結果

問題 No.950 行列累乗
コンテスト
ユーザー heno239
提出日時 2019-12-13 04:14:13
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 3,851 bytes
コンパイル時間 2,690 ms
コンパイル使用メモリ 148,752 KB
実行使用メモリ 217,272 KB
最終ジャッジ日時 2024-06-26 06:13:11
合計ジャッジ時間 9,246 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge2
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 -- * 4
other TLE * 1 -- * 56
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 # 

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define Rep(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) Rep(i,0,n)
#define stop char nyaa;cin>>nyaa;
using ll = long long;
const int inf = 1e9 + 7;
const ll INF = 1LL << 60;
//const ll mod = 1e9 + 7;

using P = pair<int, int>;

using LP = pair<ll, ll>;

ll p;
typedef vector<vector<ll>> mat;
typedef vector<ll> vec;
mat mtmul(mat &A, mat &B) {
	mat C(A.size(), vec(B[0].size()));
	rep(i, (int)A.size()) {
		rep(k, (int)B.size()) {
			rep(j, (int)B[0].size()) {
				C[i][j] = (C[i][j] + (A[i][k] * B[k][j]) % p) % p;
			}
		}
	}
	return C;
}
mat mtpow(mat A, ll n) {
	mat B(A.size(), vec(A.size()));
	rep(i, (int)A.size()) {
		B[i][i] = 1;
	}
	while (n > 0) {
		if (n&(ll)1)B = mtmul(B, A);
		A = mtmul(A, A);
		n >>= 1;
	}
	return B;
}
ll dat(mat A) {
	ll res = A[0][0] * A[1][1] - A[0][1] * A[1][0];
	res = (res%p + p) % p;
	return res;
}

ll mod_pow(ll x, ll n) {
	ll res = 1;
	while (n > 0) {
		if (n % 2)res = res * x%p;
		x = x * x%p; n >>= 1;
	}
	return res;
}
mat invmt(mat A) {
	ll b = dat(A);
	mat res = { { A[1][1],p - A[0][1] },{ p - A[1][0],A[0][0] } };
	b = mod_pow(b, p - 2);
	rep(i, 2)rep(j, 2) {
		res[i][j] = res[i][j] * b%p;
	}
	return res;
}
//a^res=b (mod p)
ll ident(ll a, ll b) {
	if (b == 1)return p - 1;
	if (p <= 100000) {
		int tmp = 1;
		ll cop = a;
		while (cop != b) {
			cop = cop * a%p;
			tmp++;
		}
		return tmp;
	}
	ll d = sqrt(p);
	map<ll, int> mp;
	ll cop = b;
	ll inva = mod_pow(a, p - 2);
	for (int y = 0; y < d; y++) {
		mp[cop] = y;
		cop = cop * inva%p;
	}
	ll spea = mod_pow(a, d);
	cop = 1;
	for (int x = 0; x <= d; x++) {
		if (mp.find(cop) != mp.end()) {
			ll res = x * d + mp[cop];
			res %= (p - 1);
			return res;
		}
		cop = cop * spea%p;
	}
	return -1;
	assert(false);
	return 0;
}
ll minp(ll a) {
	vector<ll> ds;
	for (int i = 1; i <= 100000; i++) {
		if ((p - 1) % i == 0) {
			ds.push_back(i);
			ds.push_back((p - 1) / i);
		}
	}
	sort(ds.begin(), ds.end());
	ds.erase(unique(ds.begin(), ds.end()), ds.end());
	rep(i, ds.size()) {
		ll kata = ds[i];
		if (mod_pow(a, kata) == 1)return kata;
	}
	return p - 1;
}

void solve() {
	mat A(2), B(2);
	cin >> p;
	rep(i, 2) {
		A[i].resize(2);
		rep(j, 2)cin >> A[i][j];
	}
	rep(i, 2) {
		B[i].resize(2);
		rep(j, 2)cin >> B[i][j];
	}
	ll da = dat(A), db = dat(B);
	map<mat, bool> mp;
	if (da == 0) {
		if (db > 0) {
			cout << -1 << endl; return;
		}
		mat cop = A;
		if (cop == B) {
			cout << 1 << endl; return;
		}
		mp[cop] = true;
		int tmp = 1;
		while (true) {
			tmp++;
			cop = mtmul(cop, A);
			if (mp[cop]) {
				cout << -1 << endl; return;
			}
			if (cop == B) {
				cout << tmp << endl; return;
			}
			mp[cop] = true;
		}
	}
	if (db == 0) {
		cout << -1 << endl; return;
	}
	ll k = ident(da, db);
	if (k < 0) {
		cout << -1 << endl; return;
	}
	ll res = k;

	ll e = minp(da);
	//cout << Z[0][0] << " " << Z[0][1] << " " << Z[1][0] << " " << Z[1][1] << endl;
	mat ap = mtpow(A, e);
	mat aa = mtpow(A, k); aa = invmt(aa);
	mat bp = B;
	bp = mtmul(bp, aa);
	int sz = 100000;
	map<mat, int> memo;
	ll ans = (ll)1 << 62;
	mat invap = invmt(ap);
	mat X = bp;
	for (int y = 0; y <= sz; y++) {
		if (memo.find(X) == memo.end()) {
			memo[X] = y;
		}
		X = mtmul(X, invap);
	}
	mat app = mtpow(ap, sz);
	X = { { 1,0 },{ 0,1 } };
	for (ll x = 0; x <= sz; x++) {
		if (memo.find(X) != memo.end()) {
			ll cs = x * sz + memo[X];
			ans = min(ans, cs);
		}
		X = mtmul(X, app);
	}
	if (ans == ((ll)1 << 62)) {
		cout << -1 << endl; return;
	}
	cout << ans * (p - 1) + k << endl;
}


signed main() {
	solve();
	//stop
	return 0;
}
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