[training Day8] series [matrix multiplication]

Ideas ：

We consider constructing a transition matrix
f[1] The exponent of is multiplied by b[k],f[2] The exponent of is multiplied by b[k-1]+ The original , And so on
So we can construct
0 0 0 0 …… 0 b[k]
1 0 0 0 …… 0 b[k-1]
0 1 0 0 …… 0 b[k-2]
……
0 0 0 0…… 1 b[1]

$code$

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cctype>

#define mo 998244352

#define ll long long


using namespace std;

ll mod = 998244353;

ll n, k;
ll f[210], c[210][210];

struct matrix
{
    
	ll a[210][210];
	matrix() {
     memset(a, 0, sizeof(a)); }
}A;
matrix ans;

matrix operator *(const matrix &x,const matrix &y)
{
    
	matrix z;
	for(register int i = 0; i < k; i ++)
	 for(register int j = 0; j < k; j ++)
	  for(register int t = 0; t < k; t ++)
	   z.a[i][j] = (z.a[i][j] + x.a[i][t] * y.a[t][j] % mo) % mo;
	return z;
}

inline matrix qpow(ll x) {
    
	ans = A;
	for(; x; A = A * A, x >>= 1)
		if(x & 1) ans = ans * A;
	return ans;
}

ll qpow2(ll x, ll k) {
    
	ll ans = 1;
	while(k) {
    
		if(k & 1) ans= ans * x % mod;
		x = x * x % mod;
		k >>= 1;
	}
	return ans;
}

int main() {
    
	freopen("seq.in", "r", stdin);
	freopen("seq.out", "w", stdout);
	scanf("%lld%lld", &n, &k);
	for(ll i = k - 1; i >= 0; i --)
		scanf("%lld", &A.a[i][k - 1]);
	for(ll i = 1; i < k; i ++) A.a[i][i - 1] = 1;
	for(ll i = 0; i < k; i ++) scanf("%lld", &f[i]);
	if(n <= k) {
    
		printf("%lld", f[n - 1]);
		return 0;
	}
	n -= k;
	ans = qpow(n - 1);
	ll sum = 1;
	for(ll i = 0; i < k; i ++)
		sum = sum * qpow2(f[i], ans.a[i][k - 1]) % mod;
	printf("%lld", sum);
	return 0;
}