【模板】线性递推

求

首先暴力的矩阵乘是 的
引入概念：特征多项式

为单位矩阵
然后对于这道题来讲，转移矩阵是这样的


对第一行展开然后分别求行列式

然后发现去除第一行和某一列后主对角线都是满的，消成下三角矩阵，行列式是主对角线的乘积

然后由 定理
，一个巧妙的伪证是
设 ，即
又因为 ，所以
看看求的是什么，是向量 ，乘上 的第 k 列，令 的各项系数为
即
初始向量 乘上转移矩阵 次，即为给出的
于是
即特征多项式的系数可以用 对 多项式取模得出
复杂度
主要的思想就是不必求出 ，将 转换成
并且也不求 ，而是用初始向量去乘它，得到了我们想要的答案

#include<bits/stdc++.h>
#define N 400050
using namespace std;
typedef long long ll;
const int Mod = 998244353, G = 3;
ll add(ll a, ll b){ return a + b >= Mod ? a + b - Mod : a + b;}
ll mul(ll a, ll b){ return a * b % Mod;}
ll power(ll a, ll b){ ll ans = 1;
  for(;b;b>>=1){if(b&1) ans = mul(ans, a); a = mul(a, a);}
  return ans;
}
ll inv[N]; int n, k;
#define poly vector<ll>
#define C 20
poly w[C + 1];
int up, bit, rev[N]; ll a[N];
void Init(int len){ up = 1, bit = 0;  
  while(up < len) up <<= 1, bit++;
  for(int i = 0; i < up; i++) rev[i] = (rev[i>>1]>>1)|((i&1)<<(bit-1));
}
void prework(){
  inv[0] = inv[1] = 1;
  for(int i = 2; i <= N-50; i++) inv[i] = mul(Mod-Mod/i, inv[Mod%i]);
  for(int i = 1; i <= C; i++) w[i].resize(1<<(i-1));
  ll wn = power(G, (Mod-1)/(1<<C)); w[C][0] = 1; 
  for(int i = 1; i < (1<<(C-1)); i++) w[C][i] = mul(w[C][i-1], wn);
  for(int i = C-1; i; i--)
    for(int j = 0; j < (1<<(i-1)); j++)
      w[i][j] = w[i+1][j<<1];
}
void NTT(poly &a, int flag){
  for(int i = 0; i < up; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
  for(int i = 1, l = 1; i < up; i <<= 1, l++)
    for(int j = 0; j < up; j += (i<<1))
      for(int k = 0; k < i; k++){
        ll x = a[k + j], y = mul(w[l][k], a[k + j + i]);
        a[k + j] = add(x, y); a[k + j + i] = add(x, Mod-y);
      }
  if(flag == -1){
    reverse(a.begin() + 1, a.begin() + up);
    for(int i = 0; i < up; i++) a[i] = mul(a[i], inv[up]);
  }
}
poly operator * (poly a, poly b){
  int len = a.size() + b.size() - 1;
  Init(len); a.resize(up); b.resize(up);
  NTT(a, 1); NTT(b, 1);
  for(int i = 0; i < up; i++) a[i] = mul(a[i], b[i]); 
  NTT(a, -1); a.resize(len); return a;
}
poly Inv(poly a, int len){
  poly b(1, power(a[0], Mod - 2)), c;
  for(int lim = 4; lim < (len << 2); lim <<= 1){
    Init(lim); c = a; c.resize(lim >> 1);
    c.resize(up); b.resize(up); NTT(c, 1); NTT(b, 1);
    for(int i = 0; i < up; i++) b[i] = mul(b[i], add(2, Mod - mul(b[i], c[i])));
    NTT(b, -1); b.resize(lim >> 1);
  } b.resize(len); return b;
}
poly operator - (poly a, poly b){
  poly c; int len = max(a.size(), b.size()); c.resize(len);
  a.resize(len); b.resize(len);
  for(int i = 0; i < len; i++) c[i] = add(a[i], Mod - b[i]);
  return c;
}
poly operator / (poly a, poly b){
  int lim = 1, len = a.size() - b.size() + 1;
  reverse(a.begin(), a.end()); reverse(b.begin(), b.end());
  while(lim <= len) lim <<= 1;
  b = Inv(b, lim); b.resize(len);
  a = a * b; a.resize(len);
  reverse(a.begin(), a.end()); 
  return a;
}
poly operator % (poly a, poly b){ 
  int len = a.size() - b.size() + 1;
  if(len < 0) return a;
  poly c = a - (a / b) * b;
  c.resize(b.size() - 1); return c;
}
int main(){
  scanf("%d%d", &n, &k); prework();
  poly f(k + 1);
  for(int i = 1; i <= k; i++) scanf("%lld", &f[k - i]), f[k - i] = Mod - (f[k - i] % Mod + Mod) % Mod;
  f[k] = 1;
  for(int i = 0; i < k; i++) scanf("%lld", &a[i]), a[i] = (a[i] % Mod + Mod) % Mod;
  poly res, g;
  res.resize(k + 1); res[0] = 1;
  g.resize(k + 1); g[1] = 1;
  for(;n;n>>=1, g = g * g % f) if(n & 1) res = res * g % f;
  ll ans = 0;
  for(int i = 0; i < res.size(); i++) ans = add(ans, mul(res[i], a[i])); 
  cout << ans; 
  return 0;
}