# Codeforces1106F 【BSGS】【矩阵快速幂】【exgcd】

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll N = 1e2 + 10;
const ll Mod = 998244353;

ll add(ll a, ll b, ll mod = Mod) {
return (a += b) >= mod ? a - mod : a;
}

ll sub(ll a, ll b, ll mod = Mod) {
return (a -= b) < 0 ? a + mod : a;
}

ll mul(ll a, ll b, ll mod = Mod) {
return 1ll * a * b % mod;
}

ll fast_pow(ll a, ll b, ll mod = Mod) {
ll res = 1;
for (; b; b >>= 1, a = mul(a, a, mod))
if (b & 1) res = mul(res, a, mod);
return res;
}

ll n, m, k, b[N];

struct Matrix {
ll g[N][N];

Matrix() {
memset(g, 0, sizeof(g));
}
};

Matrix operator * (const Matrix a, const Matrix b) {
Matrix c;
for (ll i = 1; i <= k; i++)
for (ll j = 1; j <= k; j++)
for (ll p = 1; p <= k; p++)
c.g[i][j] = add(c.g[i][j], mul(a.g[i][p], b.g[p][j], Mod - 1), Mod - 1);
return c;
}

Matrix fast_pow(Matrix a, ll b) {
Matrix res;
for (ll i = 1; i <= k; i++)
res.g[i][i] = 1;
for (; b; b >>= 1, a = a * a)
if (b & 1) res = res * a;
return res;
}

ll bsgs(ll a, ll b) {
map<ll, ll> mp;
mp[b] = 0;
ll cur = 1, limit = sqrt(Mod);
for (ll i = 1; i <= limit; i++) {
cur = mul(cur, a);
mp[mul(b, fast_pow(cur, Mod - 2))] = i;
}
ll now = 1;
for (ll i = 0; i <= limit; i++) {
if (mp.count(now))
return limit * i + mp[now];
now = mul(now, cur);
}
return -1;
}

ll gcd(ll a, ll b) {
return b ? gcd(b, a % b) : a;
}

void exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1, y = 0;
return;
}
exgcd(b, a % b, y, x);
y -= a / b * x;
}

ll exgcd(ll a, ll b, ll c) {
ll g = gcd(a, b);
if (c % g) return -1;
ll x, y;
exgcd(a, b, x, y);
x *= c / g;
x = (x % (b / g) + (b / g)) % (b / g);
return x;
}

int main() {
#ifdef dream_maker
freopen("input.txt", "r", stdin);
#endif
scanf("%lld", &k);
for (ll i = 1; i <= k; i++)
scanf("%lld", &b[i]);
scanf("%lld %lld", &n, &m);
Matrix tmp;
for (ll i = 1; i < k; i++)
tmp.g[i][i + 1] = 1;
for (ll i = 1; i <= k; i++)
tmp.g[k][i] = b[k - i + 1];
tmp = fast_pow(tmp, n - k);
ll ans1 = bsgs(3, m), ans2 = exgcd(tmp.g[k][k], Mod - 1, ans1);
if (ans1 == -1 || ans2 == -1)
printf("-1");
else
printf("%lld", fast_pow(3, ans2));
return 0;
}


posted @ 2019-02-10 19:18  Dream_maker_yk  阅读(306)  评论(0编辑  收藏