【数论基础】素数判定和Miller Rabin算法

判断正整数p是否是素数

方法一 朴素的判定    aaarticlea/png;base64,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" alt="">

方法二 线筛　 aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAACYAAAATBAMAAAAOpGKDAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAEHZmmasi3Ym7RO8yVM0Ml5alAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAA9UlEQVQYGT2QMU7DQBBF38bGLIkNKy4AIkKiQQRFtCjcwKKioEiDKDEHQPIFkFJRuwckQy7AERBU6cwNTJOKgr92nJF2/vt/RrvSQle7LfQaMQfjC0HaZmbi9cxx6+gVnlXfOlYLdsZ2E6h5eHbQr5txE9tPzFwU//HUBGr9FPsrTZaU8HU6vpYp2ZxJwpqM6GXOQqZmJ5cM0qgiCTIOZd7Y25ecTONK2zmXMvcMcsk7JtN2wVKmJKxgQ3fqjTuSqvBkHuBK0x84J8xHRFqww6OpsmMYEj86gpFcWzcdbLmOWON6CNFkNfb/1tVrC8FHF0ijliX/HNUrtUeIY3MAAAAASUVORK5CYII=" alt="">

对于大整数（>1e9），可以优化

方法三 利用费马小定理

假如p是质数，且gcd(a,p)=1，那么 a^(p-1) ≡1(mod p)。

即：假如a是整数，p是质数，且a,p互质，那么a的(p-1)次方除以p的余数恒等于1。

我们这样写：a*a^(p-2) ≡1(mod p)

根据定义，a^(p-2)就是a (模p)的逆元

 

那么反过来，随机整数a能使p满足a^(p-1) ≡1(mod p)的话，那么p就是质数？

当然不是，虽然很大可能是，但是会有特殊的存在。

如2^340≡1(mod 341),但341=31*11

具体来说，特殊的数是费马伪素数和卡迈克尔数，这里不介绍了

 

方法四 二次探测定理

对于任意素数p， 满足方程aaarticlea/png;base64,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" alt="">

有唯一解 x=1||p-1

证明

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方法五 Miller Rabin算法

逆用费马小定理加上二次探测，就得到了 Miller Rabin 算法。

Miller Rabin 算法有一定的出错概率，出错的情况一定是将合数判定为素数，且出错概率极低

 

对于一个奇素数（偶素数只有2嘛），显然p-1是偶数

则有推导过程如下：

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所以算法实现过程有两步：

1.枚举k，对于每一个k，检验是否满足二次探测定理（即上面的最后一步）

2.再检验aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAACYAAAAQBAMAAACIMBAtAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAdlQyIrtmq82J70QQmd0a05NiAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAsklEQVQYGWNgQAZMmReQuWC2I0Mlw7ZcVGELhlSeAHewGI/kpitJExgOAjkzmQXSwGJs/AaiDA/YHZMYeN4wzG6B6DVlqOZ5wHNvAwO7AQNDA0RMkOEN0wV2zdsMwiwMTG8gYsU8D9i3MZxk4N7ACBZgFDJhCGCewCjMwMPg//8nWCyMoQKiGEGyBQDdiQa4EhhWowkx+Dswf0AXs2dgDNiAJsjFcGJCApoY9yXlawdQxQB8VCWO20gk9wAAAABJRU5ErkJggg==" alt="">即aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAAACAAAAAQBAMAAACFLmBqAAAAMFBMVEX///8AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAv3aB7AAAAD3RSTlMAdlQyIrtmq82J70QQmd0a05NiAAAACXBIWXMAAA7EAAAOxAGVKw4bAAAAg0lEQVQYGWNgQAY8qsg8IJtJA02AwQMiwCObNgHCggqwcR0oRhFgMGWIZpw5c6YATAtDIsMaVBXVbFAz/IDijEImDM8UIQq4Z6UwMIQxVPA8gPDBJFsAQyX3BSQBrgSG1UdjkAT8HZg/IHEZGOwZGAM2IItwMZyYkIAswH1J+doBJAEAnnoYCfcgYtYAAAAASUVORK5CYII=" alt="">是否满足费马小定理aaarticlea/png;base64,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" alt="">

 

对于任意一个p，如果通过了算法检验，则它不是素数的概率是25%

那么k轮之后，p不是素数的概率就是0.25^k，就很小了

一般8-10轮就差不多了

那么对于a的选择，可以选前面几个质数：2，3，5，7，11等，也可以用随机数

时间复杂度一般是aaarticlea/png;base64,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" alt="">,k为测试的轮数

//里面有龟速乘和快速幂，防越界
inline int slow_mult(int a,int b,int mod){
    int ans=;
    while(b){
        if(b&)    ans=(ans+a)%mod;
        a=(a+a)%mod;
        b>>=;
    }
    return ans;
}

inline int quick_pow(int a,int b,int mod){
    int ans=;
    while(b){
        if(b&)    ans=slow_mult(ans,a,mod);
        a=slow_mult(a,a,mod);
        b>>=;
    }
    return ans;
}

inline bool Miller_Rabin(int p){
    if(p==)    return true;
    if(p==||p%==)    return false;
    int u=p-,t=;
    while(u%==)    u/=,t++;
    for(int i=;i<=;i++){//我测了十次
        int a=rand()%(p-)+;
        int x=quick_pow(a,u,p);
        if(x==)    continue;
        for(int j=;j<=t;j++){
            int y=slow_mult(x,x,p);
            if(y== && x!= && x!=p-)    return false;//二次探测定理
            x=y;//递推求2的幂
        }
        if(x!=)    return false;//费马小定理
    }
    return true;
}