The method of prime number, Prime Rabin test and fast power modulus

Prime number Sieve method to produce the prime list within num

public static int  makeprimes (int primes[], int num)
	{
		int i, J, CNT;
		Primes[0] = 2;
		PRIMES[1] = 3;
		for (i = 5, cnt = 2; i < num; i + = 2)
		{
			Boolean flag = true;
			for (j = 1; primes[j]*primes[j] <= i; ++j)
			{
				if (i%primes[j] = = 0)
				{
					flag = false; 
					break;
				}
			}
			if (flag) primes[cnt++] = i;
		}
		return cnt;
	}

According to the Prime number table quickly determine whether the prime, as long as the table to sqrt (n), you can quickly determine whether the numbers within N is a prime

public static Boolean  isprime (int primes[], int n)
	{
		if (n < 2) 
			return false;
		for (int i = 0; primes[i]*primes[i] <= n; i++)
			if (n%primes[i] = = 0) 
				return false;
		return true;
	}

Fast Power-Take mode

public static long Pow_mod (long A,long i,long N)
	{
		if (i==0)
			return 1%n;
		Long Temp=pow_mod (A, i>>1, n);
		temp= (temp*temp)%n;
		if ((i&1)!=0)
			temp= (temp*a)%n;
		return temp;
	}

Prime number Test

Because of the intermediate variables of the Long*long type, it is only possible to test the int range, which requires a larger range to be used Bigdecimal/biginteger

public static Boolean IsPrime (long N)
	{
		if (n<2)
			return false;
		Int[] a={2,3,5,7,11,61}; Test set for
		(int i=0;i<a.length;i++)
			if (!test (n, A[i], n-1))
				return false;
		return true;
	}
	
	public static Boolean test (long N,long a,long D)
	{
		if (n==2)
			return true;
		if (n==a)
			return true;
		if ((n&1) ==0)
			return false;
		while ((d&1) ==0)
			d=d>>1;
		Long T =pow_mod (A, D, N);
		while ((d!=n-1) && (t!=1) && (t!=n-1))
		{
			t= (t*t)%n;
			d=d<<1;
		}
		
		Return (t==n-1| | (d&1) ==1);
	}
	
	public static long Pow_mod (long A,long i,long N)
	{
		if (i==0)
			return 1%n;
		Long Temp=pow_mod (A, i>>1, n);
		temp= (temp*temp)%n;
		if ((i&1)!=0)
			temp= (temp*a)%n;
		return temp;
	}