LintCode #3 统计数字

解题思路请参考

代码（可以通过，不过很乱，需要整理）：

/// <summary>
        /// 计算n在数组[targetNum]中出现的次数
        /// 形如：[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
        /// </summary>
        /// <param name="targetNum"></param>
        /// <param name="n"></param>
        /// <returns></returns>
        public static int GetCount(int targetNum, int n)
        {
            int count = ;

            if (n >  && n <= )
            {
                int ratio = ;

                while (true)
                {
                    int tmp = targetNum/ratio * ratio;

                    if (tmp > )
                    {
                        count += tmp / ;
                    }

                    int tmpA = targetNum - tmp;

                    if (ratio == )
                    {
                        if (tmpA >= n)
                        {
                            count++;
                        }
                    }
                    else
                    {
                        int tmpB = tmpA/(ratio/);
                        if (tmpB > n)
                        {
                            count += *(ratio/);
                        }
                        else if (tmpB == n)
                        {
                            count += tmpA%(ratio/) + ;
                        }
                    }
                    ratio *= ;

                    if (tmp == )
                    {
                        break;
                    }
                }
            }
            else if (n == )
            {
                int ratio = ;

                while (true)
                {
                    int tmp = targetNum / ratio * ratio;

                    if (tmp > )
                    {
                        count += tmp / ;
                    }

                    int tmpA = targetNum - tmp;
                    if (tmpA == targetNum)
                    {
                        if (tmpA > )
                        {
                            count -= ratio//;
                        }
                        else if (ratio == )
                        {
                            count = ;
                        }
                        break;
                    }
                    if (ratio == )
                    {
                        if (tmpA >= n)
                        {
                            count++;
                        }
                    }
                    else
                    {
                        int tmpB = tmpA / (ratio / );
                        if (tmpB > n)
                        {
                            count +=  * (ratio / );
                        }
                        else if (tmpB == n)
                        {
                            count += tmpA % (ratio / ) + ;
                        }
                    }
                    ratio *= ;
                }
            }
            return count;
        }