题目大意
给出 (k,a_{1,2,...,k},f_{1,2,...,k}) ，存在：
[forall n>kwedge nin mathbb{N},exists a_n=sum_{i=1}^{k}f_ka_{n-k}
]
给出 (n) ，求出 (a_n) 。
(nle 10^9,kle 32000)
思路
以下部分借鉴了 BJpers2 的题解
- 老师我会矩阵快速幂！！！
恭喜您！您获得了 (Theta(k^3log n)) 的时间复杂度以及 (0) 分的好成绩！！！
以下是正解。我们考虑从斐波拉契数列入手，我们发现一下事情，假设 (n=5)，那么我们可以得到：
[f_5=f_3+f_4=2f_3+f_2=2f_1+3f_2
]
就直接展开就好了。（这里的 (f_i) 指的是第 (i) 位斐波拉契数列，并不是上面转移式里面的值）
然后我们就发现这个过程其实就是 (x^5) 对 (x^2-x-1) 取模的过程。
一个感性的解释就是说，这个过程就相当于将一个高次抵消，换成它的展开式。最后的结果肯定就是最后的 (k) 次多项式。
考虑拓展，然后你发现根据上面的感性解释，其实它的系数就是
[x^nmod{(x^k-f_1x^{k-1}-f_2x^{k-2}-f_3x^{k-3}-...-f_kx^0)}
]
然后你发现这个东西直接多项式快速幂即可。时间复杂度 (Theta(klog klog n)) 。
( exttt{Code})
#include <bits/stdc++.h>
using namespace std;
#define SZ(x) ((int)x.size())
#define Int register int
#define mod 998244353
#define MAXN 1000005
int mul (int a,int b){return 1ll * a * b % mod;}
int dec (int a,int b){return a >= b ? a - b : a + mod - b;}
int add (int a,int b){return a + b >= mod ? a + b - mod : a + b;}
int qkpow (int a,int k){
int res = 1;for (;k;k >>= 1,a = 1ll * a * a % mod) if (k & 1) res = 1ll * res * a % mod;
return res;
}
int inv (int x){return qkpow (x,mod - 2);}
typedef vector <int> poly;
int w[MAXN],rev[MAXN];
void init_ntt (){
int lim = 1 << 18;
for (Int i = 0;i < lim;++ i) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << 17);
int Wn = qkpow (3,(mod - 1) / lim);w[lim >> 1] = 1;
for (Int i = lim / 2 + 1;i < lim;++ i) w[i] = mul (w[i - 1],Wn);
for (Int i = lim / 2 - 1;i;-- i) w[i] = w[i << 1];
}
void ntt (poly &a,int lim,int type){
#define G 3
#define Gi 332748118
static int d[MAXN];
for (Int i = 0,z = 18 - __builtin_ctz(lim);i < lim;++ i) d[rev[i] >> z] = a[i];
for (Int i = 1;i < lim;i <<= 1)
for (Int j = 0;j < lim;j += i << 1)
for (Int k = 0;k < i;++ k){
int x = mul (w[i + k],d[i + j + k]);
d[i + j + k] = dec (d[j + k],x),d[j + k] = add (d[j + k],x);
}
for (Int i = 0;i < lim;++ i) a[i] = d[i] % mod;
if (type == -1){
reverse (a.begin() + 1,a.begin() + lim);
for (Int i = 0,Inv = inv (lim);i < lim;++ i) a[i] = mul (a[i],Inv);
}
#undef G
#undef Gi
}
poly operator + (poly a,poly b){
a.resize (max (SZ (a),SZ (b)));
for (Int i = 0;i < SZ (b);++ i) a[i] = add (a[i],b[i]);
return a;
}
poly operator - (poly a,poly b){
a.resize (max (SZ (a),SZ (b)));
for (Int i = 0;i < SZ (b);++ i) a[i] = dec (a[i],b[i]);
return a;
}
poly operator * (poly a,int b){
for (Int i = 0;i < SZ (a);++ i) a[i] = mul (a[i],b);
return a;
}
poly operator * (poly a,poly b){
int d = SZ (a) + SZ (b) - 1,lim = 1;while (lim < d) lim <<= 1;
a.resize (lim),b.resize (lim);
ntt (a,lim,1),ntt (b,lim,1);
for (Int i = 0;i < lim;++ i) a[i] = mul (a[i],b[i]);
ntt (a,lim,-1),a.resize (d);
return a;
}
poly inv (poly a,int n){
poly b(1,inv (a[0])),c;
for (Int l = 4;(l >> 2) < n;l <<= 1){
c.resize (l >> 1);
for (Int i = 0;i < (l >> 1);++ i) c[i] = i < n ? a[i] : 0;
c.resize (l),b.resize (l);
ntt (c,l,1),ntt (b,l,1);
for (Int i = 0;i < l;++ i) b[i] = mul (b[i],dec (2,mul (b[i],c[i])));
ntt (b,l,-1),b.resize (l >> 1);
}
b.resize (n);
return b;
}
poly inv (poly a){return inv (a,SZ (a));}
template <typename T> inline void read (T &t){t = 0;char c = getchar();int f = 1;while (c < '0' || c > '9'){if (c == '-') f = -f;c = getchar();}while (c >= '0' && c <= '9'){t = (t << 3) + (t << 1) + c - '0';c = getchar();} t *= f;}
template <typename T,typename ... Args> inline void read (T &t,Args&... args){read (t);read (args...);}
template <typename T> inline void write (T x){if (x < 0){x = -x;putchar ('-');}if (x > 9) write (x / 10);putchar (x % 10 + '0');}
poly operator % (poly F,poly G){
int n = F.size() - 1,m = G.size() - 1;
if (n < m) return F;
poly Q;Q.resize (m + 1);
for (Int i = 0;i <= m;++ i) Q[i] = G[i];
reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),reverse (Q.begin(),Q.end()),Q.resize (n - m + 1),Q = inv (Q) * F,Q.resize (n - m + 1),reverse (Q.begin(),Q.end());
reverse (F.begin(),F.end()),reverse (G.begin(),G.end()),Q = G * Q,Q.resize (m),Q = F - Q,Q.resize (m);
return Q;
}
poly qkpow (int b,poly C){//x^B(mod C) 的答案
poly res(1,1),a;a.push_back (0),a.push_back (1);
for (;b;b >>= 1,a = (a * a) % C) if (b & 1) res = (res * a) % C;
return res;
}
int n,k,f[MAXN],a[MAXN];
poly F,G,Q;
signed main(){
init_ntt ();
read (n,k);
for (Int i = 1;i <= k;++ i) read (f[i]),f[i] = (mod + f[i] % mod) % mod;
for (Int i = 0;i < k;++ i) read (a[i]),a[i] = (mod + a[i] % mod) % mod;
G.resize (k + 1);G[k] = 1;
for (Int i = 1;i <= k;++ i) G[k - i] = mod - f[i];
Q = qkpow (n,G);
int ans = 0;for (Int i = 0;i <= k;++ i) ans = add (ans,mul (Q[i],a[i]));
write ((ans % mod + mod) % mod),putchar ('
');
return 0;
}