結果
| 問題 | No.1322 Totient Bound |
| ユーザー |
👑  hitonanode
|
| 提出日時 | 2021-12-31 09:35:33 |
| 言語 | C++23 (gcc 12.3.0 + boost 1.83.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 11,726 bytes |
| コンパイル時間 | 4,808 ms |
| コンパイル使用メモリ | 309,332 KB |
| 実行使用メモリ | 14,016 KB |
| 最終ジャッジ日時 | 2024-04-16 20:20:08 |
| 合計ジャッジ時間 | 15,134 ms |
|
ジャッジサーバーID （参考情報） |
judge4 / 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 | 2 ms
5,248 KB |
| testcase_01 | AC | 2 ms
5,376 KB |
| testcase_02 | AC | 2 ms
5,376 KB |
| testcase_03 | AC | 2 ms
5,376 KB |
| testcase_04 | AC | 2 ms
5,376 KB |
| testcase_05 | AC | 2 ms
5,376 KB |
| testcase_06 | AC | 2 ms
5,376 KB |
| testcase_07 | AC | 2 ms
5,376 KB |
| testcase_08 | AC | 2 ms
5,376 KB |
| testcase_09 | AC | 2 ms
5,376 KB |
| testcase_10 | AC | 2 ms
5,376 KB |
| testcase_11 | AC | 2 ms
5,376 KB |
| testcase_12 | AC | 2 ms
5,376 KB |
| testcase_13 | AC | 8 ms
5,376 KB |
| testcase_14 | AC | 7 ms
5,376 KB |
| testcase_15 | AC | 4 ms
5,376 KB |
| testcase_16 | AC | 5 ms
5,376 KB |
| testcase_17 | AC | 3 ms
5,376 KB |
| testcase_18 | AC | 3,187 ms
5,376 KB |
| testcase_19 | TLE | - |
| testcase_20 | TLE | - |
| testcase_21 | -- | - |
| testcase_22 | -- | - |
| testcase_23 | -- | - |
| testcase_24 | -- | - |
| testcase_25 | -- | - |
| testcase_26 | -- | - |
| testcase_27 | -- | - |
| testcase_28 | -- | - |
| testcase_29 | -- | - |
| testcase_30 | -- | - |
| testcase_31 | -- | - |
| testcase_32 | -- | - |
| testcase_33 | -- | - |
| testcase_34 | -- | - |
| testcase_35 | -- | - |
| testcase_36 | -- | - |
| testcase_37 | -- | - |
| testcase_38 | -- | - |
ソースコード
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#include <bits/stdc++.h>
using namespace std;
using lint = long long;
using pint = pair<int, int>;
using plint = pair<lint, lint>;
struct fast_ios { fast_ios(){ cin.tie(nullptr), ios::sync_with_stdio(false), cout << fixed << setprecision(20); }; } fast_ios_;
#define ALL(x) (x).begin(), (x).end()
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
template <typename T, typename V>
void ndarray(vector<T>& vec, const V& val, int len) { vec.assign(len, val); }
template <typename T, typename V, typename... Args> void ndarray(vector<T>& vec, const V& val, int len, Args... args) { vec.resize(len), for_each(begin(vec), end(vec), [&](T& v) { ndarray(v, val, args...); }); }
template <typename T> bool chmax(T &m, const T q) { if (m < q) {m = q; return true;} else return false; }
template <typename T> bool chmin(T &m, const T q) { if (m > q) {m = q; return true;} else return false; }
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
template <typename T1, typename T2> pair<T1, T2> operator+(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first + r.first, l.second + r.second); }
template <typename T1, typename T2> pair<T1, T2> operator-(const pair<T1, T2> &l, const pair<T1, T2> &r) { return make_pair(l.first - r.first, l.second - r.second); }
template <typename T> vector<T> sort_unique(vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <typename T> istream &operator>>(istream &is, vector<T> &vec) { for (auto &v : vec) is >> v; return is; }
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T, size_t sz> ostream &operator<<(ostream &os, const array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; }
#if __cplusplus >= 201703L
template <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os; }
#endif
template <typename T> ostream &operator<<(ostream &os, const deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <typename T> ostream &operator<<(ostream &os, const set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T, typename TH> ostream &operator<<(ostream &os, const unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &pa) { os << '(' << pa.first << ',' << pa.second << ')'; return os; }
template <typename TK, typename TV> ostream &operator<<(ostream &os, const map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <typename TK, typename TV, typename TH> ostream &operator<<(ostream &os, const unordered_map<TK, TV, TH> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
#ifdef HITONANODE_LOCAL
const string COLOR_RESET = "\033[0m", BRIGHT_GREEN = "\033[1;32m", BRIGHT_RED = "\033[1;31m", BRIGHT_CYAN = "\033[1;36m", NORMAL_CROSSED = "\033[0;9;37m", RED_BACKGROUND = "\033[1;41m", NORMAL_FAINT = "\033[0;2m";
#define dbg(x) cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << endl
#else
#define dbg(x) {}
#endif
struct CountPrimes {
// Count Primes less than or equal to x (\pi(x)) for each x = N / i (i = 1, ..., N) in O(N^(2/3)) time
// Learned this algorihtm from https://old.yosupo.jp/submission/14650
// Reference: https://min-25.hatenablog.com/entry/2018/11/11/172216
using Int = long long;
Int n, n2, n3, n6;
std::vector<int> is_prime; // [0, 0, 1, 1, 0, 1, 0, 1, ...]
std::vector<Int> primes; // primes up to O(N^(1/2)), [2, 3, 5, 7, ...]
int s; // size of vs
std::vector<Int> vs; // [N, ..., n2, n2 - 1, n2 - 2, ..., 3, 2, 1]
std::vector<Int> pi; // pi[i] = (# of primes s.t. <= vs[i]) is finally obtained
std::vector<int> _fenwick;
int getidx(Int a) const { return a <= n2 ? s - a : n / a - 1; } // vs[i] >= a を満たす最大の i を返す
void _fenwick_rec_update(int i, Int cur, bool first) { // pi[n3:] に対して cur * (primes[i] 以上の素因数) の数の寄与を減じる
if (!first) {
for (int x = getidx(cur) - n3; x >= 0; x -= (x + 1) & (-x - 1)) _fenwick[x]--;
}
for (int j = i; cur * primes[j] <= vs[n3]; j++) _fenwick_rec_update(j, cur * primes[j], false);
}
CountPrimes(Int n_) : n(n_), n2((Int)sqrtl(n)), n3((Int)cbrtl(n)), n6((Int)sqrtl(n3)) {
is_prime.assign(n2 + 300, 1), is_prime[0] = is_prime[1] = 0; // `+ 300`: https://en.wikipedia.org/wiki/Prime_gap
for (size_t p = 2; p < is_prime.size(); p++) {
if (is_prime[p]) {
primes.push_back(p);
for (size_t j = p * 2; j < is_prime.size(); j += p) is_prime[j] = 0;
}
}
for (Int now = n; now; now = n / (n / now + 1))
vs.push_back(now); // [N, N / 2, ..., 1], Relevant integers (decreasing) length ~= 2sqrt(N)
s = vs.size();
// pi[i] = (# of integers x s.t. x <= vs[i], (x is prime or all factors of x >= p))
// pre = (# of primes less than p)
// 最小の素因数 p = 2, ..., について篩っていく
pi.resize(s);
for (int i = 0; i < s; i++) pi[i] = vs[i] - 1;
int pre = 0;
auto trans = [&](int i, Int p) { pi[i] -= pi[getidx(vs[i] / p)] - pre; };
size_t ip = 0;
// [Sieve Part 1] For each prime p satisfying p <= N^(1/6) (Only O(N^(1/6) / log N) such primes exist),
// O(sqrt(N)) simple operation is conducted.
// - Complexity of this part: O(N^(2/3) / logN)
for (; primes[ip] <= n6; ip++, pre++) {
const auto &p = primes[ip];
for (int i = 0; p * p <= vs[i]; i++) trans(i, p);
}
// [Sieve Part 2] For each prime p satisfying N^(1/6) < p <= N^(1/3),
// point-wise & Fenwick tree-based hybrid update is used
// - first N^(1/3) elements are simply updated by quadratic algorithm.
// - Updates of latter segments are managed by Fenwick tree.
// - Complexity of this part: O(N^(2/3)) (O(N^(2/3)/log N) operations for Fenwick tree (O(logN) per query))
_fenwick.assign(s - n3, 0); // Fenwick tree, inversed order (summation for upper region)
auto trans2 = [&](int i, Int p) {
int j = getidx(vs[i] / p);
auto z = pi[j];
if (j >= n3) {
for (j -= n3; j < int(_fenwick.size()); j += (j + 1) & (-j - 1)) z += _fenwick[j];
}
pi[i] -= z - pre;
};
for (; primes[ip] <= n3; ip++, pre++) {
const auto &p = primes[ip];
for (int i = 0; i < n3 and p * p <= vs[i]; i++)
trans2(i, p); // upto n3, total trans2 times: O(N^(2/3) / logN)
_fenwick_rec_update(ip, primes[ip], true); // total update times: O(N^(2/3) / logN)
}
for (int i = s - n3 - 1; i >= 0; i--) {
int j = i + ((i + 1) & (-i - 1));
if (j < s - n3) _fenwick[i] += _fenwick[j];
}
for (int i = 0; i < s - n3; i++) pi[i + n3] += _fenwick[i];
// [Sieve Part 3] For each prime p satisfying N^(1/3) < p <= N^(1/2), use only simple updates.
// - Complexity of this part: O(N^(2/3) / logN)
// \sum_i (# of factors of vs[i] of the form p^2, p >= N^(1/3)) = \sum_{i=1}^{N^(1/3)} \pi(\sqrt(vs[i])))
// = sqrt(N) \sum_i^{N^(1/3)} i^{-1/2} / logN = O(N^(2/3) / logN)
// (Note: \sum_{i=1}^{N} i^{-1/2} = O(sqrt N) https://math.stackexchange.com/questions/2600796/finding-summation-of-inverse-of-square-roots )
for (; primes[ip] <= n2; ip++, pre++) {
const auto &p = primes[ip];
for (int i = 0; p * p <= vs[i]; i++) trans(i, p);
}
}
};
namespace SPRP {
// http://miller-rabin.appspot.com/
const std::vector<std::vector<__int128>> bases{
{126401071349994536}, // < 291831
{336781006125, 9639812373923155}, // < 1050535501 (1e9)
{2, 2570940, 211991001, 3749873356}, // < 47636622961201 (4e13)
{2, 325, 9375, 28178, 450775, 9780504, 1795265022} // <= 2^64
};
inline int get_id(long long n) {
if (n < 291831) {
return 0;
} else if (n < 1050535501) {
return 1;
} else if (n < 47636622961201)
return 2;
else {
return 3;
}
}
} // namespace SPRP
// Miller-Rabin primality test
// https://ja.wikipedia.org/wiki/%E3%83%9F%E3%83%A9%E3%83%BC%E2%80%93%E3%83%A9%E3%83%93%E3%83%B3%E7%B4%A0%E6%95%B0%E5%88%A4%E5%AE%9A%E6%B3%95
// Complexity: O(lg n) per query
struct {
long long modpow(__int128 x, __int128 n, long long mod) noexcept {
__int128 ret = 1;
for (x %= mod; n; x = x * x % mod, n >>= 1) ret = (n & 1) ? ret * x % mod : ret;
return ret;
}
bool operator()(long long n) noexcept {
if (n < 2) return false;
if (n % 2 == 0) return n == 2;
int s = __builtin_ctzll(n - 1);
for (__int128 a : SPRP::bases[SPRP::get_id(n)]) {
if (a % n == 0) continue;
a = modpow(a, (n - 1) >> s, n);
bool may_composite = true;
if (a == 1) continue;
for (int r = s; r--; a = a * a % n) {
if (a == n - 1) may_composite = false;
}
if (may_composite) return false;
}
return true;
}
} is_prime;
int main() {
lint N;
cin >> N;
CountPrimes p(N);
vector<lint> dp(p.vs.size());
dbg(p.vs);
dp[0] = 1;
int h = int(p.vs.size()) - 1;
lint npuse = 0;
for (lint pr : p.primes) {
dbg(pr);
if (pr * (pr - 1) > N) break;
npuse++;
while (h >= 0 and (pr - 1) > p.vs[h]) {
--h;
// lint b = (h + 1 == dp.size() ? 0 : p.pi[h + 1] + is_prime(xs[h + 1] + 1));
// dp[h] = p.pi[h] + is_prime(xs[h] + 1) - b;
// h++;
}
for (int i = h; i >= 0; --i) {
auto tmp = dp[i];
for (lint x = p.vs[i] / (pr - 1); x; x /= pr) dp[p.getidx(x)] += tmp;
}
dbg(dp);
// REP(i, h) {
// lint dpnew = dp[i];
// for (lint x = xs[i] / (pr - 1); x; x /= pr) dpnew += dp[p.getidx(x)];
// dp[i] = dpnew;
// }
// dbg(dp);
}
lint ret = 0;
REP(i, dp.size()) {
ret += dp[i] * (max(0LL, p.pi[i] + is_prime(p.vs[i] + 1) - npuse) + 1);
}
cout << ret << '\n';
}