結果

問題 No.1322 Totient Bound
ユーザー hitonanodehitonanode
提出日時 2021-12-31 09:53:24
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 17,014 bytes
コンパイル時間 2,502 ms
コンパイル使用メモリ 188,580 KB
実行使用メモリ 17,772 KB
最終ジャッジ日時 2024-10-07 23:49:03
合計ジャッジ時間 13,785 ms
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 17 TLE * 1 -- * 18
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 <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

// Linear sieve algorithm for fast prime factorization
// Complexity: O(N) time, O(N) space:
// - MAXN = 10^7:  ~44 MB,  80~100 ms (Codeforces / AtCoder GCC, C++17)
// - MAXN = 10^8: ~435 MB, 810~980 ms (Codeforces / AtCoder GCC, C++17)
// Reference:
// [1] D. Gries, J. Misra, "A Linear Sieve Algorithm for Finding Prime Numbers,"
//     Communications of the ACM, 21(12), 999-1003, 1978.
// - https://cp-algorithms.com/algebra/prime-sieve-linear.html
// - https://37zigen.com/linear-sieve/
struct Sieve {
    std::vector<int> min_factor;
    std::vector<int> primes;
    Sieve(int MAXN) : min_factor(MAXN + 1) {
        for (int d = 2; d <= MAXN; d++) {
            if (!min_factor[d]) {
                min_factor[d] = d;
                primes.emplace_back(d);
            }
            for (const auto &p : primes) {
                if (p > min_factor[d] or d * p > MAXN) break;
                min_factor[d * p] = p;
            }
        }
    }
    // Prime factorization for 1 <= x <= MAXN^2
    // Complexity: O(log x)           (x <= MAXN)
    //             O(MAXN / log MAXN) (MAXN < x <= MAXN^2)
    template <class T> std::map<T, int> factorize(T x) const {
        std::map<T, int> ret;
        assert(x > 0 and
               x <= ((long long)min_factor.size() - 1) * ((long long)min_factor.size() - 1));
        for (const auto &p : primes) {
            if (x < T(min_factor.size())) break;
            while (!(x % p)) x /= p, ret[p]++;
        }
        if (x >= T(min_factor.size())) ret[x]++, x = 1;
        while (x > 1) ret[min_factor[x]]++, x /= min_factor[x];
        return ret;
    }
    // Enumerate divisors of 1 <= x <= MAXN^2
    // Be careful of highly composite numbers https://oeis.org/A002182/list
    // https://gist.github.com/dario2994/fb4713f252ca86c1254d#file-list-txt (n, (# of div. of n)):
    // 45360->100, 735134400(<1e9)->1344, 963761198400(<1e12)->6720
    template <class T> std::vector<T> divisors(T x) const {
        std::vector<T> ret{1};
        for (const auto p : factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                for (T a = 1, d = 1; d <= p.second; d++) {
                    a *= p.first;
                    ret.push_back(ret[i] * a);
                }
            }
        }
        return ret; // NOT sorted
    }
    // Euler phi functions of divisors of given x
    // Verified: ABC212 G https://atcoder.jp/contests/abc212/tasks/abc212_g
    // Complexity: O(sqrt(x) + d(x))
    template <class T> std::map<T, T> euler_of_divisors(T x) const {
        assert(x >= 1);
        std::map<T, T> ret;
        ret[1] = 1;
        std::vector<T> divs{1};
        for (auto p : factorize(x)) {
            int n = ret.size();
            for (int i = 0; i < n; i++) {
                ret[divs[i] * p.first] = ret[divs[i]] * (p.first - 1);
                divs.push_back(divs[i] * p.first);
                for (T a = divs[i] * p.first, d = 1; d < p.second; a *= p.first, d++) {
                    ret[a * p.first] = ret[a] * p.first;
                    divs.push_back(a * p.first);
                }
            }
        }
        return ret;
    }
    // Moebius function Table, (-1)^{# of different prime factors} for square-free x
    // return: [0=>0, 1=>1, 2=>-1, 3=>-1, 4=>0, 5=>-1, 6=>1, 7=>-1, 8=>0, ...] https://oeis.org/A008683
    std::vector<int> GenerateMoebiusFunctionTable() const {
        std::vector<int> ret(min_factor.size());
        for (unsigned i = 1; i < min_factor.size(); i++) {
            if (i == 1) {
                ret[i] = 1;
            } else if ((i / min_factor[i]) % min_factor[i] == 0) {
                ret[i] = 0;
            } else {
                ret[i] = -ret[i / min_factor[i]];
            }
        }
        return ret;
    }
    // Calculate [0^K, 1^K, ..., nmax^K] in O(nmax)
    // Note: **0^0 == 1**
    template <class MODINT> std::vector<MODINT> enumerate_kth_pows(long long K, int nmax) const {
        assert(nmax < int(min_factor.size()));
        assert(K >= 0);
        if (K == 0) return std::vector<MODINT>(nmax + 1, 1);
        std::vector<MODINT> ret(nmax + 1);
        ret[0] = 0, ret[1] = 1;
        for (int n = 2; n <= nmax; n++) {
            if (min_factor[n] == n) {
                ret[n] = MODINT(n).pow(K);
            } else {
                ret[n] = ret[n / min_factor[n]] * ret[min_factor[n]];
            }
        }
        return ret;
    }
};
Sieve sieve((1 << 20));


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());
    // dp[0] = 1;
    dp.back() = 1;

    int h = 0;
    lint npuse = 0;
    for (lint pr : p.primes) {
        if (pr * (pr - 1) > N) break;
        npuse++;
        while (h < int(p.vs.size()) and (pr - 1) * p.vs[h] > N) ++h;
        for (int i = h; i < int(p.vs.size()); ++i) {
            auto tmp = dp[i];
            for (lint x = p.vs[i] * (pr - 1); x <= N; x *= pr) dp[p.getidx(x)] += tmp;
        }
        dbgif(N <= 10, pr);
        dbgif(N <= 10, dp);
    }
    reverse(ALL(dp));
    lint ret = 0;
    REP(i, dp.size()) {
        auto ispr = [&](lint x) {
            if (x < sieve.min_factor.size())
                return sieve.min_factor[x] == x;
            else
                return is_prime(x);
        };
        ret += dp[i] * (max(0LL, p.pi[i] + ispr(p.vs[i] + 1) - npuse) + 1);
    }
    cout << ret << '\n';
}
0