POJ 2429 GCD & LCM Inverse Pollard_Rho大数分解+Miller_Rabin

PS：图片由作者用wps制作，使用请注明链接，O(∩_∩)O谢谢！

如对Miller-Rabin有疑问，请参考作者其他博客：Miller-Rabin素性测试算法详解

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

typedef long long int ll;

const int MAX_SIZE = 1000;
const ll INF = 0x3f3f3f3f3f;
ll fac[MAX_SIZE], ct, cnt;
ll m, n; // GCD : m, LCM : n

ll GCD(ll x, ll y)
{
    return !y ? x : GCD(y, x % y);
}

typedef long long int ll;

ll mod_mul(ll a, ll b, ll mod)
{
    ll res = 0;
    while (b)
    {
        if (b & 1)
            res = (res + a) % mod;
        a = (a + a) % mod;
        b >>= 1;
    }
    return res;
}

ll mod_pow(ll a, ll n, ll mod)
{
    ll res = 1;
    while (n)
    {
        if (n & 1)
            res = mod_mul(res, a, mod);
        a = mod_mul(a, a, mod);
        n >>= 1;
    }
    return res;
}

// Miller-Rabin随机算法检测n是否为素数
bool Miller_Rabin(ll n)
{
    if (n == 2)
        return true;
    if (n < 2 || !(n & 1))
        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;
        ll x = mod_pow(a, m, n);
        ll y;
        for (int j = 1; j <= k; j++)
        {
            y = mod_mul(x, x, n);
            if (y == 1 && x != 1 && x != n - 1)
                return false;
            x = y;
        }
        if (y != 1)
            return false;
    }
    return true;
}

// 大数因子分解
ll Pollard_Rho(ll n, ll c)
{
    ll i = 1, k = 2;
    ll x = rand() % (n - 1) + 1, y = x;
    while (1)
    {
        i++;
        x = (mod_mul(x, x, n) + c) % n;
        ll d = GCD((y - x + n) % n, n);
        if (d != 1 && d != n)  // if (d > 1 && d < n)
            return d;
        if (y == x)
            return n;
        if (i == k)
        {
            y = x;
            k <<= 1;
        }
    }
}

void find(ll n, ll c)
{
    if (n == 1)  // 1不处理，因为1的质因数分解方式有无穷种
        return;
    if (Miller_Rabin(n))
    {
        fac[ct++] = n;
        return;
    }
    ll p = n;
    ll k = c;
    while (p >= n)
        p = Pollard_Rho(p, c--);
    find(p, k);
    find(n / p, k);
}

ll ans, s;
void dfs(ll cur, ll val)
{
    if (val > ans)
        ans = val;
    for (ll i = cur; i < cnt; i++)
    {
        ll t = fac[i] * val;
        if (t <= s)
            dfs(i + 1, t);
    }
}

/*
void dfs(ll cur, ll val)
{
    if (val > ans)
        ans = val;
    if (cur == cnt)
        return;
    if (val * fac[cur] <= s)
        dfs(cur + 1, val * fac[cur]);
    dfs(cur + 1, val);
}
*/

int main()
{
    //freopen("test.txt", "r", stdin);

    while (~scanf("%I64d%I64d", &m, &n))
    {
        ct = 0;
        find(n / m, 11);  // 因式分解后不一定是递增的,有可能出现13,3,3的情况
        sort(fac, fac + ct);
        cnt = 0;
        int t = fac[0];
        for (int i = 1; i < ct; i++)
        {
            if (fac[i] == fac[i - 1])
                t *= fac[i];
            else
            {
                fac[cnt++] = t;
                t = fac[i];
            }
        }
        fac[cnt++] = t;
        ans = -INF;
        s = (ll)sqrt(n * 1.0 / m);
        dfs(0, 1);  // 代码中的两种dfs均是可行的
        printf("%I64d %I64d\n", ans * m, n / ans);
    }
    return 0;
}