結果

問題 No.2645 Sum of Divisors?
ユーザー miscalcmiscalc
提出日時 2024-02-19 23:03:57
言語 C++17
(gcc 12.3.0 + boost 1.83.0)
結果
AC  
実行時間 54 ms / 2,000 ms
コード長 2,917 bytes
コンパイル時間 3,662 ms
コンパイル使用メモリ 264,676 KB
実行使用メモリ 19,176 KB
最終ジャッジ日時 2024-02-19 23:04:04
合計ジャッジ時間 5,330 ms
ジャッジサーバーID
（参考情報） 		judge13 / judge16
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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 9 ms
19,176 KB
testcase_01 AC 8 ms
19,176 KB
testcase_02 AC 26 ms
19,176 KB
testcase_03 AC 9 ms
19,176 KB
testcase_04 AC 8 ms
19,176 KB
testcase_05 AC 8 ms
19,176 KB
testcase_06 AC 52 ms
19,176 KB
testcase_07 AC 54 ms
19,176 KB
testcase_08 AC 10 ms
19,176 KB
testcase_09 AC 9 ms
19,176 KB
testcase_10 AC 8 ms
19,176 KB
testcase_11 AC 8 ms
19,176 KB
testcase_12 AC 8 ms
19,176 KB
testcase_13 AC 8 ms
19,176 KB
testcase_14 AC 8 ms
19,176 KB
testcase_15 AC 9 ms
19,176 KB
testcase_16 AC 9 ms
19,176 KB
testcase_17 AC 9 ms
19,176 KB
testcase_18 AC 9 ms
19,176 KB
testcase_19 AC 9 ms
19,176 KB
testcase_20 AC 9 ms
19,176 KB
testcase_21 AC 10 ms
19,176 KB
testcase_22 AC 9 ms
19,176 KB
testcase_23 AC 9 ms
19,176 KB
testcase_24 AC 13 ms
19,176 KB
testcase_25 AC 9 ms
19,176 KB
testcase_26 AC 10 ms
19,176 KB
testcase_27 AC 11 ms
19,176 KB
testcase_28 AC 17 ms
19,176 KB
testcase_29 AC 50 ms
19,176 KB
testcase_30 AC 17 ms
19,176 KB
testcase_31 AC 25 ms
19,176 KB
testcase_32 AC 38 ms
19,176 KB
testcase_33 AC 19 ms
19,176 KB
testcase_34 AC 34 ms
19,176 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 # 

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pll = pair<ll, ll>;
using tlll = tuple<ll, ll, ll>;
constexpr ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll safemod(ll A, ll M) {ll res = A % M; if (res < 0) res += M; return res;}
ll divfloor(ll A, ll B) {if (B < 0) A = -A, B = -B; return (A - safemod(A, B)) / B;}
ll divceil(ll A, ll B) {if (B < 0) A = -A, B = -B; return divfloor(A + B - 1, B);}
ll pow_ll(ll A, ll B) {if (A == 0 || A == 1) {return A;} if (A == -1) {return B & 1 ? -1 : 1;} ll res = 1; for (int i = 0; i < B; i++) {res *= A;} return res;}
ll mul_limited(ll A, ll B, ll M = INF) { return B == 0 ? 0 : A > M / B ? M : A * B; }
ll pow_limited(ll A, ll B, ll M = INF) { if (A == 0 || A == 1) {return A;} ll res = 1; for (int i = 0; i < B; i++) {if (res > M / A) return M; res *= A;} return res;}
template<class T> void unique(vector<T> &V) {V.erase(unique(V.begin(), V.end()), V.end());}
template<class T> void sortunique(vector<T> &V) {sort(V.begin(), V.end()); V.erase(unique(V.begin(), V.end()), V.end());}
#define FINALANS(A) do {cout << (A) << '\n'; exit(0);} while (false)
//*
#include <atcoder/all>
using namespace atcoder;
template <class T, internal::is_static_modint_t<T> * = nullptr>
istream &operator>>(istream &is, T &a) { ll v; is >> v; a = v; return is; }
template <class T, internal::is_dynamic_modint_t<T> * = nullptr>
istream &operator>>(istream &is, T &a) { ll v; is >> v; a = v; return is; }
template <class T, internal::is_static_modint_t<T> * = nullptr>
ostream &operator<<(ostream &os, const T &a) { os << a.val(); return os; }
template <class T, internal::is_dynamic_modint_t<T> * = nullptr>
ostream &operator<<(ostream &os, const T &a) { os << a.val(); return os; }
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = modint;
//*/
template<class T> void printvec(const vector<T> &V) {int _n = V.size(); for (int i = 0; i < _n; i++) cout << V[i] << (i == _n - 1 ? "" : " "); cout << '\n';}
template<class T> void printvect(const vector<T> &V) {for (auto v : V) cout << v << '\n';}
template<class T> void printvec2(const vector<vector<T>> &V) {for (auto &v : V) printvec(v);}

int main()
{
  ll N;
  cin >> N;
  const ll M = 1000010;

  vector<ld> is(M + 1, 0);
  for (ll i = M; i >= 1; i--)
  {
    is.at(M) += 1 / ld(i);
  }
  for (ll i = M - 1; i >= 0; i--)
  {
    is.at(i) = is.at(i + 1) - 1 / ld(i + 1);
  }
  auto inv_sum = [&](ll n) -> ld
  {
    if (n <= M)
      return is.at(n);
    return is.at(M) + logl(n) - logl(M);
  };

  ld ans = 0;
  for (ll a = 1; a * a <= N; a++)
  {
    ans += 2 * inv_sum(N / a) / a;
  }
  ans -= pow(inv_sum(sqrtl(N)), 2);
  cout << fixed << setprecision(20) << ans << endl;
}
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