結果

問題 No.181 A↑↑N mod M
コンテスト
ユーザー Pachicobue
提出日時 2017-11-18 15:41:46
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 4,222 bytes
コンパイル時間 2,611 ms
コンパイル使用メモリ 209,464 KB
最終ジャッジ日時 2025-01-05 04:06:35
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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ファイルパターン 結果
sample AC * 6
other AC * 15 WA * 3 TLE * 18 MLE * 1
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ソースコード

diff # 

#include <bits/stdc++.h>

#define show(x) cerr << #x << " = " << x << endl

using namespace std;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    os << "sz=" << v.size() << "\n[";
    for (const auto& p : v) {
        os << p << ",";
    }
    os << "]\n";
    return os;
}

template <typename S, typename T>
ostream& operator<<(ostream& os, const pair<S, T>& p)
{
    os << "(" << p.first << "," << p.second
       << ")";
    return os;
}


constexpr ll MOD = 1e9 + 7;

template <typename T>
constexpr T INF = numeric_limits<T>::max() / 100;

ll A, N, M;
vector<ll> prime;
map<ll, ll> A_factor;

template <typename T>
T extgcd(const T a, const T b, T& x, T& y)  // ax+by=gcd(a,b)
{
    T d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    } else {
        x = 1;
        y = 0;
    }
    return d;
}

ll power(const ll n, const ll mod)
{
    if (n == 0) {
        return 1;
    }
    if (n % 2 == 1) {
        return (mod != -1) ? ((power(n - 1, mod) * A) % mod) : (power(n - 1, mod) * A);
    } else {
        const ll pp = power(n / 2, mod);
        return (mod != -1) ? ((pp * pp) % mod) : (pp * pp);
    }
}

template <typename T>
pair<T, T> ChineseRemainderTheorem(const pair<T, T> a1, const pair<T, T> a2)
{
    assert(gcd(a1.first, a2.first) == 1);
    const T p1 = a1.first;
    const T m1 = a1.second;
    const T p2 = a2.first;
    const T m2 = a2.second;
    if (m1 == m2) {
        return make_pair(p1 * p2, m1);
    } else {
        T x, y;
        extgcd(p1, p2, x, y);
        assert(p1 * x + p2 * y == 1);
        x *= (m2 - m1);
        y *= (m2 - m1);
        const T p = p1 * p2;
        return make_pair(p, (((p1 * x + m1) % p) + p) % p);
    }
}

ll over(const ll n, const long double x)  // A↑↑n >= x なら 0, それ以外なら A↑↑n
{
    if (n == 0) {
        return (1 >= x) ? 0 : 1;
    }
    const long double lx = log2(x) / log2(A);
    const ll p = over(n - 1, lx);
    if (p == 0) {
        return 0;
    } else {
        return power(p, -1);
    }
}

ll rem(const ll n, const ll mod)  // A↑↑n (mod m)
{
    if (n == 0) {
        return 1 % mod;
    } else if (mod == 1) {
        return 0;
    } else {
        ll num = mod;
        vector<pair<ll, ll>> fac;
        vector<ll> phi;
        vector<ll> count;
        for (const ll p : prime) {
            if (num % p == 0) {
                ll beki = 1;
                ll cnt = 0;
                while (num % p == 0) {
                    beki *= p;
                    num /= p;
                    cnt++;
                }
                fac.push_back(make_pair(beki, 1));
                phi.push_back(beki / p * (p - 1));
                count.push_back(cnt);
            }
            if (num == 1) {
                break;
            }
        }
        const ll size = fac.size();
        for (ll i = 0; i < size; i++) {
            if (A_factor[fac[i].first] > 0) {
                const ll ind = over(n - 1, (long double)A_factor[fac[i].first] / count[i]);
                if (ind == 0) {
                    fac[i].second = 0;
                } else {
                    fac[i].second = power(ind, fac[i].first);
                }
            } else {
                const ll m = rem(n - 1, phi[i]);
                fac[i].second = power(m, fac[i].first);
            }
        }

        pair<ll, ll> result = fac[0];
        for (ll i = 1; i < size; i++) {
            result = ChineseRemainderTheorem(result, fac[i]);
        }
        return result.second;
    }
}

int main()
{
    cin >> A >> N >> M;
    vector<bool> isprime(M + 1, true);
    for (ll i = 2; i <= M; i++) {
        if (isprime[i]) {
            for (ll j = 2; i * j <= M; j++) {
                isprime[i * j] = false;
            }
        }
    }
    for (ll i = 2; i <= M; i++) {
        if (isprime[i]) {
            prime.push_back(i);
        }
    }

    ll num = A;
    for (const ll p : prime) {
        ll cnt = 0;
        while (num % p == 0) {
            cnt++;
            num /= p;
        }
        A_factor[p] = cnt;
    }

    cout << rem(N, M) << endl;

    return 0;
}
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