結果

問題 No.2122 黄金比で擬似乱数生成
ユーザー tokusakuraitokusakurai
提出日時 2022-11-04 23:36:34
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 116 ms / 2,000 ms
コード長 10,566 bytes
コンパイル時間 2,403 ms
コンパイル使用メモリ 216,460 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-18 22:12:58
合計ジャッジ時間 3,672 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge1
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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 6 ms
6,816 KB
testcase_01 AC 11 ms
6,940 KB
testcase_02 AC 10 ms
6,944 KB
testcase_03 AC 12 ms
6,940 KB
testcase_04 AC 12 ms
6,940 KB
testcase_05 AC 14 ms
6,944 KB
testcase_06 AC 9 ms
6,940 KB
testcase_07 AC 10 ms
6,944 KB
testcase_08 AC 12 ms
6,944 KB
testcase_09 AC 12 ms
6,940 KB
testcase_10 AC 15 ms
6,940 KB
testcase_11 AC 8 ms
6,940 KB
testcase_12 AC 12 ms
6,944 KB
testcase_13 AC 13 ms
6,944 KB
testcase_14 AC 15 ms
6,944 KB
testcase_15 AC 21 ms
6,940 KB
testcase_16 AC 41 ms
6,944 KB
testcase_17 AC 12 ms
6,940 KB
testcase_18 AC 7 ms
6,944 KB
testcase_19 AC 13 ms
6,940 KB
testcase_20 AC 6 ms
6,940 KB
testcase_21 AC 6 ms
6,944 KB
testcase_22 AC 116 ms
6,940 KB
testcase_23 AC 113 ms
6,940 KB
testcase_24 AC 92 ms
6,940 KB
testcase_25 AC 106 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

diff # 

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define per(i, n) for (int i = (n)-1; i >= 0; i--)
#define rep2(i, l, r) for (int i = (l); i < (r); i++)
#define per2(i, l, r) for (int i = (r)-1; i >= (l); i--)
#define each(e, v) for (auto &e : v)
#define MM << " " <<
#define pb push_back
#define eb emplace_back
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;

template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;

template <typename T>
using maxheap = priority_queue<T>;

template <typename T>
bool chmax(T &x, const T &y) {
    return (x < y) ? (x = y, true) : false;
}

template <typename T>
bool chmin(T &x, const T &y) {
    return (x > y) ? (x = y, true) : false;
}

template <typename T>
int flg(T x, int i) {
    return (x >> i) & 1;
}

template <typename T>
void print(const vector<T> &v, T x = 0) {
    int n = v.size();
    for (int i = 0; i < n; i++) cout << v[i] + x << (i == n - 1 ? '\n' : ' ');
    if (v.empty()) cout << '\n';
}

template <typename T>
void printn(const vector<T> &v, T x = 0) {
    int n = v.size();
    for (int i = 0; i < n; i++) cout << v[i] + x << '\n';
}

template <typename T>
int lb(const vector<T> &v, T x) {
    return lower_bound(begin(v), end(v), x) - begin(v);
}

template <typename T>
int ub(const vector<T> &v, T x) {
    return upper_bound(begin(v), end(v), x) - begin(v);
}

template <typename T>
void rearrange(vector<T> &v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), end(v));
}

template <typename T>
vector<int> id_sort(const vector<T> &v, bool greater = false) {
    int n = v.size();
    vector<int> ret(n);
    iota(begin(ret), end(ret), 0);
    sort(begin(ret), end(ret), [&](int i, int j) { return greater ? v[i] > v[j] : v[i] < v[j]; });
    return ret;
}

template <typename S, typename T>
pair<S, T> operator+(const pair<S, T> &p, const pair<S, T> &q) {
    return make_pair(p.first + q.first, p.second + q.second);
}

template <typename S, typename T>
pair<S, T> operator-(const pair<S, T> &p, const pair<S, T> &q) {
    return make_pair(p.first - q.first, p.second - q.second);
}

template <typename S, typename T>
istream &operator>>(istream &is, pair<S, T> &p) {
    S a;
    T b;
    is >> a >> b;
    p = make_pair(a, b);
    return is;
}

template <typename S, typename T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << p.first << ' ' << p.second;
}

struct io_setup {
    io_setup() {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout << fixed << setprecision(15);
    }
} io_setup;

const int inf = (1 << 30) - 1;
const ll INF = (1LL << 60) - 1;
// const int MOD = 1000000007;
const int MOD = 10000;

template <int mod>
struct Mod_Int {
    int x;

    Mod_Int() : x(0) {}

    Mod_Int(long long y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    static int get_mod() { return mod; }

    Mod_Int &operator+=(const Mod_Int &p) {
        if ((x += p.x) >= mod) x -= mod;
        return *this;
    }

    Mod_Int &operator-=(const Mod_Int &p) {
        if ((x += mod - p.x) >= mod) x -= mod;
        return *this;
    }

    Mod_Int &operator*=(const Mod_Int &p) {
        x = (int)(1LL * x * p.x % mod);
        return *this;
    }

    Mod_Int &operator/=(const Mod_Int &p) {
        *this *= p.inverse();
        return *this;
    }

    Mod_Int &operator++() { return *this += Mod_Int(1); }

    Mod_Int operator++(int) {
        Mod_Int tmp = *this;
        ++*this;
        return tmp;
    }

    Mod_Int &operator--() { return *this -= Mod_Int(1); }

    Mod_Int operator--(int) {
        Mod_Int tmp = *this;
        --*this;
        return tmp;
    }

    Mod_Int operator-() const { return Mod_Int(-x); }

    Mod_Int operator+(const Mod_Int &p) const { return Mod_Int(*this) += p; }

    Mod_Int operator-(const Mod_Int &p) const { return Mod_Int(*this) -= p; }

    Mod_Int operator*(const Mod_Int &p) const { return Mod_Int(*this) *= p; }

    Mod_Int operator/(const Mod_Int &p) const { return Mod_Int(*this) /= p; }

    bool operator==(const Mod_Int &p) const { return x == p.x; }

    bool operator!=(const Mod_Int &p) const { return x != p.x; }

    Mod_Int inverse() const {
        assert(*this != Mod_Int(0));
        return pow(mod - 2);
    }

    Mod_Int pow(long long k) const {
        Mod_Int now = *this, ret = 1;
        for (; k > 0; k >>= 1, now *= now) {
            if (k & 1) ret *= now;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const Mod_Int &p) { return os << p.x; }

    friend istream &operator>>(istream &is, Mod_Int &p) {
        long long a;
        is >> a;
        p = Mod_Int<mod>(a);
        return is;
    }
};

using mint = Mod_Int<MOD>;

template <typename T>
struct Matrix {
    vector<vector<T>> A;

    Matrix(int m, int n) : A(m, vector<T>(n, 0)) {}

    int height() const { return A.size(); }

    int width() const { return A.front().size(); }

    inline const vector<T> &operator[](int k) const { return A[k]; }

    inline vector<T> &operator[](int k) { return A[k]; }

    static Matrix I(int l) {
        Matrix ret(l, l);
        for (int i = 0; i < l; i++) ret[i][i] = 1;
        return ret;
    }

    Matrix &operator*=(const Matrix &B) {
        int m = height(), n = width(), p = B.width();
        assert(n == B.height());
        Matrix ret(m, p);
        for (int i = 0; i < m; i++) {
            for (int k = 0; k < n; k++) {
                for (int j = 0; j < p; j++) ret[i][j] += A[i][k] * B[k][j];
            }
        }
        swap(A, ret.A);
        return *this;
    }

    Matrix operator*(const Matrix &B) const { return Matrix(*this) *= B; }

    Matrix pow(long long k) const {
        int m = height(), n = width();
        assert(m == n);
        Matrix now = *this, ret = I(n);
        for (; k > 0; k >>= 1, now *= now) {
            if (k & 1) ret *= now;
        }
        return ret;
    }

    bool eq(const T &a, const T &b) const {
        return a == b;
        // return abs(a-b) <= EPS;
    }

    // 行基本変形を用いて簡約化を行い、(rank, det) の組を返す
    pair<int, T> row_reduction(vector<T> &b) {
        int m = height(), n = width(), check = 0, rank = 0;
        T det = 1;
        assert(b.size() == m);
        for (int j = 0; j < n; j++) {
            int pivot = check;
            for (int i = check; i < m; i++) {
                if (A[i][j] != 0) pivot = i;
                // if(abs(A[i][j]) > abs(A[pivot][j])) pivot = i; // T が小数の場合はこちら
            }
            if (check != pivot) det *= T(-1);
            swap(A[check], A[pivot]), swap(b[check], b[pivot]);
            if (eq(A[check][j], T(0))) {
                det = T(0);
                continue;
            }
            rank++;
            det *= A[check][j];
            T r = T(1) / A[check][j];
            for (int k = j + 1; k < n; k++) A[check][k] *= r;
            b[check] *= r;
            A[check][j] = T(1);
            for (int i = 0; i < m; i++) {
                if (i == check) continue;
                if (!eq(A[i][j], 0)) {
                    for (int k = j + 1; k < n; k++) A[i][k] -= A[i][j] * A[check][k];
                    b[i] -= A[i][j] * b[check];
                }
                A[i][j] = T(0);
            }
            if (++check == m) break;
        }
        return make_pair(rank, det);
    }

    pair<int, T> row_reduction() {
        vector<T> b(height(), T(0));
        return row_reduction(b);
    }

    // 行基本変形を行い、逆行列を求める
    Matrix inverse() {
        if (height() != width()) return Matrix(0, 0);
        int n = height();
        Matrix ret = I(n);
        for (int j = 0; j < n; j++) {
            int pivot = j;
            for (int i = j; i < n; i++) {
                if (A[i][j] != 0) pivot = i;
                // if(abs(A[i][j]) > abs(A[pivot][j])) pivot = i; // T が小数の場合はこちら
            }
            swap(A[j], A[pivot]), swap(ret[j], ret[pivot]);
            if (eq(A[j][j], T(0))) return Matrix(0, 0);
            T r = T(1) / A[j][j];
            for (int k = j + 1; k < n; k++) A[j][k] *= r;
            for (int k = 0; k < n; k++) ret[j][k] *= r;
            A[j][j] = T(1);
            for (int i = 0; i < n; i++) {
                if (i == j) continue;
                if (!eq(A[i][j], T(0))) {
                    for (int k = j + 1; k < n; k++) A[i][k] -= A[i][j] * A[j][k];
                    for (int k = 0; k < n; k++) ret[i][k] -= A[i][j] * ret[j][k];
                }
                A[i][j] = T(0);
            }
        }
        return ret;
    }

    // Ax = b の解の 1 つと解空間の基底の組を返す
    vector<vector<T>> Gausiann_elimination(vector<T> b) {
        int m = height(), n = width();
        row_reduction(b);
        vector<vector<T>> ret;
        vector<int> p(m, n);
        vector<bool> is_zero(n, true);
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (!eq(A[i][j], T(0))) {
                    p[i] = j;
                    break;
                }
            }
            if (p[i] < n)
                is_zero[p[i]] = false;
            else if (!eq(b[i], T(0)))
                return {};
        }
        vector<T> x(n, T(0));
        for (int i = 0; i < m; i++) {
            if (p[i] < n) x[p[i]] = b[i];
        }
        ret.push_back(x);
        for (int j = 0; j < n; j++) {
            if (!is_zero[j]) continue;
            x[j] = T(1);
            for (int i = 0; i < m; i++) {
                if (p[i] < n) x[p[i]] = -A[i][j];
            }
            ret.push_back(x), x[j] = T(0);
        }
        return ret;
    }
};

using mat = Matrix<mint>;

int main() {
    int K = 10000;

    string S;
    ll M, L;
    cin >> S >> M >> L;

    vector<vector<int>> next(61, vector<int>(K));
    rep(i, K) {
        mat A(2, 2);
        A[0][0] = i;
        A[0][1] = 1;
        A[1][0] = 1;
        A = A.pow(M);
        mint x = A[1][0];
        if (M % 2 == 1) x--;
        next[0][i] = x.x;
    }

    rep(i, 60) {
        rep(j, K) {
            next[i + 1][j] = next[i][next[i][j]]; //
        }
    }

    int s = stoi(S);
    rep(i, 61) {
        if (flg(L, i)) s = next[i][s];
    }

    string T = to_string(s);
    while (sz(T) < 4) T = '0' + T;
    cout << T << '\n';
}
0