#595680 (C++17) No.1322 Totient Bound

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結果

問題	No.1322 Totient Bound
ユーザー	hitonanode
提出日時	2020-12-20 05:56:33
言語	C++17 (gcc 12.3.0 + boost 1.83.0)
結果	TLE
実行時間	-
コード長	11,403 bytes
コンパイル時間	2,992 ms
コンパイル使用メモリ	216,856 KB
実行使用メモリ	12,616 KB
最終ジャッジ日時	2024-09-21 11:32:37
合計ジャッジ時間	14,374 ms
ジャッジサーバーID （参考情報）	judge1 / judge4

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	2 ms 6,812 KB
testcase_01	AC	2 ms 6,940 KB
testcase_02	AC	2 ms 6,944 KB
testcase_03	AC	2 ms 6,944 KB
testcase_04	AC	2 ms 6,940 KB
testcase_05	AC	2 ms 6,944 KB
testcase_06	AC	2 ms 6,940 KB
testcase_07	AC	2 ms 6,944 KB
testcase_08	AC	2 ms 6,940 KB
testcase_09	AC	2 ms 6,940 KB
testcase_10	AC	2 ms 6,944 KB
testcase_11	AC	2 ms 6,940 KB
testcase_12	AC	2 ms 6,940 KB
testcase_13	AC	10 ms 6,940 KB
testcase_14	AC	9 ms 6,940 KB
testcase_15	AC	4 ms 6,940 KB
testcase_16	AC	6 ms 6,944 KB
testcase_17	AC	4 ms 6,940 KB
testcase_18	AC	4,596 ms 6,940 KB
testcase_19	TLE	-
testcase_20	--	-
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	--	-

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ソースコード

raw source code

#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 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} // <= 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;
    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)]) {
            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;
    CountPrimes p(N);

    const auto &xs = p.vs;
    auto ps = p.primes;
    while (ps.back() > N + 1 or (ps.back() - 1) * (ps.back() - 2) > 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 (true) {
        const lint d = N / (plast - 1 + 1);
        if (d == 0) break;
        lint cnt = p.pi[p.getidx(N / d)] + is_prime(N / d + 1);
        dp[p.getidx(d)] += cnt - less_cnt;
        less_cnt = cnt;
        plast = N / d + 1;
    }
    dbg(dp);
    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';
}

yukicoder

結果

テストケース

ソースコード