Codeforces 1106F（数论）

要点

• 998244353的原根g = 3，意味着对于任意$$1 <= x,y<p$$$$x\neq\ y$$$$g^{x%p\neq g}y%p$$因此可以有构造序列\(q(a)与a一一对应，g^{q(a)}\%p=a\)。那么对应到这道题上，因为\(f_i\)是%p的，所以构造\(h_i\)序列，使得$$g^{{h_i}%p=f_i=\prod_{j=1}}{k}(f_{i-j})^{b_j}%p=g{\sum_{j=1}^k{h_{i-j}\times\ b_j}}%p$$$$\because 原根的唯一对应性质且g^{p-1}%p=1$$$$\therefore h_i\equiv \sum_{j=1}^kh_{i-j}\times b_j(mod\ p-1)$$
• 以上就是本题全部关键了，接下来就是数论复习内容了。
• 首先看到这个熟悉的式子想到我们可以\(\%(p-1)\)意义下矩阵快速幂求解，往常是给前面的项求第n项，这次是有\(h_n\)求\(h_k\)。
• 其中\(h_n\)的求法是BSGS算法
• 矩阵快速幂以后，因为题面说初始除了\(f_k\)以外都是1，所以\(h_{1…k-1}\)都是0，故而有$$h_n\equiv Matrix[0][0]\times h_k(mod\ p-1)$$
• 这就变成了\(ah_x\equiv c(\%b)\)，变形为\(ax+by=c\)即可用扩展欧几里得求解，若有解，用快速幂求得\(f_k\)，否则输出-1.

#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;

typedef long long ll;
const int p = 998244353, g = 3;
int K, b[101], n, fn, hn, hk;

struct Matrix {
	int n;
	int v[101][101];

	Matrix(int n) { memset(v, 0, sizeof v); this->n = n; }

	friend Matrix operator * (Matrix A, Matrix B) {
		int n = A.n;
		Matrix ret(n);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				for (int k = 0; k < n; k++) {
					ret.v[i][j] = ((ll)ret.v[i][j] + (ll)A.v[i][k] * B.v[k][j] % (p - 1)) % (p - 1);
				}
		return ret;
	}

	friend Matrix operator ^ (Matrix A, int k) {
		int n = A.n;
		Matrix ret(n);
		for (int i = 0; i < n; i++)
			ret.v[i][i] = 1;
		for (; k; k >>= 1) {
			if (k & 1)	ret = ret * A;
			A = A * A;
		}
		return ret;
	}
};

namespace BSGS {
	const int maxm = 1e5 + 1000;
	int hash_table[maxm], val[maxm];

	int ksm(int a, int b, int mod) {
		int res = 1;
		for (; b; b >>= 1) {
			if (b & 1)	res = (ll)res * a % mod;
			a = (ll)a * a % mod;
		}
		return res;
	}

	int find(int n) {
		int id = n % maxm;
		while (hash_table[id] >= 0 && hash_table[id] != n)
			id = (id + 1) % maxm;
		return id;
	}

	int bsgs(int a, int b, int p) {
		a %= p, b %= p;
		if (!a)	return b ? -1 : 1;
		memset(hash_table, -1, sizeof hash_table);

		int m = sqrt(p) + 1;
		int now = b;
		hash_table[now % maxm] = now;
		val[now % maxm] = 0;
		for (int i = 1; i <= m; i++) {
			now = (ll)now * a % p;
			int pos = find(now);
			hash_table[pos] = now;
			val[pos] = i;
		}

		int t = ksm(a, m, p);
		now = 1;
		for (int i = 1; i <= m; i++) {
			now = (ll)now * t % p;
			int pos = find(now);
			if (hash_table[pos] >= 0) {
				return i * m - val[pos];
			}
		}

		return -1;
	}
}

namespace EXGCD {
	int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }

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

	int solve(int a, int b, int c) {//ax = c (% b)求x的解
		if (!c)	return 0;
		int q = gcd(a, b);
		if (c % q)	return -1;

		a /= q, b /= q, c /= q;
		ll ans, __;
		exgcd((ll)a, (ll)b, ans, __);
		ans = (ans * c % b + b) % b;
		return ans;
	}
}

int main() {
	scanf("%d", &K);
	for (int i = 0; i < K; i++)
		scanf("%d", &b[i]), b[i] %= p - 1;
	scanf("%d%d", &n, &fn);

	hn = BSGS::bsgs(g, fn, p);
	Matrix A(K);
	for (int i = 0; i < K; i++)
		A.v[0][i] = b[i];
	for (int j = 1; j < K; j++)
		A.v[j][j - 1] = 1;
	A = A ^ (n - K);
	hk = EXGCD::solve(A.v[0][0], p - 1, hn);

	if (hk >= 0) {
		printf("%d\n", BSGS::ksm(g, hk, p));
	} else {
		printf("-1\n");
	}
	return 0;
}