素数判断算法 - 拉宾-米勒测试定理(c++实现)

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在1000如此小的素数判断，在不考虑效率的情况下可以利用素数的定义来判断

printf("2 ");//2是唯一一个偶素数
for( int a = 3; a <= 1000; a+=2) //步进为2, 因为只有奇数才有可能是素数（已排除了2）
{
bool bDivision = false;
int _nTemp = sqrt((float)a); //只需要检测自身开平方的数
for( int i = 3; i < nTemp; i+=2)
{
if( a % i == 0 )
{
bDivision = true;
break;
}
}

if( !bDivision )
printf("%d ", a );
}

 

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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.
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// 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;
}