結果

問題 No.1810 RGB Biscuits
ユーザー tokusakurai
提出日時 2022-01-14 21:44:17
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 26 ms / 2,000 ms
コード長 10,198 bytes
コンパイル時間 2,261 ms
コンパイル使用メモリ 207,580 KB
最終ジャッジ日時 2025-01-27 11:11:30
ジャッジサーバーID
（参考情報） 		judge5 / judge3
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 * 2
other AC * 20
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 #
プレゼンテーションモードにする

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < n; i++)
#define rep2(i, x, n) for (int i = x; i <= n; i++)
#define rep3(i, x, n) for (int i = x; i >= n; i--)
#define each(e, v) for (auto &e : v)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#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>
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 = 998244353;
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;
    }
    pair<int, T> row_reduction(vector<T> &b) { // 行基本変形を用いて簡約化を行い、(階数、行列式) の組を返す
        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);
    }
    Matrix inverse() { // 行基本変形によって正方行列の逆行列を求める
        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;
    }
    vector<vector<T>> Gausiann_elimination(vector<T> b) { // Ax = b の解の 1 つと解空間の基底の組を返す
        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;
    }
};
int main() {
    mint a, b;
    cin >> a >> b;
    using mat = Matrix<mint>;
    mat A(3, 3), B(3, 3);
    rep(i, 3) A[i][i] = 1;
    A[2][0] = a, A[2][1] = b;
    B[0][2] = 1, B[1][0] = 1;
    mat C = B * A;
    int Q;
    cin >> Q;
    while (Q--) {
        ll K;
        cin >> K;
        mat x(3, 1);
        x[0][0] = 1, x[1][0] = 1;
        mat D = C.pow(K / 2);
        if (K & 1) D = A * D;
        x = D * x;
        cout << x[0][0] + x[1][0] + x[2][0] << '\n';
    }
}
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