結果

問題 No.3170 [Cherry 7th Tune KY] Even if you could say "See you ..."
ユーザー shobonvip
提出日時 2025-05-30 23:13:43
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3,266 ms / 4,000 ms
コード長 5,383 bytes
コンパイル時間 6,268 ms
コンパイル使用メモリ 272,984 KB
実行使用メモリ 22,528 KB
最終ジャッジ日時 2025-06-18 15:36:11
合計ジャッジ時間 37,726 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 40
権限があれば一括ダウンロードができます

ソースコード

diff # 

/**
	author:  shobonvip
	created: 2025.05.30 22:38:12
**/

#include<bits/stdc++.h>
using namespace std;

//* ATCODER
#include<atcoder/all>
using namespace atcoder;
typedef modint mint;
//*/

/* BOOST MULTIPRECISION
#include<boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
//*/

typedef long long ll;

#define rep(i, s, n) for (int i = (int)(s); i < (int)(n); i++)
#define rrep(i, s, n) for (int i = (int)(n)-1; i >= (int)(s); i--)
#define all(v) v.begin(), v.end()

template <typename T> bool chmin(T &a, const T &b) {
	if (a <= b) return false;
	a = b;
	return true;
}

template <typename T> bool chmax(T &a, const T &b) {
	if (a >= b) return false;
	a = b;
	return true;
}

template <typename T> T max(vector<T> &a){
	assert(!a.empty());
	T ret = a[0];
	for (int i=0; i<(int)a.size(); i++) chmax(ret, a[i]);
	return ret;
}

template <typename T> T min(vector<T> &a){
	assert(!a.empty());
	T ret = a[0];
	for (int i=0; i<(int)a.size(); i++) chmin(ret, a[i]);
	return ret;
}

template <typename T> T sum(vector<T> &a){
	T ret = 0;
	for (int i=0; i<(int)a.size(); i++) ret += a[i];
	return ret;
}

ll ceil_sqrt(ll n) {
	ll ub = 2e9;
	ll lb = -1;
	while (ub - lb > 1) {
		ll t = (ub + lb) / 2;
		if (t * t >= n) {
			ub = t;
		}else{
			lb = t;
		}
	}
	return ub;
}

ll modpow(ll n, ll m, ll mod) {
	ll v = 1;
	ll w = n;
	while (m > 0) {
		if (m & 1) {
			v = v * w % mod;
		}
		w = w * w % mod;
		m >>= 1;
	}
	return v;
}

vector<vector<mint>> prod(
	const vector<vector<mint>> &a,
	const vector<vector<mint>> &b
) {
	vector<vector<mint>> ret(2, vector<mint>(2));
	rep(i,0,2)rep(j,0,2)rep(k,0,2) ret[i][j] += a[i][k] * b[k][j];
	return ret;
}

vector<vector<mint>> matpow(
	vector<vector<mint>> &n,
	ll m
) {
	vector<vector<mint>> v(2, vector<mint>(2));
	v[0][0] = 1;
	v[1][1] = 1;
	vector<vector<mint>> w = n;
	while (m > 0) {
		if (m & 1) {
			v = prod(v, w);
		}
		w = prod(w, w);
		m >>= 1;
	}
	return v;
}

ll discrete_logarithm(ll x, ll y, ll m) {
	if (m == 1) return 0;
	if (y == 1) return 0;
	if (x == 0 && y == 0) return 1;
	ll r = ceil_sqrt(m);
	map<ll,ll> d;
	ll a = 1;
	for (ll i=0; i<r+1; i++) {
		if (a == y) return i;
		if (d.find(a * y % m) == d.end()) {
			d[a * y % m] = i;
		}
		a = a * x % m;
	}
	ll xi = modpow(x, r, m);
	ll b = xi;
	for (ll i=1; i<r+1; i++) {
		if (d.find(b) != d.end()) {
			ll ans = i * r - d[b];
			if (modpow(x, ans, m) == y) {
				return ans;
			}
			return -1;
		}
		b = b * xi % m;
	}
	return -1;
}


bool is_identity(vector<vector<mint>> &x) {
	if (x[0][0].val() != 1) return false;
	if (x[1][1].val() != 1) return false;
	if (x[1][0].val() != 0) return false;
	if (x[0][1].val() != 0) return false;
	return true;
}


typedef array<int,4> ar;
ll discrete_logarithm_matrix(vector<vector<mint>> x, vector<vector<mint>> y) {

	ll r = (ll)mint::mod();

	map<ar, ll> d;
	vector<vector<mint>> a(2,vector<mint>(2));
	a[0][0]=1;
	a[1][1]=1;

	for (ll i=0; i<r+1; i++) {
		if (a == y) return i;
		vector<vector<mint>> pr = prod(a,y);
		ar prl;
		rep(s,0,2)rep(t,0,2)prl[s*2+t]=pr[s][t].val();
		if (d.find(prl) == d.end()) {
			d[prl] = i;
		}
		a = prod(a, x);
	}

	vector<vector<mint>> xi = matpow(x, r);
	vector<vector<mint>> b = xi;
	for (ll i=1; i<r+1; i++) {
		ar bl;
		rep(s,0,2)rep(t,0,2)bl[s*2+t]=b[s][t].val();
		if (d.find(bl) != d.end()) {
			ll ans = i * r - d[bl];
			if (matpow(x, ans) == y) {
				return ans;
			}
			return -1;
		}
		b = prod(b, xi);
	}
	return -1;
}


mint det(vector<vector<mint>> &a) {
	return a[0][0] * a[1][1] - a[1][0] * a[0][1];
}

void solve() {
	int p; cin >> p;
	mint::set_mod(p);

	vector<vector<mint>> a(2, vector<mint>(2));
	vector<vector<mint>> b(2, vector<mint>(2));

	rep(i,0,2) rep(j,0,2) {
		int x; cin >> x;
		a[i][j] = x;
	}

	rep(i,0,2) rep(j,0,2) {
		int x; cin >> x;
		b[i][j] = x;
	}

	if (is_identity(b)) {
		cout << 0 << '\n';
		return;
	}

	if (a == b) {
		cout << 1 << '\n';
		return;
	}

	// https://math.stackexchange.com/questions/3278216/the-order-of-a-2-times-2-matrix-mod-p
	// the period devides p^2-1 or p^2-p
	// thus period <= p^2, so just bsgs and done in O(p sqrt(p)).

	ll ans = discrete_logarithm_matrix(a, b);
	cout << ans << '\n';
	return;

	/*
	if (det(a) == 0) {
		if (det(b) != 0) {
			cout << -1 << '\n';
			return;
		}

		bool mode = 0;
		vector<vector<mint>> c(2, vector<mint>(2));
		mint tr = a[0][0] + a[1][1];

		if (tr.val() != 0) {
			bool mode = 0;
			mint captain = 0;
			rep(i,0,2) {
				rep(j,0,2) {
					if (a[i][j].val() == 0 && b[i][j].val() != 0) {
						cout << -1 << '\n';
						return;
					}
					if (a[i][j].val() != 0 && b[i][j].val() == 0) {
						cout << -1 << '\n';
						return;
					}
					if (a[i][j].val() != 0 && b[i][j].val() != 0) {
						if (!mode) {
							captain = b[i][j] / a[i][j];
							mode = 1;
						} else {
							if (captain.val() != (b[i][j] / a[i][j]).val()) {
								cout << -1 << '\n';
								return;
							}
						}
					}
				}
			}

			ll v = discrete_logarithm(tr.val(), captain.val(), p);
			cout << v+1 << '\n';
			return;
		}else{
			rep(i,0,2) {
				rep(j,0,2) {
					if (b[i][j].val() != 0) {
						cout << -1 << '\n';
						return;
					}
				}
			}

			cout << 1 << '\n';
			return;
		}


	}else{
		if (det(b) == 0) {
			cout << -1 << '\n';
			return;
		}



	}
		*/
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int t; cin >> t;
	while(t--) solve();
}
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