結果

問題 No.1529 Constant Lcm
コンテスト
ユーザー keijak
提出日時 2021-06-04 20:46:32
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 964 ms / 3,000 ms
コード長 3,652 bytes
コンパイル時間 2,145 ms
コンパイル使用メモリ 206,272 KB
最終ジャッジ日時 2025-01-21 22:30:37
ジャッジサーバーID
（参考情報） 		judge2 / judge5
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 * 24
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 

#include <bits/stdc++.h>
#define REP_(i, a_, b_, a, b, ...) \
  for (int i = (a), END_##i = (b); i < END_##i; ++i)
#define REP(i, ...) REP_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
#define ALL(x) std::begin(x), std::end(x)
using i64 = long long;

#include <atcoder/modint>
using Mint = atcoder::modint998244353;
std::ostream &operator<<(std::ostream &os, const Mint &m) {
  return os << m.val();
}

template <typename T, typename U>
inline bool chmax(T &a, U b) {
  return a < b and ((a = std::move(b)), true);
}
template <typename T, typename U>
inline bool chmin(T &a, U b) {
  return a > b and ((a = std::move(b)), true);
}
template <typename T>
inline int ssize(const T &a) {
  return (int)std::size(a);
}

template <typename T>
std::istream &operator>>(std::istream &is, std::vector<T> &a) {
  for (auto &x : a) is >> x;
  return is;
}
template <typename T, typename U>
std::ostream &operator<<(std::ostream &os, const std::pair<T, U> &a) {
  return os << "(" << a.first << ", " << a.second << ")";
}
template <typename Container>
std::ostream &print_seq(const Container &a, std::string_view sep = " ",
                        std::string_view ends = "\n",
                        std::ostream &os = std::cout) {
  auto b = std::begin(a), e = std::end(a);
  for (auto it = std::begin(a); it != e; ++it) {
    if (it != b) os << sep;
    os << *it;
  }
  return os << ends;
}
template <typename T, typename = void>
struct is_iterable : std::false_type {};
template <typename T>
struct is_iterable<T, std::void_t<decltype(std::begin(std::declval<T>())),
                                  decltype(std::end(std::declval<T>()))>>
    : std::true_type {};

template <typename T, typename = std::enable_if_t<
                          is_iterable<T>::value &&
                          !std::is_same<T, std::string_view>::value &&
                          !std::is_same<T, std::string>::value>>
std::ostream &operator<<(std::ostream &os, const T &a) {
  return print_seq(a, ", ", "", (os << "{")) << "}";
}

#ifdef ENABLE_DEBUG
#include "debug_dump.hpp"
#else
#define DUMP(...)
#endif

using namespace std;

// Returns a vector that maps x to x's smallest prime factor (> 1).
// O(n log log n)
std::vector<int> sieve_smallest_prime_factors(int n) {
  std::vector<int> res(n + 1);
  for (int i = 1; i <= n; ++i) res[i] = i;
  for (int i = 2; i <= n; i += 2) res[i] = 2;
  for (int i = 3; i * i <= n; i += 2) {
    if (res[i] != i) continue;
    for (int j = i * i; j <= n; j += i) {
      if (res[j] == j) res[j] = i;
    }
  }
  return res;
}

// Quick factorization.
map<int, int> quick_factorize(int n, const std::vector<int> &spf) {
  assert(0 < n and n < int(spf.size()));
  map<int, int> res;
  for (;;) {
    const int p = spf[n];
    if (p == 1) break;
    int count = 0;
    do {
      n /= p;
      ++count;
    } while (n % p == 0);
    res[p] = count;
  }
  return res;
}

auto solve() {
  i64 n;
  cin >> n;

  auto spf = sieve_smallest_prime_factors(n);

  auto lf = quick_factorize(n - 1, spf);
  for (i64 i = 2; i < n; ++i) {
    auto f1 = quick_factorize(i, spf);
    auto f2 = quick_factorize(n - i, spf);
    if (f1.size() < f2.size()) {
      swap(f1, f2);
    }
    for (auto [k, v] : f2) {
      f1[k] += v;
    }
    if (lf.size() < f1.size()) {
      swap(lf, f1);
    }
    for (auto [k, v] : f1) {
      auto it = lf.find(k);
      if (it == lf.end()) {
        lf[k] = v;
      } else {
        chmax(it->second, v);
      }
    }
  }
  Mint ans = 1;
  for (auto [k, v] : lf) {
    ans *= Mint(k).pow(v);
  }
  return ans;
}

int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  cout << solve() << "\n";
}
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