結果

問題 No.1717 Levi-Civita Triangle
コンテスト
ユーザー zer0-star
提出日時 2025-07-15 18:13:03
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 17 ms / 2,000 ms
コード長 4,667 bytes
コンパイル時間 7,031 ms
コンパイル使用メモリ 354,228 KB
実行使用メモリ 7,716 KB
最終ジャッジ日時 2025-07-15 18:13:13
合計ジャッジ時間 8,750 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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 * 3
other AC * 42
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 #
raw source code 

#define USE_ACLIBRARY 0

#if __has_include("all.hpp")

#include "all.hpp"

#else

#include <bits/extc++.h>

#if __has_include(<atcoder/all>) || USE_ACLIBRARY

#include <atcoder/all>

#endif

#endif

using ll = long long int;

using pll = std::pair<ll, ll>;
using pil = std::pair<int, ll>;
using pli = std::pair<ll, int>;
using pii = std::pair<int, int>;

using namespace std::literals;

template <class T>
bool chmin(T &x, const T &val) {
  if (x > val) {
    x = val;
    return true;
  } else {
    return false;
  }
}

template <class T>
bool chmax(T &x, const T &val) {
  if (x < val) {
    x = val;
    return true;
  } else {
    return false;
  }
}

ll isqrt(ll n) {
  assert(n >= 0);
  if (n == 0) return 0;

  uint32_t c = (std::bit_width(uint64_t(n)) - 1) / 2;
  ll a = 1;
  ll d = 0;

  for (int s = std::bit_width(c) - 1; s >= 0; s--) {
    ll e = d;
    d = c >> s;
    a = (a << (d - e - 1)) + (n >> (2 * c - e - d + 1)) / a;
  }

  return a - (a * a > n);
}

#if __has_include(<atcoder/all>) || USE_ACLIBRARY

template <class mint, atcoder::internal::is_static_modint_t<mint> * = nullptr>
std::ostream &operator<<(std::ostream &os, const mint &v) {
  return os << v.val();
}

template <class mint, atcoder::internal::is_static_modint_t<mint> * = nullptr>
std::istream &operator>>(std::istream &is, mint &v) {
  int tmp;
  is >> tmp;
  v = tmp;
  return is;
}

#endif

template <class T, class U>
std::istream &operator>>(std::istream &is, std::pair<T, U> &p) {
  return is >> p.first >> p.second;
}

template <class... T>
std::istream &operator>>(std::istream &is, std::tuple<T...> &tpl) {
  std::apply([&](auto &&...args) { (is >> ... >> args); }, tpl);
  return is;
}

template <class T>
std::istream &operator>>(std::istream &is, std::vector<T> &v) {
  for (T &x : v) is >> x;
  return is;
}

template <class T>
std::ostream &operator<<(std::ostream &os, const std::vector<T> &v) {
  for (size_t i = 0; i < v.size(); i++)
    os << v[i] << (i == v.size() - 1 ? "" : " ");
  return os;
}

struct Initialization {
  Initialization() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
  }
} initialization;

constexpr std::pair<int, int> dir[] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};

template <typename T>
using infs = std::numeric_limits<T>;

template <typename T>
class factorials {
 public:
  static size_t n;
  static std::vector<T> fact, inv_fact;

  static void extend(size_t m) {
    if (m <= n) return;
    fact.resize(m + 1);
    inv_fact.resize(m + 1);
    for (size_t i = n + 1; i <= m; i++) fact[i] = fact[i - 1] * i;
    inv_fact[m] = fact[m].inv();
    for (size_t i = m; i > n + 1; i--) inv_fact[i - 1] = inv_fact[i] * i;
    n = m;
  }

  static T inv(int k) {
    extend(k);
    return inv_fact[k];
  }

  static T get(int k) {
    extend(k);
    return fact[k];
  }

  static T perm(int n, int k) {
    if (n < k) return 0;
    if (k < 0) return 0;
    extend(n);
    return fact[n] * inv_fact[n - k];
  }

  static T choose(int n, int k) {
    if (n < k) return 0;
    if (k < 0) return 0;
    extend(n);
    return fact[n] * inv_fact[n - k] * inv_fact[k];
  }

  static T catalan(int n) { return get(2 * n) * inv(n + 1) * inv(n); }
};

template <typename T>
size_t factorials<T>::n = 0;

template <typename T>
std::vector<T> factorials<T>::fact = {1};

template <typename T>
std::vector<T> factorials<T>::inv_fact = {1};

#if __has_include(<atcoder/all>) || USE_ACLIBRARY

using mint = atcoder::modint998244353;
// using mint = atcoder::modint1000000007;

using fs = factorials<mint>;

#endif

template <typename T>
using pq_rev = std::priority_queue<T, std::vector<T>, std::greater<T>>;

namespace pbds = __gnu_pbds;

template <typename T>
using tree = pbds::tree<T, pbds::null_type, std::less<T>, pbds::rb_tree_tag,
                        pbds::tree_order_statistics_node_update>;

int main() {
  int N;
  std::cin >> N;
  std::vector<int> A(2 * N + 1);
  std::cin >> A;

  auto f = [&](int i) {
    using t = std::tuple<int, int, int>;
    t a = {A[i], A[i + 1], A[i + 2]};
    if (a == t{0, 1, 2} || a == t{1, 2, 0} || a == t{2, 0, 1}) {
      return 1;
    } else if (a == t{0, 2, 1} || a == t{1, 0, 2} || a == t{2, 1, 0}) {
      return 2;
    } else {
      return 0;
    }
  };

  std::vector<int> Q(A.size() - 2);

  for (int k = 0; k < A.size() - 2; k++) {
    Q[k] = f(k);
  }

  bool isok = true;

  for (int i = 0; i < N; i++) {
    if (i & 1) {
      isok = isok && (Q[i] == 0);
    } else if (i == 0) {
    } else {
      isok = isok && Q[i] != 0 && Q[i] != Q[i - 2];
    }

    if (!isok) break;
  }

  if (isok) {
    std::cout << Q.back() << "\n";
  } else {
    std::cout << 0 << "\n";
  }
}
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