#946760 (C++23) No.2620 Sieve of Coins

提出ソース

結果

問題	No.2620 Sieve of Coins
ユーザー	hitonanode
提出日時	2024-01-26 23:02:26
言語	C++23 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	83 ms / 2,000 ms
コード長	17,779 bytes
コンパイル時間	2,348 ms
コンパイル使用メモリ	199,960 KB
実行使用メモリ	29,968 KB
最終ジャッジ日時	2024-09-28 08:51:41
合計ジャッジ時間	7,791 ms
ジャッジサーバーID （参考情報）	judge4 / judge1

このコードへのチャレンジ
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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	31 ms 29,168 KB
testcase_01	AC	31 ms 28,604 KB
testcase_02	AC	31 ms 29,968 KB
testcase_03	AC	81 ms 28,648 KB
testcase_04	AC	82 ms 28,332 KB
testcase_05	AC	32 ms 28,212 KB
testcase_06	AC	32 ms 28,340 KB
testcase_07	AC	32 ms 28,340 KB
testcase_08	AC	31 ms 28,208 KB
testcase_09	AC	31 ms 28,336 KB
testcase_10	AC	31 ms 28,336 KB
testcase_11	AC	31 ms 28,340 KB
testcase_12	AC	31 ms 28,212 KB
testcase_13	AC	31 ms 28,336 KB
testcase_14	AC	31 ms 28,344 KB
testcase_15	AC	30 ms 28,336 KB
testcase_16	AC	31 ms 28,340 KB
testcase_17	AC	34 ms 28,336 KB
testcase_18	AC	80 ms 28,208 KB
testcase_19	AC	81 ms 28,340 KB
testcase_20	AC	31 ms 28,336 KB
testcase_21	AC	81 ms 28,340 KB
testcase_22	AC	81 ms 28,332 KB
testcase_23	AC	83 ms 28,212 KB
testcase_24	AC	81 ms 28,340 KB
testcase_25	AC	83 ms 28,340 KB
testcase_26	AC	82 ms 28,264 KB
testcase_27	AC	81 ms 28,336 KB
testcase_28	AC	31 ms 28,208 KB
testcase_29	AC	31 ms 28,332 KB
testcase_30	AC	31 ms 28,332 KB
testcase_31	AC	81 ms 28,340 KB
testcase_32	AC	81 ms 28,332 KB
testcase_33	AC	81 ms 28,336 KB
testcase_34	AC	81 ms 28,336 KB
testcase_35	AC	82 ms 28,336 KB
testcase_36	AC	83 ms 28,336 KB
testcase_37	AC	81 ms 28,212 KB
testcase_38	AC	83 ms 28,332 KB
testcase_39	AC	81 ms 28,336 KB
testcase_40	AC	82 ms 28,340 KB
testcase_41	AC	83 ms 28,336 KB
testcase_42	AC	81 ms 28,208 KB
testcase_43	AC	81 ms 28,336 KB
testcase_44	AC	82 ms 28,336 KB
testcase_45	AC	32 ms 28,212 KB
testcase_46	AC	81 ms 28,340 KB
testcase_47	AC	82 ms 28,336 KB
testcase_48	AC	82 ms 28,332 KB
testcase_49	AC	82 ms 28,208 KB
testcase_50	AC	82 ms 28,472 KB
testcase_51	AC	81 ms 28,376 KB
testcase_52	AC	81 ms 28,216 KB
testcase_53	AC	82 ms 28,340 KB
testcase_54	AC	81 ms 28,244 KB
testcase_55	AC	82 ms 28,340 KB
testcase_56	AC	82 ms 28,212 KB
testcase_57	AC	82 ms 28,216 KB

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ソースコード

raw source code

#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 <memory>
#include <numeric>
#include <optional>
#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> 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; }
const std::vector<std::pair<int, int>> grid_dxs{{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int floor_lg(long long x) { return x <= 0 ? -1 : 63 - __builtin_clzll(x); }
template <class T1, class T2> T1 floor_div(T1 num, T2 den) { return (num > 0 ? num / den : -((-num + den - 1) / den)); }
template <class T1, class T2> std::pair<T1, T2> operator+(const std::pair<T1, T2> &l, const std::pair<T1, T2> &r) { return std::make_pair(l.first + r.first, l.second + r.second); }
template <class T1, class T2> std::pair<T1, T2> operator-(const std::pair<T1, T2> &l, const std::pair<T1, T2> &r) { return std::make_pair(l.first - r.first, l.second - r.second); }
template <class T> std::vector<T> sort_unique(std::vector<T> vec) { sort(vec.begin(), vec.end()), vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec; }
template <class 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 <class 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 <class IStream, class T> IStream &operator>>(IStream &is, std::vector<T> &vec) { for (auto &v : vec) is >> v; return is; }

template <class OStream, class T> OStream &operator<<(OStream &os, const std::vector<T> &vec);
template <class OStream, class T, size_t sz> OStream &operator<<(OStream &os, const std::array<T, sz> &arr);
template <class OStream, class T, class TH> OStream &operator<<(OStream &os, const std::unordered_set<T, TH> &vec);
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const pair<T, U> &pa);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::deque<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::set<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::multiset<T> &vec);
template <class OStream, class T> OStream &operator<<(OStream &os, const std::unordered_multiset<T> &vec);
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const std::pair<T, U> &pa);
template <class OStream, class TK, class TV> OStream &operator<<(OStream &os, const std::map<TK, TV> &mp);
template <class OStream, class TK, class TV, class TH> OStream &operator<<(OStream &os, const std::unordered_map<TK, TV, TH> &mp);
template <class OStream, class... T> OStream &operator<<(OStream &os, const std::tuple<T...> &tpl);

template <class OStream, class T> OStream &operator<<(OStream &os, const std::vector<T> &vec) { os << '['; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <class OStream, class T, size_t sz> OStream &operator<<(OStream &os, const std::array<T, sz> &arr) { os << '['; for (auto v : arr) os << v << ','; os << ']'; return os; }
template <class... T> std::istream &operator>>(std::istream &is, std::tuple<T...> &tpl) { std::apply([&is](auto &&... args) { ((is >> args), ...);}, tpl); return is; }
template <class OStream, class... T> OStream &operator<<(OStream &os, const std::tuple<T...> &tpl) { os << '('; std::apply([&os](auto &&... args) { ((os << args << ','), ...);}, tpl); return os << ')'; }
template <class OStream, class T, class TH> OStream &operator<<(OStream &os, const std::unordered_set<T, TH> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::deque<T> &vec) { os << "deq["; for (auto v : vec) os << v << ','; os << ']'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::set<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T> OStream &operator<<(OStream &os, const std::unordered_multiset<T> &vec) { os << '{'; for (auto v : vec) os << v << ','; os << '}'; return os; }
template <class OStream, class T, class U> OStream &operator<<(OStream &os, const std::pair<T, U> &pa) { return os << '(' << pa.first << ',' << pa.second << ')'; }
template <class OStream, class TK, class TV> OStream &operator<<(OStream &os, const std::map<TK, TV> &mp) { os << '{'; for (auto v : mp) os << v.first << "=>" << v.second << ','; os << '}'; return os; }
template <class OStream, class TK, class TV, class TH> OStream &operator<<(OStream &os, const std::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) std::cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << std::endl
#define dbgif(cond, x) ((cond) ? std::cerr << BRIGHT_CYAN << #x << COLOR_RESET << " = " << (x) << NORMAL_FAINT << " (L" << __LINE__ << ") " << __FILE__ << COLOR_RESET << std::endl : std::cerr)
#else
#define dbg(x) ((void)0)
#define dbgif(cond, x) ((void)0)
#endif

#include <cassert>
#include <map>
#include <vector>

// 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);
        }
    }
};



void expe(int n) {
    vector<int> dp(n + 1);
    dp.at(1) = 1;
    FOR(i, 1, dp.size()) {
        if (dp.at(i)) {
            dbg(sieve.factorize(i));
            for (int j = i * 2; j <= n; j += i) dp.at(j) ^= 1;
        }
    }
    dbg(dp);
}

int main() {
    // expe(211);
    lint L;
    int N;
    cin >> L >> N;

    auto moebius = sieve.GenerateMoebiusFunctionTable();
    vector<plint> nonzeros;
    FOR(i, 1, moebius.size()) {
        if (moebius.at(i)) nonzeros.emplace_back(i, moebius.at(i));
    }

    auto f = [&](lint x) {
        lint ret = 0;
        for (auto [d, m] : nonzeros) {
            if (d * d > x) break;
            ret += m * (x / (d * d));
        }
        return ret;
    };

    vector<lint> A(N);
    cin >> A;

    int D3 = 2;
    {
        lint tmp = L;
        while (tmp) {
            ++D3;
            tmp /= 3;
        }
    }

    const int D2 = floor_lg(L) + 2;
    vector dp(D2, vector<lint>(D3));
    IREP(i, D2) IREP(j, D3) {
        if (i + 1 < D2) dp.at(i).at(j) -= dp.at(i + 1).at(j);
        if (j + 1 < D3) dp.at(i).at(j) -= dp.at(i).at(j + 1);
        if (i + 1 < D2 and j + 1 < D3) dp.at(i).at(j) -= dp.at(i + 1).at(j + 1);

        lint base = 1LL << i;
        REP(_, j) base *= 3;
        dp.at(i).at(j) += f(L / base);

        dbgif(L <= 100, make_tuple(i, j, L / base, dp.at(i).at(j)));
    }

    vector acnt(D2, vector<int>(D3));
    for (lint a : A) {
        int n2 = 0, n3 = 0;
        while (a % 2 == 0) a /= 2, n2++;
        while (a % 3 == 0) a /= 3, n3++;
        REP(d, 2) REP(e, 2) acnt.at(n2 + d).at(n3 + e) ^= 1;
    }

    lint ret = 0;
    REP(n2, D2) REP(n3, D3) {
        if (!acnt.at(n2).at(n3)) continue;
        ret += dp.at(n2).at(n3);
    }

    cout << ret << '\n';
}

yukicoder

結果

テストケース

ソースコード