cplib-cpp

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:heavy_check_mark: Zeta transform / Moebius transform （約数包除）
(number/zeta_moebius_transform.hpp)

整除関係に基づくゼータ変換・メビウス変換．またこれらを使用した添字 GCD・添字 LCM 畳み込み．計算量は $O(N \log \log N)$．なお，引数の vector<> について第 0 要素（f[0]）の値は無視される．

実装されている関数

void multiple_zeta(vector<T> &f)

f[x] を f[x * j] ($j = 1, 2, \dots$) の総和で置き換える．

void multiple_moebius(vector<T> &f)

multiple_zeta() の逆関数．

void divisor_zeta(vector<T> &f)

f[x] を f[y] ($y$ は $x$ の約数) の総和で置き換える．

void divisor_moebius(vector<T> &f)

divisor_zeta() の逆関数．

vector<T> gcdconv(vector<T> f, vector<T> g)

添字 GCD 畳み込み．

\[\mathrm{ret}[k] = \sum_{k = \mathrm{GCD}(i, j)} f[i] * g[j]\]

vector<T> lcmconv(vector<T> f, vector<T> g)

添字 LCM 畳み込み（もとの配列長をはみ出す部分は無視）．

\[\mathrm{ret}[k] = \sum_{k = \mathrm{LCM}(i, j)} f[i] * g[j]\]

使用例

int N;
vector<unsigned long long> dp(N + 1);
multiple_moebius(dp);

vector<mint> f(N + 1), g(N + 1);
vector<mint> h = gcdconv(f, g);

問題例

文献・リンク

Depends on

Verified with

Code

#pragma once
#include "../number/sieve.hpp"
#include <algorithm>
#include <cassert>
#include <utility>
#include <vector>

// f[n] に対して、全ての n の倍数 n*i に対する f[n*i] の和が出てくる 計算量 O(N loglog N)
// 素数p毎に処理する高速ゼータ変換
// 使用例 https://yukicoder.me/submissions/385043
template <class T> void multiple_zeta(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; p++) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = N / p; j > 0; --j) f[j] += f[j * p];
        }
    }
}

// inverse of multiple_zeta O(N loglog N)
// 使用例 https://yukicoder.me/submissions/385120
template <class T> void multiple_moebius(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; p++) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = 1; j * p <= N; ++j) f[j] -= f[j * p];
        }
    }
}

// f[n] に関して、全ての n の約数 m に対する f[m] の総和が出てくる 計算量 O(N loglog N)
template <class T> void divisor_zeta(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; ++p) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = 1; j * p <= N; ++j) f[j * p] += f[j];
        }
    }
}
// inverse of divisor_zeta()
// Verified: https://codeforces.com/contest/1630/problem/E
template <class T> void divisor_moebius(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; ++p) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = N / p; j > 0; --j) f[j * p] -= f[j];
        }
    }
}

// GCD convolution
// ret[k] = \sum_{gcd(i, j) = k} f[i] * g[j]
template <class T> std::vector<T> gcdconv(std::vector<T> f, std::vector<T> g) {
    assert(f.size() == g.size());
    multiple_zeta(f);
    multiple_zeta(g);
    for (int i = 0; i < int(g.size()); ++i) f[i] *= g[i];
    multiple_moebius(f);
    return f;
}

// LCM convolution
// ret[k] = \sum_{lcm(i, j) = k} f[i] * g[j]
template <class T> std::vector<T> lcmconv(std::vector<T> f, std::vector<T> g) {
    assert(f.size() == g.size());
    divisor_zeta(f);
    divisor_zeta(g);
    for (int i = 0; i < int(g.size()); ++i) f[i] *= g[i];
    divisor_moebius(f);
    return f;
}

// fast_integer_moebius の高速化（登場しない素因数が多ければ計算量改善）
// Requirement: f の key として登場する正整数の全ての約数が key として登場
// Verified: https://toph.co/p/height-of-the-trees
template <typename Int, typename Val>
void sparse_fast_integer_moebius(std::vector<std::pair<Int, Val>> &f, const Sieve &sieve) {
    if (f.empty()) return;
    std::sort(f.begin(), f.end());
    assert(f.back().first < Int(sieve.min_factor.size()));
    std::vector<Int> primes;
    for (auto p : f) {
        if (sieve.min_factor[p.first] == p.first) primes.push_back(p.first);
    }
    std::vector<std::vector<int>> p2is(primes.size());
    for (int i = 0; i < int(f.size()); i++) {
        Int a = f[i].first, pold = 0;
        int k = 0;
        while (a > 1) {
            auto p = sieve.min_factor[a];
            if (p != pold) {
                while (primes[k] != p) k++;
                p2is[k].emplace_back(i);
            }
            pold = p, a /= p;
        }
    }
    for (int d = 0; d < int(primes.size()); d++) {
        Int p = primes[d];
        for (auto i : p2is[d]) {
            auto comp = [](const std::pair<Int, Val> &l, const std::pair<Int, Val> &r) {
                return l.first < r.first;
            };
            auto itr = std::lower_bound(f.begin(), f.end(), std::make_pair(f[i].first / p, 0), comp);
            itr->second -= f[i].second;
        }
    }
}
#line 2 "number/sieve.hpp"
#include <cassert>
#include <map>
#include <vector>

// CUT begin
// 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));
#line 3 "number/zeta_moebius_transform.hpp"
#include <algorithm>
#line 5 "number/zeta_moebius_transform.hpp"
#include <utility>
#line 7 "number/zeta_moebius_transform.hpp"

// f[n] に対して、全ての n の倍数 n*i に対する f[n*i] の和が出てくる 計算量 O(N loglog N)
// 素数p毎に処理する高速ゼータ変換
// 使用例 https://yukicoder.me/submissions/385043
template <class T> void multiple_zeta(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; p++) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = N / p; j > 0; --j) f[j] += f[j * p];
        }
    }
}

// inverse of multiple_zeta O(N loglog N)
// 使用例 https://yukicoder.me/submissions/385120
template <class T> void multiple_moebius(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; p++) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = 1; j * p <= N; ++j) f[j] -= f[j * p];
        }
    }
}

// f[n] に関して、全ての n の約数 m に対する f[m] の総和が出てくる 計算量 O(N loglog N)
template <class T> void divisor_zeta(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; ++p) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = 1; j * p <= N; ++j) f[j * p] += f[j];
        }
    }
}
// inverse of divisor_zeta()
// Verified: https://codeforces.com/contest/1630/problem/E
template <class T> void divisor_moebius(std::vector<T> &f) {
    int N = int(f.size()) - 1;
    std::vector<int> is_prime(N + 1, 1);
    for (int p = 2; p <= N; ++p) {
        if (is_prime[p]) {
            for (int q = p * 2; q <= N; q += p) is_prime[q] = 0;
            for (int j = N / p; j > 0; --j) f[j * p] -= f[j];
        }
    }
}

// GCD convolution
// ret[k] = \sum_{gcd(i, j) = k} f[i] * g[j]
template <class T> std::vector<T> gcdconv(std::vector<T> f, std::vector<T> g) {
    assert(f.size() == g.size());
    multiple_zeta(f);
    multiple_zeta(g);
    for (int i = 0; i < int(g.size()); ++i) f[i] *= g[i];
    multiple_moebius(f);
    return f;
}

// LCM convolution
// ret[k] = \sum_{lcm(i, j) = k} f[i] * g[j]
template <class T> std::vector<T> lcmconv(std::vector<T> f, std::vector<T> g) {
    assert(f.size() == g.size());
    divisor_zeta(f);
    divisor_zeta(g);
    for (int i = 0; i < int(g.size()); ++i) f[i] *= g[i];
    divisor_moebius(f);
    return f;
}

// fast_integer_moebius の高速化（登場しない素因数が多ければ計算量改善）
// Requirement: f の key として登場する正整数の全ての約数が key として登場
// Verified: https://toph.co/p/height-of-the-trees
template <typename Int, typename Val>
void sparse_fast_integer_moebius(std::vector<std::pair<Int, Val>> &f, const Sieve &sieve) {
    if (f.empty()) return;
    std::sort(f.begin(), f.end());
    assert(f.back().first < Int(sieve.min_factor.size()));
    std::vector<Int> primes;
    for (auto p : f) {
        if (sieve.min_factor[p.first] == p.first) primes.push_back(p.first);
    }
    std::vector<std::vector<int>> p2is(primes.size());
    for (int i = 0; i < int(f.size()); i++) {
        Int a = f[i].first, pold = 0;
        int k = 0;
        while (a > 1) {
            auto p = sieve.min_factor[a];
            if (p != pold) {
                while (primes[k] != p) k++;
                p2is[k].emplace_back(i);
            }
            pold = p, a /= p;
        }
    }
    for (int d = 0; d < int(primes.size()); d++) {
        Int p = primes[d];
        for (auto i : p2is[d]) {
            auto comp = [](const std::pair<Int, Val> &l, const std::pair<Int, Val> &r) {
                return l.first < r.first;
            };
            auto itr = std::lower_bound(f.begin(), f.end(), std::make_pair(f[i].first / p, 0), comp);
            itr->second -= f[i].second;
        }
    }
}
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