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  • 数学--数论--随机算法--Pollard Rho 大数分解算法(纯模板带输出)

    ACM常用模板合集

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    ll pr;
    ll pmod(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } //普通的快速乘会T
    ll gmod(ll a, ll b, ll p)
    {
        ll res = 1;
        while (b)
        {
            if (b & 1) res = pmod(res, a, p);
            a = pmod(a, a, p);
            b >>= 1;
        }
        return res;
    }
    inline ll gcd(ll a, ll b)
    { //听说二进制算法特快
        if (!a) return b;
        if (!b)return a;
        int t = __builtin_ctzll(a | b);
        a >>= __builtin_ctzll(a);
        do
        {
            b >>= __builtin_ctzll(b);
            if (a > b)
            {
                ll t = b;
                b = a, a = t;
            }
            b -= a;
        } while (b);
        return a << t;
    }
    bool Miller_Rabin(ll n)
    {
        if (n == 46856248255981ll || n < 2)
            return false; //强伪素数
        if (n == 2 || n == 3 || n == 7 || n == 61 || n == 24251)
            return true;
        if (!(n & 1) || !(n % 3) || !(n % 61) || !(n % 24251))
            return false;
        ll m = n - 1, k = 0;
        while (!(m & 1))
            k++, m >>= 1;
        for (int i = 1; i <= 20; ++i) // 20为Miller-Rabin测试的迭代次数
        {
            ll a = rand() % (n - 1) + 1, x = gmod(a, m, n), y;
            for (int j = 1; j <= k; ++j)
            {
                y = pmod(x, x, n);
                if (y == 1 && x != 1 && x != n - 1)
                    return 0;
                x = y;
            }
            if (y != 1)
                return 0;
        }
        return 1;
    }
    ll Pollard_Rho(ll x)
    {
        ll n = 0, m = 0, t = 1, q = 1, c = rand() % (x - 1) + 1;
        for (ll k = 2;; k <<= 1, m = n, q = 1)
        {
            for (ll i = 1; i <= k; ++i)
            {
                n = (pmod(n, n, x) + c) % x;
                q = pmod(q, abs(m - n), x);
            }
            t = gcd(x, q);
            if (t > 1)
                return t;
        }
    }
    void fid(ll n)
    {
        if (n == 1)
            return;
        if (Miller_Rabin(n))
        {
            pr = max(pr, n);
            return;
        }
        ll p = n;
        while (p >= n)
            p = Pollard_Rho(p);
        fid(p);
        fid(n / p);
    }
    int main()
    {
        int T;
        ll n;
        scanf("%d", &T);
        while (T--)
        {
            scanf("%lld", &n);
            pr = 0;
            fid(n);
            if (pr == n)
                puts("Prime");
            else
                printf("%lld
    ", pr);
        }
        return 0;
    }
    

    带输出的我也写了

    #include <bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    ll pr;
    ll pmod(ll a, ll b, ll p) { return (a * b - (ll)((long double)a / p * b) * p + p) % p; } //普通的快速乘会T
    ll gmod(ll a, ll b, ll p)
    {
        ll res = 1;
        while (b)
        {
            if (b & 1)
                res = pmod(res, a, p);
            a = pmod(a, a, p);
            b >>= 1;
        }
        return res;
    }
    inline ll gcd(ll a, ll b)
    { //听说二进制算法特快
        if (!a)
            return b;
        if (!b)
            return a;
        int t = __builtin_ctzll(a | b);
        a >>= __builtin_ctzll(a);
        do
        {
            b >>= __builtin_ctzll(b);
            if (a > b)
            {
                ll t = b;
                b = a, a = t;
            }
            b -= a;
        } while (b);
        return a << t;
    }
    bool Miller_Rabin(ll n)
    {
        if (n == 46856248255981ll || n < 2)
            return false; //强伪素数
        if (n == 2 || n == 3 || n == 7 || n == 61 || n == 24251)
            return true;
        if (!(n & 1) || !(n % 3) || !(n % 61) || !(n % 24251))
            return false;
        ll m = n - 1, k = 0;
        while (!(m & 1))
            k++, m >>= 1;
        for (int i = 1; i <= 20; ++i) // 20为Miller-Rabin测试的迭代次数
        {
            ll a = rand() % (n - 1) + 1, x = gmod(a, m, n), y;
            for (int j = 1; j <= k; ++j)
            {
                y = pmod(x, x, n);
                if (y == 1 && x != 1 && x != n - 1)
                    return 0;
                x = y;
            }
            if (y != 1)
                return 0;
        }
        return 1;
    }
    ll Pollard_Rho(ll x)
    {
        ll n = 0, m = 0, t = 1, q = 1, c = rand() % (x - 1) + 1;
        for (ll k = 2;; k <<= 1, m = n, q = 1)
        {
            for (ll i = 1; i <= k; ++i)
            {
                n = (pmod(n, n, x) + c) % x;
                q = pmod(q, abs(m - n), x);
            }
            t = gcd(x, q);
            if (t > 1)
                return t;
        }
    }
    map<long long, int> m;
    void fid(ll n)
    {
        if (n == 1)
            return;
        if (Miller_Rabin(n))
        {
            pr = max(pr, n);
            m[n]++;
            return;
        }
        ll p = n;
        while (p >= n)
            p = Pollard_Rho(p);
        fid(p);
        fid(n / p);
    }
    int main()
    {
        int T;
        ll n;
        scanf("%d", &T);
        while (T--)
        {
            m.clear();
            scanf("%lld", &n);
            pr = 0;
            fid(n);
            if (pr == n)
                puts("Prime");
            else
            {
                printf("%lld
    ", pr);
                for (map<long long, int>::iterator c = m.begin(); c != m.end();)
                {
                    printf("%lld^%d", c->first, c->second);
                    if ((++c) != m.end())
                        printf(" * ");
                }
                printf("
    ");
            }
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/lunatic-talent/p/12798453.html
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