Pollard-Rho的一些应用

P4718

https://www.luogu.com.cn/problem/P4718

要求找最大的素因子，考虑可能出现在因子的因子中，所以需要递归

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i64 max_prime(i64 n) 
{
    if (isp(n)) {return n;}
    i64 mx{std::numeric_limits<i64>::min()}; while (n != 1) {
        auto div{findDiv(n)};
        mx = std::max(mx, max_prime(div));//递归找最大素因子
        n /= div;
    }
    return mx;
}

牛客多校第八场E

https://ac.nowcoder.com/acm/contest/81603/E

尽可能快地找出所有因子

先用 Pollard-Rho 快速找出所有质因子，再用质因子组合出所有因数

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i64 mul(i64 a, i64 b, i64 m) {
    return static_cast<__int128>(a) * b % m;
}
i64 power(i64 a, i64 b, i64 m) {
    i64 res = 1 % m;
    for (; b; b >>= 1, a = mul(a, a, m))
        if (b & 1)
            res = mul(res, a, m);
    return res;
}
bool isprime(i64 n) {
    if (n < 2)
        return false;
    static constexpr int A[] = {2, 3, 5, 7, 11, 13, 17, 19, 23};
    int s = __builtin_ctzll(n - 1);
    i64 d = (n - 1) >> s;
    for (auto a : A) {
        if (a == n)
            return true;
        i64 x = power(a, d, n);
        if (x == 1 || x == n - 1)
            continue;
        bool ok = false;
        for (int i = 0; i < s - 1; ++i) {
            x = mul(x, x, n);
            if (x == n - 1) {
                ok = true;
                break;
            }
        }
        if (!ok)
            return false;
    }
    return true;
}

std::vector<i64> factorize(i64 n) {//目的是进行因式分解，得出这个数的所有质因数
    std::vector<i64> p;
    std::function<void(i64)> f = [&](i64 n) {
        if (n <= 10000) {
            for (int i = 2; i * i <= n; ++i)
                for (; n % i == 0; n /= i)
                    p.push_back(i);
            if (n > 1)
                p.push_back(n);
            return;
        }
        if (isprime(n)) {
            p.push_back(n);
            return;
        }
        auto g = [&](i64 x) {
            return (mul(x, x, n) + 1) % n;
        };
        i64 x0 = 2;
        while (true) {
            i64 x = x0;
            i64 y = x0;
            i64 d = 1;
            i64 power = 1, lam = 0;
            i64 v = 1;
            while (d == 1) {
                y = g(y);
                ++lam;
                v = mul(v, std::abs(x - y), n);
                if (lam % 127 == 0) {
                    d = std::gcd(v, n);
                    v = 1;
                }
                if (power == lam) {
                    x = y;
                    power *= 2;
                    lam = 0;
                    d = std::gcd(v, n);
                    v = 1;
                }
            }
            if (d != n) {
                f(d);
                f(n / d);
                return;
            }
            ++x0;
        }
    };
    f(n);
    std::sort(p.begin(), p.end());
    return p;
}

//通过质因数组合出所有因数
using factor = std::pair<i64, int>;//(质因数，有几个这个质因数)

std::vector<i64> GetDivisors(const std::vector<i64>& factors) {
    std::unordered_map<i64, int> cnt;
    for (auto fi : factors) {cnt[fi] += 1;}
    std::vector<factor> fac_cnt(cnt.begin(), cnt.end());
    std::vector<i64> divisors = {1};
    for (auto &p : fac_cnt) {
        int sz = divisors.size();
        for (int i = 0; i < sz; i++) {
            i64 cur = divisors[i];
            for (int j = 0; j < p.second; j++) {
                cur *= p.first;
                divisors.push_back(cur);
            }
        }
    }
    //sort(divisors.begin(), divisors.end());
    return divisors;
}