結果

問題 No.1255 ハイレーツ・オブ・ボリビアン
コンテスト
ユーザー packer_jp
提出日時 2020-10-09 23:05:34
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,410 bytes
コンパイル時間 2,478 ms
コンパイル使用メモリ 200,924 KB
最終ジャッジ日時 2025-01-15 05:43:30
ジャッジサーバーID
（参考情報） 		judge3 / 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 AC * 1
other AC * 2 WA * 11 TLE * 2
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 <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i, n) for (int i = 0; i < (int) (n); i++)
#define reps(i, n) for (int i = 1; i <= (int) (n); i++)
#define all(a) (a).begin(), (a).end()
#define uniq(a) (a).erase(unique(all(a)), (a).end())
#define bit(n) (1LL << (n))
#define dump(a) cerr << #a " = " << (a) << endl
using vint = vector<int>;
using pint = pair<int, int>;
using vpint = vector<pint>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
constexpr double PI = 3.1415926535897932384626433832795028;
constexpr int DY[9] = {0, 1, 0, -1, 1, 1, -1, -1, 0};
constexpr int DX[9] = {1, 0, -1, 0, 1, -1, -1, 1, 0};
int sign(int a) { return (a > 0) - (a < 0); }
int cdiv(int a, int b) { return (a - 1 + b) / b; }
template<typename T> void fin(T a) {
    cout << a << endl;
    exit(0);
}
template<typename T> T sq(T a) { return a * a; }
template<typename T, typename U> bool chmax(T &a, const U &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> bool chmin(T &a, const U &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &a) {
    os << "(" << a.first << ", " << a.second << ")";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const vector<T> &a) {
    os << "{";
    for (auto itr = a.begin(); itr != a.end(); itr++) {
        os << *itr << (next(itr) != a.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const deque<T> &a) {
    os << "{";
    for (auto itr = a.begin(); itr != a.end(); itr++) {
        os << *itr << (next(itr) != a.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const set<T> &a) {
    os << "{";
    for (auto itr = a.begin(); itr != a.end(); itr++) {
        os << *itr << (next(itr) != a.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const multiset<T> &a) {
    os << "{";
    for (auto itr = a.begin(); itr != a.end(); itr++) {
        os << *itr << (next(itr) != a.end() ? ", " : "");
    }
    os << "}";
    return os;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const map<T, U> &a) {
    os << "{";
    for (auto itr = a.begin(); itr != a.end(); itr++) {
        os << *itr << (next(itr) != a.end() ? ", " : "");
    }
    os << "}";
    return os;
}
struct setup {
    static constexpr int PREC = 20;
    setup() {
        cout << fixed << setprecision(PREC);
        cerr << fixed << setprecision(PREC);
    };
} setup;

// a^b
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) { res = res * a % mod; }
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}

// a^-1
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0) { u += m; }
    return u;
}

// a^x ≡ b (mod. m) となる最小の正の整数 x を求める
long long modlog(long long a, long long b, int m) {
    a %= m, b %= m;

    // calc sqrt{M}
    long long lo = -1, hi = m;
    while (hi - lo > 1) {
        long long mid = (lo + hi) / 2;
        if (mid * mid >= m) { hi = mid; }
        else { lo = mid; }
    }
    long long sqrtM = hi;

    // {a^0, a^1, a^2, ..., a^sqrt(m)}
    map<long long, long long> apow;
    long long amari = 1;
    for (long long r = 0; r < sqrtM; ++r) {
        if (!apow.count(amari)) { apow[amari] = r; }
        (amari *= a) %= m;
    }

    // check each A^p
    long long A = modpow(modinv(a, m), sqrtM, m);
    amari = b;
    for (long long q = 0; q < sqrtM; ++q) {
        if (apow.count(amari)) {
            long long res = q * sqrtM + apow[amari];
            if (res > 0) { return res; }
        }
        (amari *= A) %= m;
    }

    // no solutions
    return -1;
}

signed main() {
    int T;
    cin >> T;
    while (T--) {
        int N;
        cin >> N;
        if (N == 1) {
            cout << 1 << endl;
            continue;
        }
        cout << modlog(2, 1, 2 * N - 1) << endl;
    }
}
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