結果
問題 | No.1491 銀将 |
ユーザー | hitonanode |
提出日時 | 2021-04-30 22:43:32 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 29 ms / 1,000 ms |
コード長 | 18,203 bytes |
コンパイル時間 | 3,863 ms |
コンパイル使用メモリ | 246,208 KB |
最終ジャッジ日時 | 2025-01-21 04:03:39 |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 16 |
ソースコード
#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 >= 201703Ltemplate <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); returnis; }template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);},tpl); return os; }#endiftemplate <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_LOCALconst 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)#endiftemplate <int mod> struct ModInt {#if __cplusplus >= 201402L#define MDCONST constexpr#else#define MDCONST#endifusing 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;} elseC_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/2017/11/18/233936template <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/21837815template <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 mtemplate <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/gtemplate <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 < m0Int 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;elseret = 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';}