已優化到 22 秒內，不知道世界最快演算法在同樣環境下執行速度是多少，下面貼出程式碼，C#

using System;

namespace prime
{
    class Program
    {
        static void Main(string[] args)
        {
            DateTime dt = DateTime.Now;
            int n = 1000000000;
            Console.WriteLine(Get_Prime_Count(n));
            DateTime dt2 = DateTime.Now;
            Console.WriteLine(dt2-dt);
        }
        static int Get_Prime_Count(int n)
        {
            int count = 0;  //後面用於統計素數個數
            bool[] big = new bool[n + 1]; //用於標記是否為素數

            int sl = 2 * 3 * 5 * 7 * 11 * 13 * 17 * 19;
            bool[] small = new bool[sl + 1]; //用於篩選掉13及以內質數的倍數

            int i;

            for (i = 2; i <= sl; i += 2)
                small[i] = true;
            for (i = 3; i <= sl; i += 3)
                small[i] = true;
            for (i = 5; i <= sl; i += 5)
                small[i] = true;
            for (i = 7; i <= sl; i += 7)
                small[i] = true;
            for (i = 11; i <= sl; i += 11)
                small[i] = true;
            for (i = 13; i <= sl; i += 13)
                small[i] = true;
            for (i = 17; i <= sl; i += 17)
                small[i] = true;
            for (i = 19; i <= sl; i += 19)
                small[i] = true;

            //以下通過拷貝去掉big中所有13及以內質數的倍數
            int remainder = n % sl; //餘數
            int p = 1;
            int q = p + n - remainder;
            while (p < q)
            {
                Array.Copy(small, 1, big, p, sl);
                p += sl;
            }
            if (remainder > 0)
            {
                Array.Copy(small, 1, big, p, remainder);
            }

            //下面這幾個不能漏掉
            big[2] = false;
            big[3] = false;
            big[5] = false;
            big[7] = false;
            big[11] = false;
            big[13] = false;
            big[17] = false;
            big[19] = false;

            int sqrt = (int)Math.Sqrt(n);//平方根
            int step,j;
            sqrt -= 2;
            for (i = 23; i <= sqrt; i += 6)
            {
                if (!big[i])
                {
                    step = i * 6;
                    for (j = i * i; j <= n; j += step)
                    {
                        big[j] = true;
                    }
                    for (j = i * i + 2 * i; j <= n; j += step)
                    {
                        big[j] = true;
                    }
                }
                int mm = i + 2;
                if (!big[mm])
                {
                    step = mm * 6;
                    for (j = mm * mm; j <= n; j += step)
                    {
                        big[j] = true;
                    }
                    for (j = mm * mm + 4 * mm; j <= n; j += step)
                    {
                        big[j] = true;
                    }
                }
            }

            //統計素數個數
            for (i = 7; i <= n-4; i += 6)   //100010  從7開始
            {
                if (!big[i]) count++;
                if (!big[i+4]) count++;
            }
            count+=3; //還有2、3、5
            return count;
        }
    }
}