結果
問題 | No.1514 Squared Matching |
ユーザー | stoq |
提出日時 | 2021-05-21 22:00:12 |
言語 | C++17 (gcc 12.3.0 + boost 1.83.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 3,400 bytes |
コンパイル時間 | 2,343 ms |
コンパイル使用メモリ | 202,968 KB |
実行使用メモリ | 233,212 KB |
最終ジャッジ日時 | 2024-10-10 08:40:32 |
合計ジャッジ時間 | 11,332 ms |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
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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 | 1,681 ms
233,212 KB |
testcase_01 | TLE | - |
testcase_02 | -- | - |
testcase_03 | -- | - |
testcase_04 | -- | - |
testcase_05 | -- | - |
testcase_06 | -- | - |
testcase_07 | -- | - |
testcase_08 | -- | - |
testcase_09 | -- | - |
testcase_10 | -- | - |
testcase_11 | -- | - |
testcase_12 | -- | - |
testcase_13 | -- | - |
testcase_14 | -- | - |
testcase_15 | -- | - |
testcase_16 | -- | - |
testcase_17 | -- | - |
testcase_18 | -- | - |
testcase_19 | -- | - |
testcase_20 | -- | - |
testcase_21 | -- | - |
testcase_22 | -- | - |
testcase_23 | -- | - |
testcase_24 | -- | - |
testcase_25 | -- | - |
ソースコード
#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] = {}; int sq[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; } for (int i = 1; i * i < MAX_N; i++) { sq[i * i] = i; } } struct init_prime_ { init_prime_() { init_prime(); }; } init_prime_; int f(int n) { bool cnt = 0; int res = n; int prev = 1; while (n > 1) { assert(can_div[n] > 1); if (prev == can_div[n]) { cnt ^= 1; } else { if (cnt and prev != -1) res /= prev; cnt = 1; prev = can_div[n]; } n /= can_div[n]; } if (cnt) { res /= prev; } return sq[res]; } void solve() { int n; cin >> n; ll sum = 0; for (int i = 2; i <= n; i++) { sum += f(i) - 1; } ll ans = sum * 2 + n; cout << ans << "\n"; } int main() { solve(); }