結果

問題 No.1491 銀将
ユーザー 👑 hitonanodehitonanode
提出日時 2021-04-30 22:43:32
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 7 ms / 1,000 ms
コード長 18,203 bytes
コンパイル時間 4,146 ms
コンパイル使用メモリ 252,532 KB
実行使用メモリ 4,384 KB
最終ジャッジ日時 2023-09-26 07:14:48
合計ジャッジ時間 4,909 ms
ジャッジサーバーID
(参考情報)
judge12 / judge11
このコードへのチャレンジ(β)

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
4,380 KB
testcase_01 AC 6 ms
4,380 KB
testcase_02 AC 6 ms
4,380 KB
testcase_03 AC 6 ms
4,376 KB
testcase_04 AC 6 ms
4,380 KB
testcase_05 AC 6 ms
4,376 KB
testcase_06 AC 6 ms
4,380 KB
testcase_07 AC 6 ms
4,376 KB
testcase_08 AC 6 ms
4,380 KB
testcase_09 AC 6 ms
4,380 KB
testcase_10 AC 6 ms
4,376 KB
testcase_11 AC 6 ms
4,380 KB
testcase_12 AC 6 ms
4,380 KB
testcase_13 AC 6 ms
4,384 KB
testcase_14 AC 6 ms
4,380 KB
testcase_15 AC 6 ms
4,380 KB
testcase_16 AC 6 ms
4,380 KB
testcase_17 AC 7 ms
4,376 KB
testcase_18 AC 6 ms
4,376 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;
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; }
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
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> sort_unique(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; }
template <typename T, size_t sz> ostream &operator<<(ostream &os, const array<T, sz> &arr) { os << '['; for (auto v : arr) 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
const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";
#define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl
#else
#define dbg(x) (x)
#endif

template <int mod> struct ModInt {
#if __cplusplus >= 201402L
#define MDCONST constexpr
#else
#define MDCONST
#endif
    using lint = long long;
    MDCONST static int get_mod() { return mod; }
    static int get_primitive_root() {
        static int primitive_root = 0;
        if (!primitive_root) {
            primitive_root = [&]() {
                std::set<int> fac;
                int v = mod - 1;
                for (lint i = 2; i * i <= v; i++)
                    while (v % i == 0) fac.insert(i), v /= i;
                if (v > 1) fac.insert(v);
                for (int g = 1; g < mod; g++) {
                    bool ok = true;
                    for (auto i : fac)
                        if (ModInt(g).pow((mod - 1) / i) == 1) {
                            ok = false;
                            break;
                        }
                    if (ok) return g;
                }
                return -1;
            }();
        }
        return primitive_root;
    }
    int val;
    MDCONST ModInt() : val(0) {}
    MDCONST ModInt &_setval(lint v) { return val = (v >= mod ? v - mod : v), *this; }
    MDCONST ModInt(lint v) { _setval(v % mod + mod); }
    MDCONST explicit operator bool() const { return val != 0; }
    MDCONST ModInt operator+(const ModInt &x) const { return ModInt()._setval((lint)val + x.val); }
    MDCONST ModInt operator-(const ModInt &x) const { return ModInt()._setval((lint)val - x.val + mod); }
    MDCONST ModInt operator*(const ModInt &x) const { return ModInt()._setval((lint)val * x.val % mod); }
    MDCONST ModInt operator/(const ModInt &x) const { return ModInt()._setval((lint)val * x.inv() % mod); }
    MDCONST ModInt operator-() const { return ModInt()._setval(mod - val); }
    MDCONST ModInt &operator+=(const ModInt &x) { return *this = *this + x; }
    MDCONST ModInt &operator-=(const ModInt &x) { return *this = *this - x; }
    MDCONST ModInt &operator*=(const ModInt &x) { return *this = *this * x; }
    MDCONST ModInt &operator/=(const ModInt &x) { return *this = *this / x; }
    friend MDCONST ModInt operator+(lint a, const ModInt &x) { return ModInt()._setval(a % mod + x.val); }
    friend MDCONST ModInt operator-(lint a, const ModInt &x) { return ModInt()._setval(a % mod - x.val + mod); }
    friend MDCONST ModInt operator*(lint a, const ModInt &x) { return ModInt()._setval(a % mod * x.val % mod); }
    friend MDCONST ModInt operator/(lint a, const ModInt &x) { return ModInt()._setval(a % mod * x.inv() % mod); }
    MDCONST bool operator==(const ModInt &x) const { return val == x.val; }
    MDCONST bool operator!=(const ModInt &x) const { return val != x.val; }
    MDCONST bool operator<(const ModInt &x) const { return val < x.val; } // To use std::map<ModInt, T>
    friend std::istream &operator>>(std::istream &is, ModInt &x) {
        lint t;
        return is >> t, x = ModInt(t), is;
    }
    MDCONST friend std::ostream &operator<<(std::ostream &os, const ModInt &x) { return os << x.val; }
    MDCONST ModInt pow(lint n) const {
        lint ans = 1, tmp = this->val;
        while (n) {
            if (n & 1) ans = ans * tmp % mod;
            tmp = tmp * tmp % mod, n /= 2;
        }
        return ans;
    }

    static std::vector<long long> facs, invs;
    MDCONST static void _precalculation(int N) {
        int l0 = facs.size();
        if (N <= l0) return;
        facs.resize(N), invs.resize(N);
        for (int i = l0; i < N; i++) facs[i] = facs[i - 1] * i % mod;
        long long facinv = ModInt(facs.back()).pow(mod - 2).val;
        for (int i = N - 1; i >= l0; i--) {
            invs[i] = facinv * facs[i - 1] % mod;
            facinv = facinv * i % mod;
        }
    }
    MDCONST lint inv() const {
        if (this->val < std::min(mod >> 1, 1 << 21)) {
            while (this->val >= int(facs.size())) _precalculation(facs.size() * 2);
            return invs[this->val];
        } else {
            return this->pow(mod - 2).val;
        }
    }
    MDCONST ModInt fac() const {
        while (this->val >= int(facs.size())) _precalculation(facs.size() * 2);
        return facs[this->val];
    }

    MDCONST ModInt doublefac() const {
        lint k = (this->val + 1) / 2;
        return (this->val & 1) ? ModInt(k * 2).fac() / (ModInt(2).pow(k) * ModInt(k).fac())
                               : ModInt(k).fac() * ModInt(2).pow(k);
    }
    MDCONST ModInt nCr(const ModInt &r) const {
        return (this->val < r.val) ? 0 : this->fac() / ((*this - r).fac() * r.fac());
    }

    ModInt sqrt() const {
        if (val == 0) return 0;
        if (mod == 2) return val;
        if (pow((mod - 1) / 2) != 1) return 0;
        ModInt b = 1;
        while (b.pow((mod - 1) / 2) == 1) b += 1;
        int e = 0, m = mod - 1;
        while (m % 2 == 0) m >>= 1, e++;
        ModInt x = pow((m - 1) / 2), y = (*this) * x * x;
        x *= (*this);
        ModInt z = b.pow(m);
        while (y != 1) {
            int j = 0;
            ModInt t = y;
            while (t != 1) j++, t *= t;
            z = z.pow(1LL << (e - j - 1));
            x *= z, z *= z, y *= z;
            e = j;
        }
        return ModInt(std::min(x.val, mod - x.val));
    }
};
template <int mod> std::vector<long long> ModInt<mod>::facs = {1};
template <int mod> std::vector<long long> ModInt<mod>::invs = {0};


// Berlekamp–Massey algorithm
// https://en.wikipedia.org/wiki/Berlekamp%E2%80%93Massey_algorithm
// Complexity: O(N^2)
// input: S = sequence from field K
// return: L          = degree of minimal polynomial,
//         C_reversed = monic min. polynomial (size = L + 1, reversed order, C_reversed[0] = 1))
// Formula: convolve(S, C_reversed)[i] = 0 for i >= L
// Example:
// - [1, 2, 4, 8, 16]   -> (1, [1, -2])
// - [1, 1, 2, 3, 5, 8] -> (2, [1, -1, -1])
// - [0, 0, 0, 0, 1]    -> (5, [1, 0, 0, 0, 0, 998244352]) (mod 998244353)
// - []                 -> (0, [1])
// - [0, 0, 0]          -> (0, [1])
// - [-2]               -> (1, [1, 2])
template <typename Tfield> std::pair<int, std::vector<Tfield>> linear_recurrence(const std::vector<Tfield> &S) {
    int N = S.size();
    using poly = std::vector<Tfield>;
    poly C_reversed{1}, B{1};
    int L = 0, m = 1;
    Tfield b = 1;

    // adjust: C(x) <- C(x) - (d / b) x^m B(x)
    auto adjust = [](poly C, const poly &B, Tfield d, Tfield b, int m) -> poly {
        C.resize(std::max(C.size(), B.size() + m));
        Tfield a = d / b;
        for (unsigned i = 0; i < B.size(); i++) C[i + m] -= a * B[i];
        return C;
    };

    for (int n = 0; n < N; n++) {
        Tfield d = S[n];
        for (int i = 1; i <= L; i++) d += C_reversed[i] * S[n - i];

        if (d == 0)
            m++;
        else if (2 * L <= n) {
            poly T = C_reversed;
            C_reversed = adjust(C_reversed, B, d, b, m);
            L = n + 1 - L;
            B = T;
            b = d;
            m = 1;
        } else
            C_reversed = adjust(C_reversed, B, d, b, m++);
    }
    return std::make_pair(L, C_reversed);
}


// Calculate ^N \bmod f(x)$
// Known as `Kitamasa method`
// Input: f_reversed: monic, reversed (f_reversed[0] = 1)
// Complexity: (K^2 \log N)$ ($: deg. of $)
// Example: (4, [1, -1, -1]) -> [2, 3]
//          ( x^4 = (x^2 + x + 2)(x^2 - x - 1) + 3x + 2 )
// Reference: http://misawa.github.io/others/fast_kitamasa_method.html
//            http://sugarknri.hatenablog.com/entry/2017/11/18/233936
template <typename Tfield>
std::vector<Tfield> monomial_mod_polynomial(long long N, const std::vector<Tfield> &f_reversed) {
    assert(!f_reversed.empty() and f_reversed[0] == 1);
    int K = f_reversed.size() - 1;
    if (!K) return {};
    int D = 64 - __builtin_clzll(N);
    std::vector<Tfield> ret(K, 0);
    ret[0] = 1;
    auto self_conv = [](std::vector<Tfield> x) -> std::vector<Tfield> {
        int d = x.size();
        std::vector<Tfield> ret(d * 2 - 1);
        for (int i = 0; i < d; i++) {
            ret[i * 2] += x[i] * x[i];
            for (int j = 0; j < i; j++) ret[i + j] += x[i] * x[j] * 2;
        }
        return ret;
    };
    for (int d = D; d--;) {
        ret = self_conv(ret);
        for (int i = 2 * K - 2; i >= K; i--) {
            for (int j = 1; j <= K; j++) ret[i - j] -= ret[i] * f_reversed[j];
        }
        ret.resize(K);
        if ((N >> d) & 1) {
            std::vector<Tfield> c(K);
            c[0] = -ret[K - 1] * f_reversed[K];
            for (int i = 1; i < K; i++) { c[i] = ret[i - 1] - ret[K - 1] * f_reversed[K - i]; }
            ret = c;
        }
    }
    return ret;
}

// Find k-th element of the sequence, assuming linear recurrence
// initial_elements: 0-ORIGIN
// Verify: abc198f https://atcoder.jp/contests/abc198/submissions/21837815
template <typename Tfield> Tfield find_kth_element(const std::vector<Tfield> &initial_elements, long long k) {
    assert(k >= 0);
    if (k < static_cast<long long>(initial_elements.size())) return initial_elements[k];
    const auto f = linear_recurrence<Tfield>(initial_elements).second;
    const auto g = monomial_mod_polynomial<Tfield>(k, f);
    Tfield ret = 0;
    for (unsigned i = 0; i < g.size(); i++) ret += g[i] * initial_elements[i];
    return ret;
}

constexpr int mod1 = 1000000009;
constexpr int mod2 = 1000000007;
constexpr int mod3 = 998244353;
using mint = ModInt<mod1>;
using mint2 = ModInt<mod2>;
using mint3 = ModInt<mod3>;

template <typename Int> Int power(Int x, Int n, Int MOD) {
    Int ans = 1;
    while (n > 0) {
        if (n & 1) (ans *= x) %= MOD;
        (x *= x) %= MOD;
        n >>= 1;
    }
    return ans;
}


// Solve ax+by=gcd(a, b)
template <typename Int> Int extgcd(Int a, Int b, Int &x, Int &y) {
    Int d = a;
    if (b != 0) {
        d = extgcd(b, a % b, y, x), y -= (a / b) * x;
    } else {
        x = 1, y = 0;
    }
    return d;
}
// Calculate a^(-1) (MOD m) s if gcd(a, m) == 1
// Calculate x s.t. ax == gcd(a, m) MOD m
template <typename Int> Int mod_inverse(Int a, Int m) {
    Int x, y;
    extgcd<Int>(a, m, x, y);
    x %= m;
    return x + (x < 0) * m;
}

// Require: 1 <= b
// return: (g, x) s.t. g = gcd(a, b), xa = g MOD b, 0 <= x < b/g
template <typename Int> constexpr std::pair<Int, Int> inv_gcd(Int a, Int b) {
    a %= b;
    if (a < 0) a += b;
    if (a == 0) return {b, 0};
    Int s = b, t = a, m0 = 0, m1 = 1;
    while (t) {
        Int u = s / t;
        s -= t * u, m0 -= m1 * u;
        auto tmp = s;
        s = t, t = tmp, tmp = m0, m0 = m1, m1 = tmp;
    }
    if (m0 < 0) m0 += b / s;
    return {s, m0};
}

template <typename Int> constexpr std::pair<Int, Int> crt(const std::vector<Int> &r, const std::vector<Int> &m) {
    assert(r.size() == m.size());
    int n = int(r.size());
    // Contracts: 0 <= r0 < m0
    Int r0 = 0, m0 = 1;
    for (int i = 0; i < n; i++) {
        assert(1 <= m[i]);
        Int r1 = r[i] % m[i], m1 = m[i];
        if (r1 < 0) r1 += m1;
        if (m0 < m1) {
            std::swap(r0, r1);
            std::swap(m0, m1);
        }
        if (m0 % m1 == 0) {
            if (r0 % m1 != r1) return {0, 0};
            continue;
        }
        Int g, im;
        std::tie(g, im) = inv_gcd<Int>(m0, m1);

        Int u1 = m1 / g;
        if ((r1 - r0) % g) return {0, 0};

        Int x = (r1 - r0) / g % u1 * im % u1;
        r0 += x * m0;
        m0 *= u1;
        if (r0 < 0) r0 += m0;
    }
    return {r0, m0};
}

// 蟻本 P.262
// 中国剰余定理を利用して,色々な素数で割った余りから元の値を復元
// 連立線形合同式 A * x = B mod M の解
// Requirement: M[i] > 0
// Output: x = first MOD second (if solution exists), (0, 0) (otherwise)
template <typename Int>
std::pair<Int, Int> linear_congruence(const std::vector<Int> &A, const std::vector<Int> &B, const std::vector<Int> &M) {
    Int r = 0, m = 1;
    assert(A.size() == M.size());
    assert(B.size() == M.size());
    for (int i = 0; i < (int)A.size(); i++) {
        assert(M[i] > 0);
        const Int ai = A[i] % M[i];
        Int a = ai * m, b = B[i] - ai * r, d = std::__gcd(M[i], a);
        if (b % d != 0) {
            return std::make_pair(0, 0); // 解なし
        }
        Int t = b / d * mod_inverse<Int>(a / d, M[i] / d) % (M[i] / d);
        r += m * t;
        m *= M[i] / d;
    }
    return std::make_pair((r < 0 ? r + m : r), m);
}


__int128 str2i128(std::string str) {
    __int128 ret = 0;
    bool minus = false;
    for (auto c : str) {
        if (c == '-')
            minus = true;
        else
            ret = ret * 10 + c - '0';
    }
    return minus ? -ret : ret;
}
std::istream &operator>>(std::istream &is, __int128 &x) {
    std::string s;
    return is >> s, x = str2i128(s), is;
}
std::ostream &operator<<(std::ostream &os, const __int128 &x) {
    __int128 tmp = x;
    if (tmp == 0) return os << 0;
    std::vector<int> ds;
    if (tmp < 0) {
        os << '-';
        while (tmp) {
            int d = tmp % 10;
            if (d > 0) d -= 10;
            ds.emplace_back(-d), tmp = (tmp - d) / 10;
        }
    } else {
        while (tmp) ds.emplace_back(tmp % 10), tmp /= 10;
    }
    std::reverse(ds.begin(), ds.end());
    for (auto i : ds) os << i;
    return os;
}


int main() {
    lint K;
    cin >> K;
    constexpr int W = 200;
    using BS = bitset<W * 2 + 1>;
    vector<BS> bs(W * 2 + 1);
    bs[W][W] = 1;
    vector<mint> seq;
    vector<mint2> seq2;
    vector<mint3> seq3;
    REP(t, 100) {
        vector<BS> nxt = bs;
        FOR(i, 1 ,bs.size()) {
            nxt[i - 1] |= bs[i] | (bs[i] << 1) | (bs[i] >> 1);
            nxt[i + 1] |= (bs[i] << 1) | (bs[i] >> 1);
        }
        bs = nxt;
        int cnt = 0;
        for (auto b : bs) cnt += b.count();
        seq.emplace_back(cnt);
        seq2.emplace_back(cnt);
        seq3.emplace_back(cnt);
    }
    __int128 m1 = find_kth_element(seq, K - 1).val;
    __int128 m2 = find_kth_element(seq2, K - 1).val;
    __int128 m3 = find_kth_element(seq3, K - 1).val;
    cout << linear_congruence<__int128>({1, 1, 1}, {m1, m2, m3}, {mod1, mod2, mod3}).first << '\n';
}
0