結果
問題 | No.1322 Totient Bound |
ユーザー |
![]() |
提出日時 | 2020-12-20 05:45:17 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 11,422 bytes |
コンパイル時間 | 2,487 ms |
コンパイル使用メモリ | 212,116 KB |
最終ジャッジ日時 | 2025-01-17 04:30:39 |
ジャッジサーバーID (参考情報) |
judge5 / judge3 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 17 WA * 2 TLE * 17 |
ソースコード
#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 >= 201703Ltemplate <typename... T> istream &operator>>(istream &is, tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); returnis; }template <typename... T> ostream &operator<<(ostream &os, const tuple<T...> &tpl) { std::apply([&os](auto &&... args) { ((os << args << ','), ...);},tpl); return os; }#endiftemplate <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_LOCALconst 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) {}#endifstruct CountPrimes {// Count Primes less than 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 vsstd::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 obtainedstd::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} // <= 1e64};inline int get_id(long long n) {if (n < 291831)return 0;else if (n < 1050535501)return 1;else if (n < 47636622961201)return 2;elsereturn 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 querystruct {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)]) {a = modpow(a, (n - 1) >> s, n);bool may_composite = true;if (!a) return 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("1");return 0;}CountPrimes p(N);const auto &xs = p.vs;auto ps = p.primes;while (ps.back() > N + 1 or (ps.back() - 1) * ps.back() > N) ps.pop_back();reverse(ps.begin(), ps.end());vector<lint> dp(xs.size());lint plast = ps[0], less_cnt = CountPrimes(plast).pi[0];while (plast <= N) {const lint d = N / (plast - 1 + 1);lint cnt = p.pi[d - 1] + is_prime(N / d + 1);dp[p.getidx(N / plast)] += cnt - less_cnt;less_cnt = cnt;plast = N / d + 1;}dp[0] += 1;for (auto pr : ps) {IREP(i, dp.size()) {const auto v = dp[i];for (int now = p.getidx(xs[i] / (pr - 1)); now < p.s; now = p.getidx(xs[now] / pr)) dp[now] += v;}}cout << accumulate(dp.begin(), dp.end(), 0LL) << '\n';}