結果

問題 No.950 行列累乗
コンテスト
ユーザー やむなく
提出日時 2019-12-13 02:53:47
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,201 bytes
コンパイル時間 5,532 ms
コンパイル使用メモリ 201,208 KB
実行使用メモリ 67,840 KB
最終ジャッジ日時 2024-06-26 05:14:05
合計ジャッジ時間 31,938 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
このコードへのチャレンジ
（要ログイン）
ファイルパターン 結果
sample AC * 4
other AC * 27 WA * 22 TLE * 8
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 # 

//
// Created by yamunaku on 2019/12/13.
//

#include <bits/stdc++.h>

using namespace std;

#define rep(i, n) for(int i = 0; i < (n); i++)
#define repl(i, l, r) for(int i = (l); i < (r); i++)
#define per(i, n) for(int i = ((n)-1); i >= 0; i--)
#define perl(i, l, r) for(int i = ((r)-1); i >= (l); i--)
#define all(x) (x).begin(),(x).end()
#define MOD9 998244353
#define MOD1 1000000007
#define IINF 1000000000
#define LINF 1000000000000000000
#define SP <<" "<<
#define CYES cout<<"Yes"<<endl
#define CNO cout<<"No"<<endl
#define CFS cin.tie(0);ios::sync_with_stdio(false)
#define CST(x) cout<<fixed<<setprecision(x)

using ll = long long;
using ld = long double;
using vi = vector<int>;
using mti = vector<vector<int>>;
using vl = vector<ll>;
using mtl = vector<vector<ll>>;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
template<typename T>
using heap = priority_queue<T, vector<T>, function<bool(const T, const T)>>;

ll p;
mtl a, b;

ll det(mtl &v, ll m){
    return ((v[0][0] * v[1][1] % m - v[0][1] * v[1][0] % m) % m + m) % m;
}

mtl prod(mtl a, mtl b, ll m){
    return {
            {(a[0][0] * b[0][0] + a[0][1] * b[1][0]) % m, (a[0][0] * b[0][1] + a[0][1] * b[1][1]) % m},
            {(a[1][0] * b[0][0] + a[1][1] * b[1][0]) % m, (a[1][0] * b[0][1] + a[1][1] * b[1][1]) % m}
    };
}

mtl mtpow(mtl v, ll x, ll m){
    mtl ans = {
            {1, 0},
            {0, 1}
    };
    while(x){
        if(x & 1) ans = prod(ans, v, m);
        x >>= 1;
        v = prod(v, v, m);
    }
    return ans;
}

ll modpow(ll x, ll a, ll m){
    ll ans = 1;
    while(a){
        if(a & 1) ans = ans * x % m;
        a >>= 1;
        x = x * x % m;
    }
    return ans;
}

ll inv(ll x, ll m){
    return modpow(x, m - 2, m);
}

ll nlog(ll a, ll b, ll p){
    // a^(i*r+j)=b
    // a^j = b/a^(i*r)
    ll r = sqrt(p) + 10;
    ll ar = inv(modpow(a, r, p), p);
    map<ll, ll> mp;
    ll tmp = b;
    rep(i, r){
        auto itr = mp.find(tmp);
        if(itr == mp.end()) mp[tmp] = i;
        tmp = tmp * ar % p;
    }
    ll ans = LINF;
    tmp = 1;
    rep(i, r){
        auto itr = mp.find(tmp);
        if(itr != mp.end()) ans = min(ans, (itr->second * r + i) % (p - 1));
        tmp = tmp * a % p;
    }
    return ans;
}

mtl mtinv(mtl a, ll p){
    ll d = inv(det(a, p), p);
    return {
            {a[1][1] * d % p,            (-a[0][1] * d % p + p) % p},
            {(-a[1][0] * d % p + p) % p, a[0][0] * d % p},
    };
}

ll mtlog(mtl a, mtl b, ll p){
    // a^(i*r+j)=b
    // a^j = b/a^(i*r)
    ll r = sqrt(1e11);
    mtl ar = mtinv(mtpow(a, r, p), p);
    map<mtl, ll> mp;
    mtl tmp = b;
    rep(i, r){
        auto itr = mp.find(tmp);
        if(itr == mp.end()) mp[tmp] = i;
        tmp = prod(tmp, ar, p);
    }
    ll ans = LINF;
    tmp = {
            {1, 0},
            {0, 1}
    };
    rep(i, r){
        auto itr = mp.find(tmp);
        if(itr != mp.end()) ans = min(ans, itr->second * r + i);
        tmp = prod(tmp, a, p);
    }
    return ans;
}

void answer(ll x, ll y, ll z){
    ll x1 = x % IINF, x2 = x / IINF;
    ll y1 = y % IINF, y2 = y / IINF;
    ll ans1 = x1 * y1 + z;
    ll c = ans1 / IINF;
    ans1 %= IINF;
    ll ans2 = x1 * y2 + x2 * y1 + c;
    c = ans2 / IINF;
    ans2 %= IINF;
    ll ans3 = x2 * y2 + c;
    if(ans3 > 0) cout << ans3;
    if(ans2 > 0) cout << ans2;
    cout << ans1 << endl;
}

int main(){
    cin >> p;
    a = b = mtl(2, vl(2));
    rep(i, 2) rep(j, 2) cin >> a[i][j];
    rep(i, 2) rep(j, 2) cin >> b[i][j];
    if(a == b){
        cout << 1 << endl;
        return 0;
    }

    ll x = det(a, p), y = det(b, p);
    if(x == 0){
        if(y > 0){
            cout << -1 << endl;
            return 0;
        }
        ll ans = mtlog(a, b, p);
        if(ans == LINF) cout << -1 << endl;
        else cout << ans << endl;
        return 0;
    }
    if(y == 0){
        cout << -1 << endl;
        return 0;
    }

    ll l = nlog(x, y, p);

    mtl tmp = mtpow(a, l, p);
    if(tmp == b){
        cout << l << endl;
        return 0;
    }
    a = mtpow(a, p - 1, p);
    b = prod(b, mtinv(tmp, p), p);
    ll ans = mtlog(a, b, p);

    if(ans == LINF) cout << -1 << endl;
    else answer(ans, p - 1, l);
    return 0;
}
0