結果

問題 No.1514 Squared Matching
ユーザー stoqstoq
提出日時 2021-05-21 22:16:48
言語 C++17
(gcc 13.2.0 + boost 1.83.0)
結果
AC  
実行時間 2,277 ms / 4,000 ms
コード長 3,246 bytes
コンパイル時間 2,332 ms
コンパイル使用メモリ 196,240 KB
実行使用メモリ 394,300 KB
最終ジャッジ日時 2024-04-18 15:47:59
合計ジャッジ時間 53,908 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge4
このコードへのチャレンジ
（要ログイン）

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 1,667 ms
394,176 KB
testcase_01 AC 1,998 ms
394,240 KB
testcase_02 AC 1,681 ms
394,172 KB
testcase_03 AC 1,689 ms
394,172 KB
testcase_04 AC 1,638 ms
394,168 KB
testcase_05 AC 1,659 ms
394,044 KB
testcase_06 AC 1,728 ms
394,112 KB
testcase_07 AC 1,706 ms
394,112 KB
testcase_08 AC 2,006 ms
394,168 KB
testcase_09 AC 1,958 ms
394,112 KB
testcase_10 AC 1,944 ms
393,984 KB
testcase_11 AC 2,050 ms
393,984 KB
testcase_12 AC 2,050 ms
394,112 KB
testcase_13 AC 2,053 ms
394,044 KB
testcase_14 AC 2,025 ms
394,176 KB
testcase_15 AC 2,017 ms
394,240 KB
testcase_16 AC 2,064 ms
394,112 KB
testcase_17 AC 1,983 ms
394,112 KB
testcase_18 AC 2,026 ms
394,040 KB
testcase_19 AC 2,016 ms
394,112 KB
testcase_20 AC 2,018 ms
394,112 KB
testcase_21 AC 2,277 ms
394,172 KB
testcase_22 AC 1,705 ms
394,112 KB
testcase_23 AC 1,787 ms
394,300 KB
testcase_24 AC 1,863 ms
394,112 KB
testcase_25 AC 1,935 ms
394,240 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

#pragma region Macros

#include <bits/stdc++.h>
using namespace std;

#if 0
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
using Int = boost::multiprecision::cpp_int;
using lld = boost::multiprecision::cpp_dec_float_100;
#endif
#if 1
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#endif
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>>;

constexpr ll MOD = (MOD_TYPE == 1 ? (ll)(1e9 + 7) : 998244353);
constexpr int INF = (int)1e9 + 10;
constexpr ll LINF = (ll)4e18;
constexpr ld PI = acos(-1.0);
constexpr ld EPS = 1e-7;
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 MP make_pair
#define MT make_tuple
#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";

struct io_init
{
  io_init()
  {
    cin.tie(0);
    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, 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;
}
#pragma endregion

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

const int MAX_N = 5e7 + 10;
int can_div[MAX_N] = {};

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

struct init_prime_
{
  init_prime_() { init_prime(); };
} init_prime_;

int dp[MAX_N];

int f(int n)
{
  if (n == 1)
    return 1;
  if (dp[n] != -1)
    return dp[n];
  if ((n / can_div[n]) % can_div[n] == 0)
  {
    return dp[n] = f(n / (can_div[n] * can_div[n])) * can_div[n];
  }
  else
  {
    return dp[n] = f(n / can_div[n]);
  }
}

void solve()
{
  Fill(dp, -1);
  int n;
  cin >> n;
  int sum = 0;
  for (int i = 2; i <= n; i++)
  {
    sum += f(i) - 1;
  }
  int ans = sum * 2 + n;
  cout << ans << "\n";
}

int main()
{
  solve();
}
0