結果

問題 No.2365 Present of good number
ユーザー stoqstoq
提出日時 2023-06-30 22:15:24
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 7,079 bytes
コンパイル時間 4,276 ms
コンパイル使用メモリ 284,980 KB
実行使用メモリ 13,696 KB
最終ジャッジ日時 2024-07-07 10:04:26
合計ジャッジ時間 5,947 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge5
このコードへのチャレンジ
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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 24 ms
13,364 KB
testcase_01 AC 23 ms
13,408 KB
testcase_02 WA -
testcase_03 WA -
testcase_04 AC 23 ms
13,344 KB
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 WA -
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 WA -
testcase_14 WA -
testcase_15 WA -
testcase_16 AC 23 ms
13,504 KB
testcase_17 AC 23 ms
13,332 KB
testcase_18 AC 25 ms
13,440 KB
testcase_19 AC 24 ms
13,496 KB
testcase_20 AC 24 ms
13,344 KB
testcase_21 AC 24 ms
13,388 KB
testcase_22 AC 24 ms
13,440 KB
testcase_23 AC 24 ms
13,244 KB
testcase_24 AC 24 ms
13,440 KB
testcase_25 AC 24 ms
13,236 KB
testcase_26 AC 23 ms
13,440 KB
testcase_27 AC 23 ms
13,272 KB
testcase_28 AC 23 ms
13,344 KB
testcase_29 AC 24 ms
13,440 KB
testcase_30 AC 25 ms
13,228 KB
testcase_31 WA -
testcase_32 WA -
testcase_33 WA -
testcase_34 WA -
testcase_35 WA -
testcase_36 AC 24 ms
13,268 KB
testcase_37 AC 24 ms
13,316 KB
testcase_38 AC 24 ms
13,488 KB
testcase_39 AC 24 ms
13,380 KB
testcase_40 AC 24 ms
13,372 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 # 

#define MOD_TYPE 1

#include <bits/stdc++.h>
using namespace std;
#include <atcoder/all>
// #include <atcoder/lazysegtree>
// #include <atcoder/modint>
// #include <atcoder/segtree>
using namespace atcoder;
#if 0
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif
#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/rope>
using namespace __gnu_pbds;
using namespace __gnu_cxx;
template <typename T>
using extset =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
#endif
#if 0
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
#pragma region Macros
using ll = long long int;
using ld = long double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using pld = pair<ld, ld>;
template <typename Q_type>
using smaller_queue = priority_queue<Q_type, vector<Q_type>, greater<Q_type>>;
#if MOD_TYPE == 1
constexpr ll MOD = ll(1e9 + 7);
#else
#if MOD_TYPE == 2
constexpr ll MOD = 998244353;
#else
constexpr ll MOD = 1000003;
#endif
#endif
using mint = static_modint<MOD>;
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
const double PI = acos(-1.0);
constexpr ld EPS = 1e-10;
constexpr int Dx[] = {0, 0, -1, 1, -1, 1, -1, 1, 0};
constexpr int Dy[] = {1, -1, 0, 0, -1, -1, 1, 1, 0};
#define REP(i, m, n) for (ll i = m; i < (ll)(n); ++i)
#define rep(i, n) REP(i, 0, n)
#define REPI(i, m, n) for (int i = m; i < (int)(n); ++i)
#define repi(i, n) REPI(i, 0, n)
#define RREP(i, m, n) for (ll i = n - 1; i >= m; i--)
#define rrep(i, n) RREP(i, 0, n)
#define YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define all(v) v.begin(), v.end()
#define NP(v) next_permutation(all(v))
#define dbg(x) cerr << #x << ":" << x << "\n";
#define UNIQUE(v) v.erase(unique(all(v)), v.end())
struct io_init {
  io_init() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    cout << setprecision(30) << setiosflags(ios::fixed);
  };
} io_init;
template <typename T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
inline ll floor(ll a, ll b) {
  if (b < 0) a *= -1, b *= -1;
  if (a >= 0) return a / b;
  return -((-a + b - 1) / b);
}
inline ll ceil(ll a, ll b) { return floor(a + b - 1, b); }
template <typename A, size_t N, typename T>
inline void Fill(A (&array)[N], const T &val) {
  fill((T *)array, (T *)(array + N), val);
}
template <typename T>
vector<T> compress(vector<T> &v) {
  vector<T> val = v;
  sort(all(val)), val.erase(unique(all(val)), val.end());
  for (auto &&vi : v) vi = lower_bound(all(val), vi) - val.begin();
  return val;
}
template <typename T, typename U>
constexpr istream &operator>>(istream &is, pair<T, U> &p) noexcept {
  is >> p.first >> p.second;
  return is;
}
template <typename T, typename U>
constexpr ostream &operator<<(ostream &os, pair<T, U> p) noexcept {
  os << p.first << " " << p.second;
  return os;
}
ostream &operator<<(ostream &os, mint m) {
  os << m.val();
  return os;
}
ostream &operator<<(ostream &os, modint m) {
  os << m.val();
  return os;
}
template <typename T>
constexpr istream &operator>>(istream &is, vector<T> &v) noexcept {
  for (int i = 0; i < v.size(); i++) is >> v[i];
  return is;
}
template <typename T>
constexpr ostream &operator<<(ostream &os, vector<T> &v) noexcept {
  for (int i = 0; i < v.size(); i++)
    os << v[i] << (i + 1 == v.size() ? "" : " ");
  return os;
}
template <typename T>
constexpr void operator--(vector<T> &v, int) noexcept {
  for (int i = 0; i < v.size(); i++) v[i]--;
}
random_device seed_gen;
mt19937_64 engine(seed_gen());
inline ll randInt(ll l, ll r) { return engine() % (r - l + 1) + l; }
struct BiCoef {
  vector<mint> fact_, inv_, finv_;
  BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
    fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
    for (int i = 2; i < n; i++) {
      fact_[i] = fact_[i - 1] * i;
      inv_[i] = -inv_[MOD % i] * (MOD / i);
      finv_[i] = finv_[i - 1] * inv_[i];
    }
  }
  mint C(ll n, ll k) const noexcept {
    if (n < k || n < 0 || k < 0) return 0;
    return fact_[n] * finv_[k] * finv_[n - k];
  }
  mint P(ll n, ll k) const noexcept { return C(n, k) * fact_[k]; }
  mint H(ll n, ll k) const noexcept { return C(n + k - 1, k); }
  mint Ch1(ll n, ll k) const noexcept {
    if (n < 0 || k < 0) return 0;
    mint res = 0;
    for (int i = 0; i < n; i++)
      res += C(n, i) * mint(n - i).pow(k) * (i & 1 ? -1 : 1);
    return res;
  }
  mint fact(ll n) const noexcept {
    if (n < 0) return 0;
    return fact_[n];
  }
  mint inv(ll n) const noexcept {
    if (n < 0) return 0;
    return inv_[n];
  }
  mint finv(ll n) const noexcept {
    if (n < 0) return 0;
    return finv_[n];
  }
};
BiCoef bc(200010);
#pragma endregion

// -------------------------------

const int MAX_N = 2e6;
int can_div[MAX_N] = {};

struct init_prime {
  init_prime() {
    can_div[1] = -1;
    for (int i = 2; i < MAX_N; i++) {
      if (can_div[i] != 0) continue;
      for (int j = i + i; j < MAX_N; j += i) can_div[j] = i;
    }
  }
} init_prime;

inline bool is_prime(int n) {
  if (n <= 1) return false;
  return !can_div[n];
}

void factorization(int n, map<int, ll> &res) {
  if (n <= 1) return;
  if (!can_div[n]) {
    ++res[n];
    return;
  }
  ++res[can_div[n]];
  factorization(n / can_div[n], res);
}

using mint2 = static_modint<MOD - 1>;
using Matrix = vector<vector<mint2>>;

Matrix E(int n) {
  Matrix res(n, vector<mint2>(n));
  rep(i, n) rep(j, n) {
    if (i == j)
      res[i][j] = 1;
    else
      res[i][j] = 0;
  }
  return res;
}

Matrix operator*(Matrix A, Matrix B) {
  assert(A[0].size() == B.size());
  int l = A.size(), m = A[0].size(), n = B[0].size();
  Matrix C(l, vector<mint2>(n, 0));
  rep(i, l) rep(j, n) { rep(k, m) C[i][j] += A[i][k] * B[k][j]; }
  return C;
}

Matrix matpow(Matrix A, ll n) {
  assert(n >= 0);
  Matrix res = E(A.size()), P = A;
  while (n > 0) {
    if (n & 1) res = res * P;
    P = P * P;
    n >>= 1;
  }
  return res;
}

void solve() {
  int n, k;
  cin >> n >> k;
  map<int, ll> F;
  factorization(n, F);

  rep(_, 50) {
    map<int, ll> F2;
    for (auto [p, e] : F) {
      map<int, ll> Fi;
      factorization(p + 1, Fi);
      for (auto [pi, ei] : Fi) F2[pi] += e * ei;
    }
    swap(F, F2);

    if (_ == k - 1) {
      mint ans = 1;
      for (auto [p, e] : F) {
        ans *= mint(p).pow(e);
      }
      cout << ans << "\n";
      return;
    }
  }
  k -= 50;
  Matrix A = {{0, 2}, {1, 0}};
  A = matpow(A, k);
  Matrix b = {{F[2]}, {F[3]}};
  b = A * b;
  ll x = b[0][0].val();
  ll y = b[1][0].val();
  mint ans = 1;
  ans *= mint(2).pow(x);
  ans *= mint(3).pow(y);
  cout << ans << "\n";
}

int main() { solve(); }
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