欧拉筛

void sd()
{
    int sum=0;
    for (int i = 2; i <= 100000; i++)
    {
        if (p[i] == false)
            s[sum++] = i;
        for (int j = 0; j < sum; j++)
        {
            if (i * s[j] > 100000)
                break;
            p[i * s[j]] = true;
            if (i % s[j] == 0)
                break;
        }
    }
}