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

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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;}

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