hdu 4910 Problem about GCD（Miller Rabin大素数检测）

hdu 4910 Problem about GCD

表示完全给跪了，打表找规律+大数值的素数检测。可参见这篇博客：HDU 4910 Problem about GCD(米勒拉宾）

Miller Rabin算法就是基于概率的素数测试算法，利用费马小定理，a^(p-1)%p = 1 {gcd(a,p) = 1 && prime(p)=true}，随机多个小于p且大于0的a值，若均检测出p为素数，则可认为p是素数

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 1000005;
typedef __int64 LL;
int vis[MAXN], nprm;
LL prim[MAXN];
void init_prime()
{
    for (LL i = 2; i<=1000000LL; ++i)
    {
        if (!vis[i]) prim[nprm++] = i;
        for (int j = 0; j< nprm && prim[j]*i <= 1000000LL; ++j)
        {
            vis[i*prim[j]] = 1;
            if (i%prim[j] == 0) break;
        }
    }
}
/* 
    a*b = a*(2^n*wn + 2^(n-1)*w(n-1) + 2^(n-2)*w(n-2)+ ... +2^0*w0)
*/
LL mul_mod(LL a, LL b, LL mod)
{
    LL res = 0LL;
    while (b)
    {
        if (b & 1) res = (res+a)%mod;
        a = (a+a)%mod;
        b >>= 1;
    }
    return res;
}
LL mpow(LL x, LL nm, LL mod)
{
    LL res = 1;
    while (nm)
    {
        if (nm & 1) res = mul_mod(res, x, mod);
        x = mul_mod(x, x, mod);
        nm >>= 1;
    }
    return res;
}
int miller_rabin(LL m)
{
    if (m < 2) return 0;
    if (m == 2) return 1;
    if (m % 2 == 0) return 0;
    for (int i = 0; i< 20; ++i)
        if (mpow(rand()%(m-1)+1, m-1, m) != 1) 
            return 0;
    return 1;
}
int solve(LL n)
{
    if (n % 4 == 0) return 0;
    if (n % 2 == 0) n /= 2;
    for (int i = 1; i< nprm; ++i)
    {
        LL pp = n;
        if (pp % prim[i] == 0)
        {
            while (pp % prim[i] == 0)
                pp /= prim[i];
            if (pp == 1) return 1;
            else return 0;
        }
    }
    if (miller_rabin(n)) return 1;
    LL x = (LL)sqrt(double(n));
    if (x*x == n) return 1;
    return 0;
}
int main()   
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    init_prime();
    LL n;
    while (~scanf("%I64d", &n) && n!=-1)
    {
        LL res;
        if (n <= 5 || solve(n)) printf("%I64d\n", n-1);
        else printf("1\n");
    }
    return 0;
}