定义
若数列 \(\{a_i\}\) 满足 \(a_n=\sum_{i=1}^kf_i \times a_{n-i}\) ，则该数列为 k 阶齐次线性递推数列
可以利用多项式的知识做到 \(O(k\log k \log n)\) 求第 n 项。
如果给出前 k 项，想知道 \(f_i\) ，可以在 \(O(k^2)\) 的时间内求出。
求 \(f_i\)
有 Berlekamp Massey 算法和 Reeds Sloane 算法，具体算法思想是啥咱也不知道，咱只知道这东西放进去就能跑。
前者需要满足数域内每个非零元有乘法逆元，后者不需要。
求 \(a_n\)
由矩阵乘法谈起。假设构造的 \(k\cdot k\) 的方阵 A 为矩乘的转移矩阵，前 k 项组成的列向量为 y （从上到下依次是 \(a_0,a_1...a_{k-1}\)）。我们实际要求 \(a_n=(A^ny)_0\)。
注意到 \(a_n\) 可以写成前 k 项乘以系数的累和形式，换句话说，存在数列 \(c_i\) 使得 \(a_n=\sum_{i=0}^{k-1}c_i \times a_i\) 。我们的目标是求出 \(c_i\) 。
把 \(a_n=(A^ny)_0\) 代入 \(a_n=\sum_{i=0}^{k-1}c_ia_i\) ，得 \((A^ny)_0=\sum_{i=0}^{k-1}c_i \times a_i=\sum_{i=0}^{k-1}c_i \times (A^iy)_0\) 。
实际上可以把下角标去掉，即 \(A^ny=(\sum_{i=0}^{k-1}c_i \times A^i)y\) 。
求一下 A 的特征方程，即 \(f(\lambda)=\det |\lambda E - A| = \lambda^k-f_1\lambda^{k-1}-f_2\lambda^{k-2}-...-f_k\) 。根据 Cayley-Hamilton 定理，有 \(f(A)=0\) 。
构造多项式 \(F(x)=x^n,G(x)=x^k-f_1x^{k-1}-f_2x^{k-2}-...-f_k,H(x)=\sum_{i=0}^{k-1}c_i x^i\) 。注意到 F 是 n 阶多项式，H 是 k-1 阶多项式。由多项式的知识可知，必存在多项式 \(I(x)\) ，使得 \(F=GI+H\) 。即 \(H=F \bmod G\) 。于是我们得到了 \(c_i\) 。由 Cayley-Hamilton 定理知， \(F(A)=0I(A)+H(A)=H(A)\) ，那么 \(F(A)y=H(A)y\) ，保证了正确性。
计算 \(F \bmod G\) ，可以计算 \(x^n \bmod G\) ，即快速幂+多项式乘法+多项式取模，复杂度 \(O(k\log k \log n)\)
代码
Berlekamp Massey 算法和 Reeds Sloane 算法：
// given first m items init[0..m-1] and coefficents trans[0..m-1] or
// given first 2 *m items init[0..2m-1], it will compute trans[0..m-1]
// for you. trans[0..m] should be given as that
// init[m] = sum_{i=0}^{m-1} init[i] * trans[i]
struct LinearRecurrence
{
using int64 = long long;
using vec = std::vector<int64>;
static void extand(vec& a, size_t d, int64 value = 0)
{
if (d <= a.size()) return;
a.resize(d, value);
}
static vec BerlekampMassey(const vec& s, int64 mod)
{
std::function<int64(int64)> inverse = [&](int64 a) {
return a == 1 ? 1 : (int64)(mod - mod / a) * inverse(mod % a) % mod;
};
vec A = {1}, B = {1};
int64 b = s[0];
for (size_t i = 1, m = 1; i < s.size(); ++i, m++)
{
int64 d = 0;
for (size_t j = 0; j < A.size(); ++j)
{
d += A[j] * s[i - j] % mod;
}
if (!(d %= mod)) continue;
if (2 * (A.size() - 1) <= i)
{
auto temp = A;
extand(A, B.size() + m);
int64 coef = d * inverse(b) % mod;
for (size_t j = 0; j < B.size(); ++j)
{
A[j + m] -= coef * B[j] % mod;
if (A[j + m] < 0) A[j + m] += mod;
}
B = temp, b = d, m = 0;
}
else
{
extand(A, B.size() + m);
int64 coef = d * inverse(b) % mod;
for (size_t j = 0; j < B.size(); ++j)
{
A[j + m] -= coef * B[j] % mod;
if (A[j + m] < 0) A[j + m] += mod;
}
}
}
return A;
}
static void exgcd(int64 a, int64 b, int64& g, int64& x, int64& y)
{
if (!b)
x = 1, y = 0, g = a;
else
{
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
}
static int64 crt(const vec& c, const vec& m)
{
int n = c.size();
int64 M = 1, ans = 0;
for (int i = 0; i < n; ++i) M *= m[i];
for (int i = 0; i < n; ++i)
{
int64 x, y, g, tm = M / m[i];
exgcd(tm, m[i], g, x, y);
ans = (ans + tm * x * c[i] % M) % M;
}
return (ans + M) % M;
}
static vec ReedsSloane(const vec& s, int64 mod)
{
auto inverse = [](int64 a, int64 m) {
int64 d, x, y;
exgcd(a, m, d, x, y);
return d == 1 ? (x % m + m) % m : -1;
};
auto L = [](const vec& a, const vec& b) {
int da = (a.size() > 1 || (a.size() == 1 && a[0])) ? a.size() - 1 : -1000;
int db = (b.size() > 1 || (b.size() == 1 && b[0])) ? b.size() - 1 : -1000;
return std::max(da, db + 1);
};
auto prime_power = [&](const vec& s, int64 mod, int64 p, int64 e) {
// linear feedback shift register mod p^e, p is prime
std::vector<vec> a(e), b(e), an(e), bn(e), ao(e), bo(e);
vec t(e), u(e), r(e), to(e, 1), uo(e), pw(e + 1);
;
pw[0] = 1;
for (int i = pw[0] = 1; i <= e; ++i) pw[i] = pw[i - 1] * p;
for (int64 i = 0; i < e; ++i)
{
a[i] = {pw[i]}, an[i] = {pw[i]};
b[i] = {0}, bn[i] = {s[0] * pw[i] % mod};
t[i] = s[0] * pw[i] % mod;
if (t[i] == 0)
{
t[i] = 1, u[i] = e;
}
else
{
for (u[i] = 0; t[i] % p == 0; t[i] /= p, ++u[i])
;
}
}
for (size_t k = 1; k < s.size(); ++k)
{
for (int g = 0; g < e; ++g)
{
if (L(an[g], bn[g]) > L(a[g], b[g]))
{
ao[g] = a[e - 1 - u[g]];
bo[g] = b[e - 1 - u[g]];
to[g] = t[e - 1 - u[g]];
uo[g] = u[e - 1 - u[g]];
r[g] = k - 1;
}
}
a = an, b = bn;
for (int o = 0; o < e; ++o)
{
int64 d = 0;
for (size_t i = 0; i < a[o].size() && i <= k; ++i)
{
d = (d + a[o][i] * s[k - i]) % mod;
}
if (d == 0)
{
t[o] = 1, u[o] = e;
}
else
{
for (u[o] = 0, t[o] = d; t[o] % p == 0; t[o] /= p, ++u[o])
;
int g = e - 1 - u[o];
if (L(a[g], b[g]) == 0)
{
extand(bn[o], k + 1);
bn[o][k] = (bn[o][k] + d) % mod;
}
else
{
int64 coef = t[o] * inverse(to[g], mod) % mod * pw[u[o] - uo[g]] % mod;
int m = k - r[g];
extand(an[o], ao[g].size() + m);
extand(bn[o], bo[g].size() + m);
for (size_t i = 0; i < ao[g].size(); ++i)
{
an[o][i + m] -= coef * ao[g][i] % mod;
if (an[o][i + m] < 0) an[o][i + m] += mod;
}
while (an[o].size() && an[o].back() == 0) an[o].pop_back();
for (size_t i = 0; i < bo[g].size(); ++i)
{
bn[o][i + m] -= coef * bo[g][i] % mod;
if (bn[o][i + m] < 0) bn[o][i + m] -= mod;
}
while (bn[o].size() && bn[o].back() == 0) bn[o].pop_back();
}
}
}
}
return std::make_pair(an[0], bn[0]);
};
std::vector<std::tuple<int64, int64, int>> fac;
for (int64 i = 2; i * i <= mod; ++i)
{
if (mod % i == 0)
{
int64 cnt = 0, pw = 1;
while (mod % i == 0) mod /= i, ++cnt, pw *= i;
fac.emplace_back(pw, i, cnt);
}
}
if (mod > 1) fac.emplace_back(mod, mod, 1);
std::vector<vec> as;
size_t n = 0;
for (auto&& x : fac)
{
int64 mod, p, e;
vec a, b;
std::tie(mod, p, e) = x;
auto ss = s;
for (auto&& x : ss) x %= mod;
std::tie(a, b) = prime_power(ss, mod, p, e);
as.emplace_back(a);
n = std::max(n, a.size());
}
vec a(n), c(as.size()), m(as.size());
for (size_t i = 0; i < n; ++i)
{
for (size_t j = 0; j < as.size(); ++j)
{
m[j] = std::get<0>(fac[j]);
c[j] = i < as[j].size() ? as[j][i] : 0;
}
a[i] = crt(c, m);
}
return a;
}
LinearRecurrence(const vec& s, const vec& c, int64 mod) : init(s), trans(c), mod(mod), m(s.size()) {}
LinearRecurrence(const vec& s, int64 mod, bool is_prime = true) : mod(mod)
{
vec A;
if (is_prime)
A = BerlekampMassey(s, mod);
else
A = ReedsSloane(s, mod);
if (A.empty()) A = {0};
m = A.size() - 1;
trans.resize(m);
for (int i = 0; i < m; ++i)
{
trans[i] = (mod - A[i + 1]) % mod;
}
std::reverse(trans.begin(), trans.end());
init = {s.begin(), s.begin() + m};
}
int64 calc(int64 n)
{
if (mod == 1) return 0;
if (n < m) return init[n];
vec v(m), u(m << 1);
int msk = !!n;
for (int64 m = n; m > 1; m >>= 1) msk <<= 1;
v[0] = 1 % mod;
for (int x = 0; msk; msk >>= 1, x <<= 1)
{
std::fill_n(u.begin(), m * 2, 0);
x |= !!(n & msk);
if (x < m)
u[x] = 1 % mod;
else
{ // can be optimized by fft/ntt
for (int i = 0; i < m; ++i)
{
for (int j = 0, t = i + (x & 1); j < m; ++j, ++t)
{
u[t] = (u[t] + v[i] * v[j]) % mod;
}
}
for (int i = m * 2 - 1; i >= m; --i)
{
for (int j = 0, t = i - m; j < m; ++j, ++t)
{
u[t] = (u[t] + trans[j] * u[i]) % mod;
}
}
}
v = {u.begin(), u.begin() + m};
}
int64 ret = 0;
for (int i = 0; i < m; ++i)
{
ret = (ret + v[i] * init[i]) % mod;
}
return ret;
}
vec init, trans;
int64 mod;
int m;
};
int n, m;
const int mod = 1e9;
ll calc(ll x, ll y) {
ll z = 1;
while (y){
if (y & 1) (z *= x) %= mod;
(x *= x) %= mod, y /= 2;
}
return z;
}
inline int update(int x) {
return x < mod ? x : x - mod;
}
vector<long long>f, fm;
int main(){
//freopen("input", "r", stdin);
scanf("%d %d", &n, &m);
f.push_back(0);
f.push_back(1);
for (int i = 2; i <= m + m ; i ++){
f.push_back(update(f[i - 1] + f[i - 2]));
}
fm.push_back(0);
for (int i = 1; i <= m; i ++)
fm.push_back(update(fm[i - 1] + calc(f[i], m)));
LinearRecurrence sol(fm, mod, false);
printf("%d\n", sol.calc(n));
return 0;
}
快速求线性递推数列第 n 项：
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef long double LD;
typedef pair<int,int> pii;
typedef pair<LL,int> pli;
typedef pair<LL,LL> pll;
const int SZ = 1e6 + 10;
const int INF = 1e9 + 10;
const int mod = 998244353;
const LD eps = 1e-8;
LL read() {
LL n = 0;
char a = getchar();
bool flag = 0;
while(a > '9' || a < '0') { if(a == '-') flag = 1; a = getchar(); }
while(a <= '9' && a >= '0') { n = n * 10 + a - '0',a = getchar(); }
if(flag) n = -n;
return n;
}
LL ksm(LL a,LL b) {
LL ans = 1;
while(b) {
if(b&1) ans = a * ans % mod;
a = a * a % mod;
b >>= 1;
}
return ans;
}
struct NTTranform {
const int g = 3;
void Transform(int *a,int n,int opt) {
for(int i = 0,j = 0;i < n;i ++) {
if(i < j) swap(a[i],a[j]);
for(int k = n >> 1;(j ^= k) < k;k >>= 1);
}
for(int l = 2;l <= n;l *= 2) {
int m = l / 2;
int wn = ksm(g,(mod-1)/l);
if(opt == -1) wn = ksm(wn,mod - 2);
for(int *p = a;p != a + n;p += l) {
for(int i = 0,w = 1;i < m;i ++,w=1ll*w*wn%mod) {
int t = 1ll * w * p[m + i] % mod;
p[m + i] = (p[i] - t + mod) % mod;
(p[i] += t) %= mod;
}
}
}
}
void dft(int *a,const int n) {
Transform(a,n,1);
}
void idft(int *a,const int n) {
Transform(a,n,-1);
int t = ksm(n,mod - 2);
for(int i = 0;i < n;i ++) a[i] = 1ll * a[i] * t % mod;
}
}ntt;
void multiply(int *a,int n,int *b,int m,int *ans) { /// need 4 times memory
static int c1[SZ],c2[SZ];
int len = 1;
while(len < n + m) len *= 2;
for(int i = 0;i < len;i ++) c1[i] = c2[i] = 0;
for(int i = 0;i < n;i ++) c1[i] = a[i];
for(int i = 0;i < m;i ++) c2[i] = b[i];
ntt.dft(c1,len); ntt.dft(c2,len);
for(int i = 0;i < len;i ++) c1[i] = 1ll * c1[i] * c2[i] % mod;
ntt.idft(c1,len);
for(int i = 0;i < n + m - 1;i ++) ans[i] = (c1[i] + mod) % mod;
}
void inverse(int *a,int n,int *b) { /// need 4 times memory
static int A[SZ];
b[0] = ksm(a[0],mod-2);
for(int l = 2;l < n*2;l <<= 1) {
for(int i = 0;i < l;i ++) A[i] = a[i];
for(int i = l;i < l*2;i ++) A[i] = 0;
for(int i = l/2;i < l*2;i ++) b[i] = 0;
ntt.dft(A,l*2); ntt.dft(b,l*2);
for(int i = 0;i < l*2;i ++) b[i] = (b[i]*2-1ll*A[i]*b[i]%mod*b[i]%mod+mod)%mod;
ntt.idft(b,l*2);
}
}
void divide(int *a,int n,int *b,int m,int *D) { /// need 4 times memory
static int bni[SZ];
if(n<m) { D[0]=0; return ; }
reverse(a,a+n); reverse(b,b+m);
inverse(b,n-m+1,bni);
multiply(a,n-m+1,bni,n-m+1,D);
reverse(D,D+n-m+1);
reverse(a,a+n); reverse(b,b+m);
}
void modular(int *a,int n,int *b,int m,int *R) { /// need 4 times memory
static int D[SZ],c[SZ];
if(n<m) { for(int i = 0;i < n;i ++) R[i] = a[i]; return ; }
divide(a,n,b,m,D);
multiply(b,m,D,n-m+1,c);
for(int i = 0;i < m-1;i ++) R[i] = (a[i] - c[i] + mod) % mod;
R[m-1] = 0;
}
// f:长度为k的线性递推式，a：初始k项。求第n项
int get_nth(int f[],int a[],int k,int n) {
if(n<k) return a[n];
static int ff[SZ],G[SZ],Ans[SZ];
for(int i = 1;i <= k;i ++) ff[k-i] = (mod - f[i])%mod;
ff[k] = 1;
G[1] = 1; Ans[0] = 1;
while(n) {
if(n&1) {
multiply(Ans,k,G,k,Ans),modular(Ans,2*k-1,ff,k+1,Ans);
}
multiply(G,k,G,k,G);
modular(G,2*k-1,ff,k+1,G);
n >>= 1;
}
int ans = 0;
for(int i = 0;i < k;i ++) (ans += 1ll * a[i] * Ans[i] % mod) %= mod;
ans += mod; ans %= mod;
return ans;
}
int n, k;
int f[SZ],a[SZ];
int main(){
//freopen("testdata (4).in","r",stdin);
n = read(), k = read();
for(int i = 1;i <= k;i ++) f[i] = read();
for(int i = 0;i < k;i ++) a[i] = read();
cout << get_nth(f,a,k,n);
return 0;
}