#567232 (C++17) No.1255 ハイレーツ・オブ・ボリビアン

提出ソース

結果

問題	No.1255 ハイレーツ・オブ・ボリビアン
ユーザー	hitonanode
提出日時	2020-10-16 20:47:47
言語	C++17 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	1,231 ms / 2,000 ms
コード長	5,737 bytes
コンパイル時間	2,226 ms
コンパイル使用メモリ	209,240 KB
実行使用メモリ	6,944 KB
最終ジャッジ日時	2024-07-23 08:49:08
合計ジャッジ時間	7,976 ms
ジャッジサーバーID （参考情報）	judge3 / judge1

このコードへのチャレンジ
（要ログイン）

テストケース

テストケース表示

入力	結果	実行時間実行使用メモリ
testcase_00	AC	2 ms 6,812 KB
testcase_01	AC	2 ms 6,940 KB
testcase_02	AC	3 ms 6,944 KB
testcase_03	AC	2 ms 6,944 KB
testcase_04	AC	2 ms 6,940 KB
testcase_05	AC	6 ms 6,940 KB
testcase_06	AC	7 ms 6,940 KB
testcase_07	AC	7 ms 6,940 KB
testcase_08	AC	613 ms 6,940 KB
testcase_09	AC	640 ms 6,940 KB
testcase_10	AC	647 ms 6,944 KB
testcase_11	AC	629 ms 6,940 KB
testcase_12	AC	655 ms 6,944 KB
testcase_13	AC	55 ms 6,944 KB
testcase_14	AC	1,231 ms 6,940 KB
testcase_15	AC	621 ms 6,944 KB

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

raw source code

#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }
template <typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template <typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> srtunq(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os; }
#endif
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T, typename TH> ostream &operator<<(ostream &os, const unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <typename TK, typename TV, typename TH> ostream &operator<<(ostream &os, const unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
#define dbg(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ") " << __FILE__ << endl
#else
#define dbg(x) {}
#endif

// Calculate log_A B (MOD M) (baby-step gian-step)
// DiscreteLogarithm dl(M, A);
// lint ans = dl.log(B);
// Complexity: O(M^(1/2)) for each query
// Verified: <https://judge.yosupo.jp/problem/discrete_logarithm_mod>
// Constraints: 0 <= A < M, B < M, 1 <= M <= 1e9  (M is not limited to prime)
struct DiscreteLogarithm
{
    using lint = long long;
    int M, stepsize;
    lint baby_a, giant_a, g;
    std::unordered_map<lint, int> baby_log_dict;

    lint inverse(lint a) {
        lint b = M / g, u = 1, v = 0;
        while (b) {
            lint t = a / b;
            a -= t * b; std::swap(a, b);
            u -= t * v; std::swap(u, v);
        }
        u %= M / g;
        return u >= 0 ? u : u + M / g;
    }

    DiscreteLogarithm(int mod, int a_new) : M(mod), baby_a(a_new % mod), giant_a(1) {
        g = 1;
        while (std::__gcd(baby_a, M / g) > 1) g *= std::__gcd(baby_a, M / g);
        stepsize = 32;  // lg(MAX_M)
        while (stepsize * stepsize < M / g) stepsize++;

        lint now = 1 % (M / g), inv_g = inverse(baby_a);
        for (int n = 0; n < stepsize; n++) {
            if (!baby_log_dict.count(now)) baby_log_dict[now] = n;
            (now *= baby_a) %= M / g;
            (giant_a *= inv_g) %= M / g;
        }
    }

    // log(): returns the smallest nonnegative x that satisfies a^x = b mod M, or -1 if there's no solution
    lint log(lint b) {
        b %= M;
        lint acc = 1 % M;
        for (int i = 0; i < stepsize; i++) {
            if (acc == b) return i;
            (acc *= baby_a) %= M;
        }
        if (b % g) return -1;  // No solution
        lint now = b * giant_a % (M / g);
        for (lint q = 1; q <= M / stepsize + 1; q++) {
            if (baby_log_dict.count(now)) return q * stepsize + baby_log_dict[now];
            (now *= giant_a) %= M / g;
        }
        return -1;
    }
};
lint solve(lint N)
{
    return DiscreteLogarithm(N * 2 + 1, 2).log(N + 1) + 1;
}

int main()
{
    int T;
    cin >> T;
    while (T--) {
        lint N;
        cin >> N;
        cout << solve(N - 1) << '\n';
    }
}

yukicoder

結果

テストケース

ソースコード