結果

問題 No.1322 Totient Bound
ユーザー hitonanode
提出日時 2022-01-01 15:40:12
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 926 ms / 5,000 ms
コード長 13,900 bytes
コンパイル時間 2,242 ms
コンパイル使用メモリ 188,336 KB
実行使用メモリ 11,836 KB
最終ジャッジ日時 2024-10-10 05:36:54
合計ジャッジ時間 17,788 ms
ジャッジサーバーID
(参考情報)
judge4 / judge2
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 36
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 #
プレゼンテーションモードにする

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <chrono>
#include <cmath>
#include <complex>
#include <deque>
#include <forward_list>
#include <fstream>
#include <functional>
#include <iomanip>
#include <ios>
#include <iostream>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <tuple>
#include <type_traits>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
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) { return m < q ? (m = q, true) : false; }
template <typename T> bool chmin(T &m, const T q) { return m > q ? (m = q, true) : 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> int arglb(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::lower_bound(v.begin(), v.end(), x)); }
template <typename T> int argub(const std::vector<T> &v, const T &x) { return std::distance(v.begin(), std::upper_bound(v.begin(), v.end(), x)); }
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) { os << '('; 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
#define dbgif(cond, x) ((cond) ? cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ <<
    COLOR_RESET << endl : cerr)
#else
#define dbg(x) (x)
#define dbgif(cond, x) 0
#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 i
void _fenwick_rec_update(int i, Int cur, bool first) { // pi[n3:] cur * (primes[i] )
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)
// p = 2, ...,
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;
if (N == 1) {
puts("2");
return 0;
}
CountPrimes p(N);
lint npall = p.pi.front() + is_prime(N + 1);
vector<int> p1p;
for (auto x : p.vs) p1p.push_back(is_prime(x + 1));
vector<lint> dp(p.vs.size(), 1);
// dp.back() = 1;
int h = 0;
lint npuse = 0;
vector<int> nxt(p.vs.size(), -1);
vector<int> done(dp.size());
done.back() = 1;
dbgif(N <= 100, p.pi);
dbgif(N <= 100, p.vs);
dbgif(N <= 100, dp);
int plo = 0;
for (int j = p.primes.size() - 1; j >= 0; --j) {
const lint pr = p.primes[j];
const lint pprv = p.primes[j + 1];
if (pr * (pr - 1) > N) continue;
npuse++;
chmax(plo, j + 1);
dbgif(N <= 100, pr);
// int i = dp.size() - 1;
// while (i >= 0 and p.vs[i] * (pr - 1) * pr <= N) --i;
// ++i;
// if (j == 0) i = 0;
// for (int i = int(dp.size()) - 1; i >= 0; --i) {
for (int i = 0; i < int(dp.size()); ++i) {
// if (j and p.vs[i] * (pr - 1) * pr > N) break;
if (j and p.vs[i] < (pr - 1) * pr) break;
if (chmax(done[i], 1)) {
// chmax(np2, lint(plo));
dp[i] += max(p.pi[i] + p1p[i] - j - 1, 0LL);
}
dbgif(N <= 100, i);
dbgif(N <= 100, done);
dbgif(N <= 100, dp);
lint q = p.vs[i] / (pr - 1);
// lint add = dp[i];
lint sum = dp[i];
while (q) {
int idx = p.getidx(q);
if (done[idx]) {
sum += dp[p.getidx(q)];
} else {
sum += max(0LL, p.pi[idx] + p1p[idx] - j - 1) + 1;
}
// dp[p.getidx(q)] += add;
q /= pr;
}
dp[i] = sum;
dbgif(N <= 100, dp);
}
dbgif(N <= 100, dp);
}
dbg(plo);
cout << dp.front() << endl;
// dp.back() = 2;
// dbgif(dp.size() <= 100, dp);
// reverse(ALL(dp));
// cout << accumulate(ALL(dp), 0LL) << endl;
// lint ret = 0;
// REP(i, dp.size()) {
// ret += dp[i] * (max(0LL, p.pi[i] + p1p[i] - npuse) + 1);
// }
// cout << ret << '\n';
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0