結果

問題 No.1581 Multiple Sequence
ユーザー stoqstoq
提出日時 2022-02-26 14:42:35
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 97 ms / 2,000 ms
コード長 5,455 bytes
コンパイル時間 4,852 ms
コンパイル使用メモリ 267,504 KB
実行使用メモリ 13,824 KB
最終ジャッジ日時 2024-07-04 10:51:07
合計ジャッジ時間 6,471 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge2
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 7 ms
6,016 KB
testcase_01 AC 8 ms
6,016 KB
testcase_02 AC 74 ms
12,544 KB
testcase_03 AC 83 ms
13,056 KB
testcase_04 AC 25 ms
8,448 KB
testcase_05 AC 93 ms
13,312 KB
testcase_06 AC 39 ms
9,728 KB
testcase_07 AC 24 ms
8,320 KB
testcase_08 AC 31 ms
8,960 KB
testcase_09 AC 80 ms
12,544 KB
testcase_10 AC 52 ms
10,752 KB
testcase_11 AC 15 ms
7,424 KB
testcase_12 AC 33 ms
9,216 KB
testcase_13 AC 8 ms
6,144 KB
testcase_14 AC 84 ms
13,056 KB
testcase_15 AC 60 ms
11,392 KB
testcase_16 AC 17 ms
7,424 KB
testcase_17 AC 94 ms
13,696 KB
testcase_18 AC 14 ms
7,168 KB
testcase_19 AC 73 ms
12,288 KB
testcase_20 AC 77 ms
12,416 KB
testcase_21 AC 83 ms
13,056 KB
testcase_22 AC 47 ms
10,368 KB
testcase_23 AC 97 ms
13,824 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/modint>
//#include <atcoder/lazysegtree>
//#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 1
#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;
constexpr double PI = acos(-1.0);
constexpr double EPS = 1e-11;
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 YES(n) cout << ((n) ? "YES" : "NO") << "\n"
#define Yes(n) cout << ((n) ? "Yes" : "No") << "\n"
#define possible(n) cout << ((n) ? "possible" : "impossible") << "\n"
#define Possible(n) cout << ((n) ? "Possible" : "Impossible") << "\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 CEIL(ll a, ll b) { return (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());

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

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

mint dp[100010];
bool used[100010];

mint f(int m) {
  if (m < 0) return 0;
  if (m == 0) return 1;
  if (used[m]) return dp[m];

  mint res = 0;
  for (int i = 1; i * i <= m; i++) {
    if (m % i != 0) continue;
    res += f(i - 1);
    if (i * i != m) res += f(m / i - 1);
  }
  used[m] = true;
  return dp[m] = res;
}

void solve() {
  int m;
  cin >> m;
  cout << f(m) << "\n";
}

int main() { solve(); }
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