米勒-拉宾素数测试

本文主要是介绍米勒-拉宾素数测试，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

由于卡米歇尔数的存在，导致费马小定理无法判断一个数是否是素数。

费马小定理: 设p是素数, a是任意整数且 a!三0( mod p )，则

a^(p-1) 三 1(mod p)

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卡米歇尔数:它是合数, 当 1<=a<=n, 都有 a^n 三 a(mod n)

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卡米歇尔数的考塞特判别法: 设n是合数，则n是卡米歇尔数当且仅当它是奇数，且整除n的每个素数p满足下述两个条件:

1)p^2 不整除 n

2)p-1 整除 n-1

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如果要判断相当大的素数最好使用合数的拉宾-米勒测试定理

合数的拉宾-米勒测试定理: 设n是奇素数, 记 n-1 = 2^k * q , q 是奇数, 对不被n整除的某个a, 如果下述两个条件都成立，则n是合数.

a) a^q !三 1（mod n);

b) 对所有 i = 0, 1, 2, ...., k-1, a^((2^i)*q) !三 -1(mod n);

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这里给出了合数的拉宾-米勒测试定理 a 的取值:

if n < 1,373,653, it is enough to test a = 2 and 3.
if n < 9,080,191, it is enough to test a = 31 and 73.
if n < 4,759,123,141, it is enough to test a = 2, 7, and 61.
if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.

// montgomery快速幂模算法 (n ^ p) % m, 与power算法极类似
unsigned __int64 montgomery(unsigned __int64 n, unsigned __int64 p, unsigned __int64 m)
{ unsigned __int64 r = n % m;unsigned __int64 tmp = 1;while (p > 1){if ((p & 1)!=0){tmp = (tmp * r) % m;}r = (r * r) % m;p >>= 1;}return (r * tmp) % m;
}//返回true:n是合数, 返回false:n是素数
bool R_M_Help(unsigned __int64 a, unsigned __int64 k, unsigned __int64 q, unsigned __int64 n)
{if ( 1 != montgomery( a, q, n ) ){int e = 1;for ( int i = 0; i < k; ++i ){if ( n - 1 == montgomery( a, q * e, n ) ) return false;e <<= 1;}return true;}return false;
}//拉宾-米勒测试 返回true:n是合数, 返回false:n是素数  
bool R_M( unsigned __int64 n ) 
{if( n < 2 )throw 0;if ( n == 2 || n == 3 ){return false;}if( (n & 1) == 0 )return true;// 找到k和q, n = 2^k * q + 1;unsigned __int64 k = 0, q = n - 1;while( 0 == ( q & 1 ) ){q >>= 1;k++;}/*if n < 1,373,653, it is enough to test a = 2 and 3. if n < 9,080,191, it is enough to test a = 31 and 73. if n < 4,759,123,141, it is enough to test a = 2, 7, and 61. if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.*/if( n < 1373653 ){if( R_M_Help(2, k, q, n ) || R_M_Help(3, k, q, n ) )return true;}else if( n < 9080191 ){if( R_M_Help(31, k, q, n ) || R_M_Help(73, k, q, n ) )return true;}  else if( n < 4759123141 ){if( R_M_Help(2, k, q, n ) || R_M_Help(3, k, q, n )|| R_M_Help(5, k, q, n )|| R_M_Help(11, k, q, n ) )return true;}else if( n < 2152302898747 ){if( R_M_Help(2, k, q, n ) || R_M_Help(3, k, q, n )|| R_M_Help(5, k, q, n )|| R_M_Help(7, k, q, n )|| R_M_Help(11, k, q, n ) )return true;}else {if( R_M_Help(2, k, q, n ) || R_M_Help(3, k, q, n )|| R_M_Help(5, k, q, n )|| R_M_Help(7, k, q, n )|| R_M_Help(11, k, q, n )|| R_M_Help(31, k, q, n )|| R_M_Help(61, k, q, n )|| R_M_Help(73, k, q, n ) )return true;}return false;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<time.h>
#include<stdlib.h>
long long prime[]={2,3,5,7,11,13,17,19,23,29};
#define TIME 10    //Miller测试次数
long long gcd(long long a,long long b){return b==0?a:gcd(b,a%b);}
long long mod_mult(long long a, long long b, long long n) //计算(a*b) mod n
{long long s=0; a=a%n;while(b){if (b&1){s += a;if(s>=n) s-= n;}a=a<<1;if(a>=n)a-=n; b=b>>1;}return s;
}
long long mod_exp(long long a, long long b, long long n) //计算(a^b) mod n
{long long d=1; a=a%n;while(b>=1) {if(b&1)d=mod_mult(d,a,n);a=mod_mult(a,a,n); b=b>>1;}return d;
}bool Wintess(long long a, long long n) //以a为基对n进行Miller测试并实现二次探测
{long long m, x, y;int i, j = 0; m = n-1;while (!(m&1)) //计算(n-1)=m*(2^j)中的j和m,j=0时m=n-1,不断的除以2直至m为奇数{m=m>>1; j++;}x = mod_exp(a, m, n);for (i = 1; i <= j; i++) {y = mod_mult(x, x, n);if ((y == 1) && (x != 1) && (x != n - 1)) //二次探测return true; //返回true时,n是合数x = y;}if (y!=1) return true;return false;
}
bool miller_rabin(int times, long long n) //对n进行times次的Miller测试
{long long a;int i;if (n == 2) return true;if (n<2||(n&1)==0) return false;for (i = 1; i <= times; i++) {a =rand() % (n - 2) + 2;if (Wintess(a, n)) return false;}return true;
}

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