CF1106F Lunar New Year and a Recursive Sequence

又傻掉了呢
看到连乘显然直接转原根变成线性齐次递推式。
矩阵乘法求一发。
然后分析一下发现是个x^k=m的形式。
按照套路解一下高次方程就好了。
需要用到exgcd和bsgs。

#include<iostream>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<ctime>
#include<map>
#include<cstdlib>
#include<algorithm>
#define N 120
#define L 110
#define eps 1e-7
#define inf 1e9+7
#define db double
#define ll long long
#define ldb long double
using namespace std;
inline ll read()
{
	char ch=0;
	ll x=0,flag=1;
	while(!isdigit(ch)){ch=getchar();if(ch=='-')flag=-1;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=getchar();}
	return x*flag;
}
const ll g=3;
const ll p=998244352;
const ll mo=998244353;
struct matrix 
{
	ll s[N][N];
	void clear(){memset(s,0,sizeof(s));}
};
matrix operator*(matrix a,matrix b)
{
	matrix ans;
	ans.clear();
	for(ll i=0;i<=L;i++)
	  for(ll j=0;j<=L;j++)
	    for(ll k=0;k<=L;k++)
	    ans.s[i][j]=(ans.s[i][j]+(a.s[i][k]*b.s[k][j]%p))%p;
	return ans;
}
matrix ksm(matrix x,ll k)
{
	matrix ans;
	ans.clear();
	for(ll i=0;i<=L;i++)ans.s[i][i]=1;
	while(k)
	{
		if(k&1)ans=ans*x;
		k>>=1;
		x=x*x; 
	}
	return ans;
}
matrix f;
ll qpow(ll x,ll k)
{
	ll ans=1;
	while(k)
	{
		if(k&1)ans=ans*x%mo;
		k>>=1;
		x=x*x%mo;
	}
	return ans;
}
ll X,Y;
ll exgcd(ll a,ll b)
{
	if(!b){X=1,Y=0;return a;}
	ll d=exgcd(b,a%b);
	ll t=X;X=Y,Y=t-(a/b)*Y;
	return d;
}
map<ll,ll>S;
map<ll,ll>::iterator it;
ll bsgs(ll a,ll n)
{
	S.clear();
	ll x=1,k=1,len=sqrt(mo);
	for(ll i=0;i<len;i++,x=(x*a)%mo)
	if(S.find(x)==S.end())S.insert(pair<ll,ll>{x,i});
	for(ll i=0;i<mo;i+=len,k=(k*x)%mo)
	{
		exgcd(k,mo),X=((X%mo+mo)%mo*n)%mo;
		it=S.find(X);
		if(it!=S.end())return i+(it->second);
	}
	return 0;
}
int main()
{
	ll k=read();
	f.clear();
	for(ll i=1;i<k;i++)f.s[i+1][i]=1;
	for(ll i=1;i<=k;i++)f.s[k-i+1][k]=read();
	ll n=read(),m=read();
	f=ksm(f,n-k);f.s[k][k]=(f.s[k][k]%p+p)%p;
	ll o=bsgs(g,m),d=exgcd(f.s[k][k],p);
	if(o%d==0)
	{
		X=X%p*(o/d)%p;
		printf("%lld",qpow(g,(X%p+p)%p));
	}
	else printf("-1");
	return 0;
}