bzoj4128 Matrix 矩阵 BSGS

题目传送门

https://lydsy.com/JudgeOnline/problem.php?id=4128

题解

想了十分钟没有任何思路。

然后一眼瞥见一句话“数据保证在 \(p\) 内有解”，还有 \(p \leq 19997\)...

那么这道题不就是把同余类 BSGS 里面的数换成矩阵嘛。

问题就是怎么快速判断两个矩阵是否相等。哈希呗。

然后就没有然后了。

---

这里推荐写哪种不需要求逆的版本的 BSGS。

#include<bits/stdc++.h>
#include<tr1/unordered_map>

#define fec(i, x, y) (int i = head[x], y = g[i].to; i; i = g[i].ne, y = g[i].to)
#define dbg(...) fprintf(stderr, __VA_ARGS__)
#define File(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define fi first
#define se second
#define pb push_back

template<typename A, typename B> inline char smax(A &a, const B &b) {return a < b ? a = b, 1 : 0;}
template<typename A, typename B> inline char smin(A &a, const B &b) {return b < a ? a = b, 1 : 0;}

typedef long long ll; typedef unsigned long long ull; typedef std::pair<int, int> pii;

template<typename I> inline void read(I &x) {
	int f = 0, c;
	while (!isdigit(c = getchar())) c == '-' ? f = 1 : 0;
	x = c & 15;
	while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + (c & 15);
	f ? x = -x : 0;
}

const int N = 70 + 7;
const int base = 1997;

int n, P;
std::tr1::unordered_map<ull, int> mp;

inline int smod(int x) { return x >= P ? x - P : x; }
inline void sadd(int &x, const int &y) { x += y; x >= P ? x -= P : x; }
inline int fpow(int x, int y) {
	int ans = 1;
	for (; y; y >>= 1, x = (ll)x * x % P) if (y & 1) ans = (ll)ans * x % P;
	return ans;
}

struct Matrix {
	int a[N][N];

	inline Matrix() { memset(a, 0, sizeof(a)); }
	inline Matrix(const int &x) {
		memset(a, 0, sizeof(a));
		for (int i = 1; i <= n; ++i) a[i][i] = x;
	}

	inline Matrix operator * (const Matrix &b) {
		Matrix c;
		for (int k = 1; k <= n; ++k)
			for (int i = 1; i <= n; ++i)
				for (int j = 1; j <= n; ++j) sadd(c.a[i][j], (ll)a[i][k] * b.a[k][j] % P);
		return c;
	}
	inline ull hash() {
		ull ha = 0;
		for (int i = 1; i <= n; ++i)
			for (int j = 1; j <= n; ++j)
				ha = ha * base + (a[i][j] + 1);
		return ha;
	}
} A, B;

inline Matrix fpow(Matrix x, int y) {
	Matrix ans(1);
	for (; y; y >>= 1, x = x * x) if (y & 1) ans = ans * x;
	return ans;
}

inline int bsgs() {
	int m = sqrt(P);
	Matrix C = B, e;
	for (int i = 0; i < m; ++i, C = C * A) mp[C.hash()] = i;
	e = C = fpow(A, m);
	for (int i = 1; ; ++i, C = C * e) if (mp.count(C.hash())) return i * m - mp[C.hash()];
}

inline void work() {
	printf("%d\n", bsgs());
}

inline void init() {
	read(n), read(P);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			read(A.a[i][j]);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= n; ++j)
			read(B.a[i][j]);
}

int main() {
#ifdef hzhkk
	freopen("hkk.in", "r", stdin);
#endif
	init();
	work();
	fclose(stdin), fclose(stdout);
	return 0;
}