tot = 1; while (tot <= n + n) tot <<= 1;
inv[1] = 1;
for (int i = 2; i <= tot; ++i) inv[i] = (ll)inv[mod % i] * (mod - mod / i) % mod;
pwrt[0][tot] = Pw(rt，(mod - 1) / tot);
pwrt[1][tot] = Pw(pwrt[0][tot]，mod - 2);
for (int len = (tot >> 1); len; len >>= 1) {
pwrt[0][len] = (ll)pwrt[0][len << 1] * pwrt[0][len << 1] % mod;
pwrt[1][len] = (ll)pwrt[1][len << 1] * pwrt[1][len << 1] % mod;
}
}
void NTT(int *f，int t，bool ty) {
for (int i = 1; i < t; ++i) {
num[i] = (num[i >> 1] >> 1) | ((i & 1) ? t >> 1 : 0);
if (i < num[i]) swap(f[i]，f[num[i]]);
}
for (int len = 2; len <= t; len <<= 1) {
int gap = len >> 1，w1 = pwrt[ty][len]，w，tmp;
for (int i = 0; i < t; i += len) {
w = 1;
for (int j = i; j < i + gap; ++j) {
tmp = (ll)w * f[j + gap] % mod;
f[j + gap] = (f[j] - tmp + mod) % mod;
f[j] = (f[j] + tmp) % mod;
w = (ll)w * w1 % mod;
}
}
}
if (ty) for (int i = 0; i < t; ++i) f[i] = (ll)f[i] * inv[t] % mod;
}
void Mul(int *f，int *g，int *h) {
for (int i = 0; i < tot; ++i) tmp[2][i] = f[i]，tmp[3][i] = g[i];
NTT(tmp[2]，tot，0)，NTT(tmp[3]，tot，0);
for (int i = 0; i < tot; ++i) h[i] = (ll)tmp[2][i] * tmp[3][i] % mod;
NTT(h，tot，1);
}
void Inv(int *f，int *h) {
for (int i = 0; i < tot; ++i) h[i] = tmp[1][i] = 0;
h[0] = Pw(f[0]，mod - 2)，tmp[1][0] = f[0]，tmp[1][1] = f[1];
for (int len = 2，t = 4; len < tot; len <<= 1，t = (len << 1)) {
NTT(h，t，0)，NTT(tmp[1]，t，0);
for (int i = 0; i < t; ++i) h[i] = (ll)h[i] * (2 - (ll)h[i] * tmp[1][i] % mod + mod) % mod;
NTT(h，t，1)，NTT(tmp[1]，t，1);
for (int i = len; i < t; ++i) tmp[1][i] = f[i]，h[i] = 0;
}
}
void Deriv(int *f，int *h) { for (int i = 0; i < tot - 1; ++i) h[i] = (ll)f[i + 1] * (i + 1) % mod; }
void Integ(int *f，int *h) { for (int i = tot - 1; i > 0; --i) h[i] = (ll)f[i - 1] * Pw(i，mod - 2) % mod; h[0] = 0; }
void Ln(int *f，int *h) {
for (int i = 0; i < tot; ++i) tmp[4][i] = f[i];
Inv(f，tmp[4]);
Deriv(f，f);
Mul(f，tmp[4]，f);
Integ(f，h);
}
}
int gi() {
int x = 0; char c = getchar();
while (!isdigit(c)) c = getchar();
while (isdigit(c)) x = (x << 3) + (x << 1) + c - '0'，c = getchar();
return x;
}
int main() {
n = gi();
for (int i = 0; i < n; ++i) f[i] = gi();
POLY::Init();
POLY::Ln(f，f);
for (int i = 0; i < n; ++i) printf("%d "，f[i]); putchar('\n');
return 0;
}
\(exp\)（普通方法）
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int _ = (1 << 18) + 7;
const int mod = 998244353，rt = 3;
int n，g[_]，f[_];
int Pw(int a，int p) {
int res = 1;
while (p) {
if (p & 1) res = (ll)res * a % mod;
a = (ll)a * a % mod;
p >>= 1;
}
return res;
}
namespace POLY {
int tot，num[_]，pwrt[2][_]，inv[_]，tmp[5][_];
void Init() {
tot = 1; while (tot <= n + n) tot <<= 1;
inv[1] = 1;
for (int i = 2; i <= tot; ++i) inv[i] = (ll)inv[mod % i] * (mod - mod / i) % mod;
pwrt[0][tot] = Pw(rt，(mod - 1) / tot);
pwrt[1][tot] = Pw(pwrt[0][tot]，mod - 2);
for (int len = (tot >> 1); len; len >>= 1) {
pwrt[0][len] = (ll)pwrt[0][len << 1] * pwrt[0][len << 1] % mod;
pwrt[1][len] = (ll)pwrt[1][len << 1] * pwrt[1][len << 1] % mod;
}
}
void NTT(int *f，int t，bool ty) {
for (int i = 1; i < t; ++i) {
num[i] = (num[i >> 1] >> 1) | ((i & 1) ? t >> 1 : 0);
if (i < num[i]) swap(f[i]，f[num[i]]);
}
for (int len = 2; len <= t; len <<= 1) {
int gap = len >> 1，w1 = pwrt[ty][len];
for (int i = 0，w = 1，tmp; i < t; i += len，w = 1)
for (int j = i; j < i + gap; ++j) {
tmp = (ll)w * f[j + gap] % mod;
f[j + gap] = (f[j] - tmp + mod) % mod;
f[j] = (f[j] + tmp) % mod;
w = (ll)w * w1 % mod;
}
}
if (ty) for (int i = 0; i < t; ++i) f[i] = (ll)f[i] * inv[t] % mod;
}
void Mul(int *f，int *g，int *h，int t) {
memcpy(tmp[1]，f，t << 2);
memcpy(tmp[2]，g，t << 2);
NTT(tmp[1]，t，0)，NTT(tmp[2]，t，0);
for (int i = 0; i < (t << 1); ++i) h[i] = (ll)tmp[1][i] * tmp[2][i] % mod;
NTT(h，t，1);
}
void dcNTT(int *f，int *g，int t，int l，int r) {
if (t == 1) { f[0] = l ? (ll)f[0] * inv[l] % mod : f[0]; return; }
dcNTT(f，g，t >> 1，l，(l + r) >> 1);
memset(tmp[0] + (t >> 1)，0，t << 1);
memcpy(tmp[0]，f，t << 1);
Mul(tmp[0]，g，tmp[0]，t);
for (int i = (t >> 1); i < t; ++i) f[i] = (f[i] + tmp[0][i - 1]) % mod;
dcNTT(f + (t >> 1)，g，t >> 1，(l + r) >> 1，r);
}
void Exp(int *f，int *g) { dcNTT(f，g，tot >> 1，1，tot >> 1); }
void Deriv(int *f，int *h) { for (int i = 0; i < tot - 1; ++i) h[i] = (ll)f[i + 1] * (i + 1) % mod; }
}
int main() {
scanf("%d"，&n);
for (int i = 0; i < n; ++i) scanf("%d"，&g[i]);
POLY::Init();
POLY::Deriv(g，g);
f[0] = 1;
POLY::Exp(f，g);
for (int i = 0; i < n; ++i) printf("%d "，f[i]); putchar('\n');
return 0;
}
\(exp\) （牛顿迭代）
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
const int _ = (1 << 18) + 7;
const int mod = 998244353，rt = 3;
int n，f[_]，g[_];
int Pw(int a，int p) {
int res = 1;
while (p) {
if (p & 1) res = (ll)res * a % mod;
a = (ll)a * a % mod;
p >>= 1;
}
return res;
}
namespace POLY {
int tot，num[_]，pwrt[2][_]，inv[_];
void Init() {
tot = 1; while (tot <= n + n) tot <<= 1;
inv[1] = 1; for (int i = 2; i <= tot; ++i) inv[i] = (ll)inv[mod % i] * (mod - mod / i) % mod;
pwrt[0][tot] = Pw(rt，(mod - 1) / tot);
pwrt[1][tot] = Pw(pwrt[0][tot]，mod - 2);
for (int len = (tot >> 1); len; len >>= 1) {
pwrt[0][len] = (ll)pwrt[0][len << 1] * pwrt[0][len << 1] % mod;
pwrt[1][len] = (ll)pwrt[1][len << 1] * pwrt[1][len << 1] % mod;
}
}
void Clear(int *f，int L) { memset(f，0，L << 3); }
void NTT(int *f，int L，bool ty) {
for (int i = 1; i < L; ++i) {
num[i] = (num[i >> 1] >> 1) | ((i & 1) ? L >> 1 : 0);
if (i < num[i]) swap(f[i]，f[num[i]]);
}
for (int len = 2; len <= L; len <<= 1) {
int gap = len >> 1，w1 = pwrt[ty][len];
for (int i = 0，w = 1，tmp; i < L; i += len，w = 1)
for (int j = i; j < i + gap; ++j) {
tmp = (ll)w * f[j + gap] % mod;
f[j + gap] = (f[j] - tmp + mod) % mod;
f[j] = (f[j] + tmp) % mod;
w = (ll)w * w1 % mod;
}
}
if (ty) for (int i = 0; i < L; ++i) f[i] = (ll)f[i] * inv[L] % mod;
}
void Cpy(int *h，int *f，int L) { memcpy(h，f，L << 2); }
void Inv(int *h，int *f，int L) {
int a[_]，b[_];
Clear(h，L)，Clear(a，L)，Clear(b，L);
h[0] = Pw(f[0]，mod - 2);
for (int len = 2，t = 4; len <= L; len <<= 1，t <<= 1) {
Cpy(a，f，len)，Cpy(b，h，len)，NTT(b，t，0)，NTT(a，t，0);
for (int i = 0; i < t; ++i) b[i] = (ll)b[i] * (2 - (ll)a[i] * b[i] % mod + mod) % mod;
NTT(b，t，1);
for (int i = (len >> 1); i < len; ++i) h[i] = b[i];
}
}
void Deriv(int *h，int *f，int L) { for (int i = 0; i < L - 1; ++i) h[i] = (ll)f[i + 1] * (i + 1) % mod; }
void Integ(int *h，int *f，int L) { for (int i = L - 1; i; --i) h[i] = (ll)f[i - 1] * inv[i] % mod; h[0] = 0; }
void Ln(int *h，int *f，int L) {
int a[_]，b[_];
Clear(h，L)，Clear(a，L)，Clear(b，L);
Deriv(a，f，L)，Inv(b，f，L);
NTT(a，L << 1，0)，NTT(b，L << 1，0);
for (int i = 0; i < (L << 1); ++i) h[i] = (ll)a[i] * b[i] % mod;
NTT(h，L << 1，1);
Integ(h，h，L);
}
void Exp(int *h，int *f，int L) {
int a[_]，b[_]，c[_];
Clear(h，L)，Clear(a，L)，Clear(b，L)，Clear(c，L);
h[0] = 1，a[0] = f[0]，a[1] = f[1];
for (int len = 2，t = 4; len <= L; len <<= 1，t <<= 1) {
Cpy(c，h，len)，Ln(b，h，len)，Cpy(a，f，len);
NTT(c，len，0)，NTT(b，len，0)，NTT(a，len，0);
for (int i = 0; i < len; ++i) c[i] = (ll)c[i] * (1ll - b[i] + a[i] + mod) % mod;
NTT(c，len，1);
for (int i = (len >> 1); i < len; ++i) h[i] = c[i];
}
}
}
int main() {
scanf("%d"，&n);
for (int i = 0; i < n; ++i) scanf("%d"，&g[i]);
POLY::Init();
POLY::Exp(f，g，POLY::tot >> 1);
for (int i = 0; i < n; ++i) printf("%d "，f[i]); putchar('\n');
return 0;
}
