[原创]判断质数的改进算法

参考了redraiment的 求质数的方法
转载请注明出处
基本原理：判断N是否为质数，给定一个初始化数组p[]={2,3,5}，总数total=3.
若N<=p[total]*p[total],从p中找N的因子，如果找不到，则N为质数。
若N>p[total]*p[total], 首先从p中找N的因子，如果找不到。扩展N直到N<=p[total]*p[total]，扩展的同时，计算新增的质数p[total]是否为N的因子。

代码如下：

        public bool IsPrime(long[] prime, long n, ref int total)
        {
            if (n == 1)
                return false;
            long i, j;
            long gab = 2; //6n+1, 6n+5.for example: 7 11 13 17 (4,2,4,2)
            bool flag;
            i = n;
            for (j = 0; prime[j] * prime[j] <= i; j++)
            {
                if (j > total - 1)
                {
                    break;
                }
                if(i % prime[j] == 0)
                {
                    //Console.WriteLine(n+"  Number：" + prime[j]);
                    return false;
                }
            }
            if (n > prime[total] * prime[total - 1])
            {//产生更长的素数表
                //reset gab,for example: 7 11 13. 7%6=1
                i = prime[total - 1];
                if (prime[total - 1] % 6 > 1)
                    gab = 2;
                else
                    gab = 4;

                while (n > prime[total - 1] * prime[total - 1])
                {
                    flag = true;
                    i = i + gab;
                    gab = 6 - gab;
                    for (j = 0; prime[j] * prime[j] <= i; j++)
                    {
                        if (i % prime[j] == 0)
                        {
                            flag = false;
                            break;
                        }
                    }
                    if (flag)
                    {
                        prime[total] = i;
                        total++;
                    }
                    if (n % prime[total - 1] == 0)
                    {
                        //Console.WriteLine("Number：" + prime[total - 1]);
                        return false;
                    }
                }
            }
            return true;
        }