題目大意：給定一個大整數（2^54內），判斷是否為素數：若為素數，輸出primes；否則找出該數的最小質因子。

實現程式碼如下：

#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <stdio.h>

const int Times = 10;
const int N = 5500;

using namespace std;
typedef long long LL;

LL ct, cnt;
LL fac[N], num[N];

LL gcd(LL a, LL b)
{
    return b? gcd(b, a % b) : a;
}

LL multi(LL a, LL b, LL m)
{
    LL ans = 0;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = (ans + a) % m;
            b--;
        }
        b >>= 1;
        a = (a + a) % m;
    }
    return ans;
}

LL quick_mod(LL a, LL b, LL m)
{
    LL ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)
        {
            ans = multi(ans, a, m);
            b--;
        }
        b >>= 1;
        a = multi(a, a, m);
    }
    return ans;
}

bool Miller_Rabin(LL n)
{
    if(n == 2) return true;
    if(n < 2 || !(n & 1)) return false;
    LL m = n - 1;
    int k = 0;
    while((m & 1) == 0)
    {
        k++;
        m >>= 1;
    }
    for(int i=0; i<Times; i++)
    {
        LL a = rand() % (n - 1) + 1;
        LL x = quick_mod(a, m, n);
        LL y = 0;
        for(int j=0; j<k; j++)
        {
            y = multi(x, x, n);
            if(y == 1 && x != 1 && x != n - 1) return false;
            x = y;
        }
        if(y != 1) return false;
    }
    return true;
}

LL pollard_rho(LL n, LL c)
{
    LL i = 1, k = 2;
    LL x = rand() % (n - 1) + 1;
    LL y = x;
    while(true)
    {
        i++;
        x = (multi(x, x, n) + c) % n;
        LL d = gcd((y - x + n) % n, n);
        if(1 < d && d < n) return d;
        if(y == x) return n;
        if(i == k)
        {
            y = x;
            k <<= 1;
        }
    }
}

void find(LL n, int c)
{
    if(n == 1) 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);
}

int main()
{
    int t;
    cin>>t;
    LL n;
    while(t--)
    {
        cin>>n;
        if(Miller_Rabin(n))
        {
            puts("Prime");
            continue;
        }
        ct = 0;
        find(n, 120);
        sort(fac, fac + ct);
        num[0] = 1;
        int k = 1;
        for(int i=1; i<ct; i++)
        {
            if(fac[i] == fac[i-1])
                ++num[k-1];
            else
            {
                num[k] = 1;
                fac[k++] = fac[i];
            }
        }
        cnt = k;
        printf("%lld\n",fac[0]);
    }
    return 0;
}