BSGS/exBSGS

BSGS

(y^x = z(mod p)) [(p in prime)]
令
(y^{(x+k)} = z imes y^k(mod p))
然后预处理 (k = [1, sqrt p])
BSGS 就是大块枚举 ((x+k))，就做完了。

exBSGS

(p otin prime)
(（A, p) ot| B) (and) (B != 1) 则无解。
令 (G = (A, p))
使得 (A^{x-1}frac{A}{G} = frac{B}{G} (mod frac{p}{G}))
(A^{x-1} = AB(mod frac{p}{G})
递归即可。

#include <bits/stdc++.h>
const int mod = 233333;
struct Hash_Table {
	Hash_Table() { clear(); }
	struct List { int id, val, nxt; } e[233333];
	int head[mod], cnt = 0;
	void add(int u, int v, int w) { e[++cnt] = {v, w, head[u]}; head[u] = cnt; }
	void clear() { memset(head, 0, sizeof(head)); cnt = 0; }
	void insert(int x, int i) { int k = x % mod; add(k, x, i); } // mp[x] = i
	int query(int x) {
		for (int i = head[x % mod]; i; i = e[i].nxt) if (e[i].id == x) return e[i].val;
		return -1;
	}
} hash;
int qpow(int x, int y, int mod) {
	int result = 1;
	while (y) {
		if (y & 1) { result = 1ll * result * x % mod; }
		x = 1ll * x * x % mod; y >>= 1;
	}
	return result;
}
void NoAnswer() { std::cout << "No Solution" << '
'; return; }
void exbsgs(int y, int z, int p) {
	if (z == 1) { std::cout << 0 << '
'; return; }
	int k = 0, qwq = 1; hash.clear();
	for (int g = std::__gcd(y, p); g != 1; g = std::__gcd(y, p)) {
		if (z % g) { NoAnswer(); return; }
		z /= g; p /= g; ++k;
		qwq = 1ll * qwq * (y / g) % p;
		if (qwq == z) { std::cout << k << '
'; return; }
	}
	int t = sqrt(p) + 1, s = z, base = y;
	hash.insert(s, 0);
	for (int i = 1; i <= t; i++) { s = 1ll * s * base % p; hash.insert(s, i); }
	base = qpow(y, t, p); s = qwq;
	for (int i = 1; i <= t; i++) {
		s = 1ll * s * base % p;
		if (!~hash.query(s)) { continue; }
		qwq = hash.query(s);
		std::cout << i * t - qwq + k << '
';
		return;
	}
	NoAnswer();
}
int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(NULL);
	int a, p, b;
	while (std::cin >> a >> p >> b && a && b && p) { exbsgs(a, b, p); }
	return 0;
}