素数判定--米勒测试

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

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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 );

}

//========================================================

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)

//========================================================

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

//========================================================

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

//========================================================

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;

}

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