【笔记】从零开始的多项式学习
【模板】多项式乘法
板子，基础。
#define N 2097160
inline int ck(int x){return x >= P ? x - P : x; }
int t, rev[N];
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
void mul(int *u,int *v,int *w,int n,int m){
t = 1; while(t <= n + m)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
ntt(u, 1), ntt(v, 1);
rep(i, 0, t)w[i] = u[i] * (LL)v[i] % P;
ntt(w, 0); LL iv = Pow(t, P - 2);
rep(i, 0, t)w[i] = w[i] * iv % P;
}
int n, m, u[N], v[N], w[N];
int main() {
read(n, m);
rep(i, 0, n)read(u[i]);
rep(i, 0, m)read(v[i]);
mul(u, v, w, n, m);
rep(i, 0, n + m)printf("%d ", ck(w[i] + P)); el;
return 0;
}
【模板】多项式求逆
倍增解决。
#define N 270005
int rev[N], t;
inline int ck(int x){return x >= P ? x - P : x; }
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
int p[N];
void solve(int *u,int *v,int n){
if(1 == n){ v[0] = Pow(u[0], P - 2); return ;}
int m = (n + 1) / 2;
solve(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
rep(i, 0, t - 1)p[i] = 0;
rep(i, 0, n - 1)p[i] = u[i];
ntt(p, 1), ntt(v, 1);
rep(i, 0, t - 1)v[i] = v[i] * (2 - p[i] * (LL)v[i] % P + P) % P;
ntt(v, 0); LL iv = Pow(t, P - 2);
rep(i, 0, n - 1)v[i] = v[i] * iv % P;
rep(i, n, t - 1)v[i] = 0;
}
int n, a[N], b[N];
int main() {
read(n);
rep(i, 0, n - 1)read(a[i]);
solve(a, b, n);
rep(i, 0, n - 1)printf("%d ", b[i]);
return 0;
}
【模板】多项式对数函数
求 \(\ln(F(x))\)，先求导再积分，得到 \(\ln(F(x)) = \int \dfrac{F'(x)}{F(x)}\)，套一个多项式求逆板子即可。
#define N 270005
int rev[N], t;
inline int ck(int x){return x >= P ? x - P : x; }
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
void mul(int *u,int *v,int *w,int lm){
t = 1; while(t <= lm)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
ntt(u, 1), ntt(v, 1);
rep(i, 0, t)w[i] = u[i] * (LL)v[i] % P;
ntt(w, 0); LL iv = Pow(t, P - 2);
rep(i, 0, t)w[i] = w[i] * iv % P;
}
int p[N];
void solve(int *u,int *v,int n){
if(1 == n){ v[0] = Pow(u[0], P - 2); return ;}
int m = (n + 1) / 2;
solve(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
rep(i, 0, t - 1)p[i] = 0;
rep(i, 0, n - 1)p[i] = u[i];
ntt(p, 1), ntt(v, 1);
rep(i, 0, t - 1)v[i] = v[i] * (2 - p[i] * (LL)v[i] % P + P) % P;
ntt(v, 0); LL iv = Pow(t, P - 2);
rep(i, 0, n - 1)v[i] = v[i] * iv % P;
rep(i, n, t - 1)v[i] = 0;
}
int n, a[N], b[N], c[N], g[N];
int main() {
read(n);
rep(i, 0, n - 1)read(a[i]);
solve(a, b, n);
rp(i, n - 1)c[i - 1] = a[i] * (LL)i % P;
mul(b, c, g, n + n - 2);
pc('0'), pc(' ');
rp(i, n - 1)printf("%lld ", g[i - 1] * (LL)Pow(i, P - 2) % P);
return 0;
}
泰勒展开
对于可导函数 \(f\)，有
\[f(x) = \sum\limits_{i \ge 0}\frac{f^{(i)}(x_0)}{i!}(x - x_0)^i
\]
牛顿迭代
已知多项式 \(g(x)\)，和条件：
\[g(f(x))\equiv 0\pmod {x^n}
\]
另 \(m = \left\lceil\frac{n + 1}{2}\right\rceil\)，我们求出在 \(\bmod x^m\) 意义下的解 \(f_0(x)\)，然后再 \(f_0(x)\) 处泰勒展开。由于 \(f(x)\) 和 \(f_0(x)\) 最低 \(m\) 项是相同的，所以在 \(i\ge 2\) 时 \((f(x) - f_0(x))^i \equiv 0\pmod{x^n}\)。所以我们得到：
\[f(x)\equiv f_0(x) - \dfrac{g(f_0(x))}{g'(f_0(x))}\pmod{x^n}
\]
【模板】多项式指数函数
求 \(G(x) = \exp(F(x))\)，构造函数 \(g(f(x)) = \ln f(x) - F(x)\)，有 \(g(G(x)) \equiv0\)，直接牛顿迭代得到 \(G(x)\equiv G_0(x) - \dfrac{\ln(G_0(x)) - F(x)}{1/(G_0(x))}\)，化简后得到 ：
\[G(x) = G_0(x)(1-\ln G_0(x) + F(x))
\]
倍增即可，套一个多项式 ln 的板子。
#define N 270005
int rev[N], t;
inline int ck(int x){return x >= P ? x - P : x; }
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
void mul(int *u,int *v,int *w,int lm){
t = 1; while(t <= lm)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
ntt(u, 1), ntt(v, 1);
rep(i, 0, t)w[i] = u[i] * (LL)v[i] % P;
ntt(w, 0); LL iv = Pow(t, P - 2);
rep(i, 0, t)w[i] = w[i] * iv % P;
}
int p[N];
void solve(int *u,int *v,int n){
if(1 == n){ v[0] = Pow(u[0], P - 2); return ;}
int m = (n + 1) / 2;
solve(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
rep(i, 0, t - 1)p[i] = 0;
rep(i, 0, n - 1)p[i] = u[i];
ntt(p, 1), ntt(v, 1);
rep(i, 0, t - 1)v[i] = v[i] * (2 - p[i] * (LL)v[i] % P + P) % P;
ntt(v, 0); LL iv = Pow(t, P - 2);
rep(i, 0, n - 1)v[i] = v[i] * iv % P;
rep(i, n, t - 1)v[i] = 0;
}
int b[N], c[N], iv[N];
void polyln(int *a,int *g,int n){
t = 1; while(t <= n + n)t <<= 1;
rep(i, 0, t - 1)b[i] = c[i] = 0;
solve(a, b, n);
rp(i, n - 1)c[i - 1] = a[i] * (LL)i % P;
mul(b, c, g, n + n - 2);
pr(i, n - 1)g[i] = g[i - 1] * (LL)iv[i] % P;
g[0] = 0;
}
int w[N], f[N];
void polyexp(int *a,int *u,int n){
if(1 == n){u[0] = 1; return;}
int m = (n + 1) / 2;
polyexp(a, u, m), polyln(u, f, n);
rep(i, 0, n - 1)f[i] = ck(a[i] + (i == 0) - f[i] + P);
mul(f, u, w, n + n - 2);
rep(i, 0, n - 1)u[i] = w[i];
rep(i, n, t - 1)u[i] = 0;
}
int n, a[N], g[N];
int main() {
read(n);
rep(i, 0, n - 1)read(a[i]);
iv[1] = 1; rep(i, 2, n)iv[i] = P - (P / i) * (LL)iv[P % i] % P;
polyexp(a, g, n);
rep(i, 0, n - 1)printf("%d ", g[i]);
return 0;
}
【模板】多项式快速幂
求 \(F^k(x)\)，取多项式 \(\ln\) 得到 \(\ln F^k(x) = k\ln F(x)\)，然后再 \(\exp\) 回去即可。
#define N 270005
int rev[N], t;
inline int ck(int x){return x >= P ? x - P : x; }
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
void mul(int *u,int *v,int *w,int lm){
t = 1; while(t <= lm)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
ntt(u, 1), ntt(v, 1);
rep(i, 0, t)w[i] = u[i] * (LL)v[i] % P;
ntt(w, 0); LL iv = Pow(t, P - 2);
rep(i, 0, t)w[i] = w[i] * iv % P;
}
int p[N];
void solve(int *u,int *v,int n){
if(1 == n){ v[0] = Pow(u[0], P - 2); return ;}
int m = (n + 1) / 2;
solve(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
rep(i, 0, t - 1)p[i] = 0;
rep(i, 0, n - 1)p[i] = u[i];
ntt(p, 1), ntt(v, 1);
rep(i, 0, t - 1)v[i] = v[i] * (2 - p[i] * (LL)v[i] % P + P) % P;
ntt(v, 0); LL iv = Pow(t, P - 2);
rep(i, 0, n - 1)v[i] = v[i] * iv % P;
rep(i, n, t - 1)v[i] = 0;
}
int b[N], c[N], iv[N];
void polyln(int *a,int *g,int n){
t = 1; while(t <= n + n)t <<= 1;
rep(i, 0, t - 1)b[i] = c[i] = 0;
solve(a, b, n);
rp(i, n - 1)c[i - 1] = a[i] * (LL)i % P;
mul(b, c, g, n + n - 2);
pr(i, n - 1)g[i] = g[i - 1] * (LL)iv[i] % P;
g[0] = 0;
}
int w[N], f[N];
void polyexp(int *a,int *u,int n){
if(1 == n){u[0] = 1; return;}
int m = (n + 1) / 2;
polyexp(a, u, m), polyln(u, f, n);
rep(i, 0, n - 1)f[i] = ck(a[i] + (i == 0) - f[i] + P);
mul(f, u, w, n + n - 2);
rep(i, 0, n - 1)u[i] = w[i];
rep(i, n, t - 1)u[i] = 0;
}
int n, k, a[N], g[N], h[N];
int main() {
read(n, k);
rep(i, 0, n - 1)read(a[i]);
iv[1] = 1; rep(i, 2, n)iv[i] = P - (P / i) * (LL)iv[P % i] % P;
polyln(a, h, n);
rep(i, 0, n - 1)h[i] = h[i] * (LL)k % P;
polyexp(h, g, n);
rep(i, 0, n - 1)printf("%d ", g[i]);
return 0;
}
【模板】多项式开根
构造 \(g(f(x)) = f^2(x) - A(x)\)，有 \(g(B(x)) \equiv 0\)，直接牛顿迭代即可，套一个多项式求逆的板子。
\[B(x) = \dfrac{B_0^2(x) + A(x)}{2B_0(x)}
\]
#define N 270005
int rev[N], t;
inline int ck(int x){return x >= P ? x - P : x; }
void ntt(int *a,int op){
rp(i, t - 1)if(rev[i] < i)swap(a[i], a[rev[i]]);
for(int k = 2, l = 1; k <= t; k <<= 1, l <<= 1){
int g = Pow(op ? 3 : Q, (P - 1) / k);
for(int i = 0; i < t; i += k){
LL cur = 1;
rep(j, 0, l - 1){
int x = a[i + j], y = a[i + l + j] * cur % P;
a[i + j] = ck(x + y), a[i + l + j] = ck(x - y + P);
cur = cur * g % P;
}
}
}
}
void mul(int *u,int *v,int *w,int lm){
t = 1; while(t <= lm)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
ntt(u, 1), ntt(v, 1);
rep(i, 0, t)w[i] = u[i] * (LL)v[i] % P;
ntt(w, 0); LL iv = Pow(t, P - 2);
rep(i, 0, t)w[i] = w[i] * iv % P;
}
int p[N];
void solve(int *u,int *v,int n){
if(1 == n){ v[0] = Pow(u[0], P - 2); return ;}
int m = (n + 1) / 2;
solve(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rp(i, t - 1)rev[i] = (rev[i >> 1] >> 1) | ((i & 1) ? (t >> 1) : 0);
rep(i, 0, t - 1)p[i] = 0;
rep(i, 0, n - 1)p[i] = u[i];
ntt(p, 1), ntt(v, 1);
rep(i, 0, t - 1)v[i] = v[i] * (2 - p[i] * (LL)v[i] % P + P) % P;
ntt(v, 0); LL iv = Pow(t, P - 2);
rep(i, 0, n - 1)v[i] = v[i] * iv % P;
rep(i, n, t - 1)v[i] = 0;
}
int l[N], c[N], d[N];
void sqr(int *u,int *v,int n){
if(1 == n){v[0] = 1; return ;}
int m = (n + 1) / 2;
sqr(u, v, m), t = 1;
while(t <= n + n)t <<= 1;
rep(i, 0, t - 1)l[i] = c[i] = d[i] = 0;
rep(i, 0, n - 1)l[i] = v[i];
solve(v, c, n),
mul(l, v, d, n + n);
rep(i, 0, n - 1)ad(d[i], u[i]);
rep(i, n, t - 1)d[i] = 0;
mul(d, c, v, n + n);
rep(i, 0, n - 1)v[i] = v[i] * (LL)I % P;
rep(i, n, t - 1)v[i] = 0;
}
int n, a[N], b[N];
int main() {
read(n);
rep(i, 0, n - 1)read(a[i]);
sqr(a, b, n);
rep(i, 0, n - 1)printf("%d ", b[i]);
return 0;
}
