string ss = "a1a2a3a4a5a6a7a8a9";
            MatchCollection mc = Regex.Matches(ss, @"a\d");
            for (int i = 0; i < mc.Count; i++)
            {
                Console.WriteLine(mc[i].Value);
            }

            string[] r = new string[mc.Count];
            for (int i = 0; i < mc.Count; i++)
            {
                r[i] = mc[i].Value;
            }

        private string[] SplitStringArray()
        {
            System.Collections.ArrayList list = new System.Collections.ArrayList();
            string ss = "a1a2a3a4a5a6a7a8a9";
            System.Text.RegularExpressions.MatchCollection mc = System.Text.RegularExpressions.Regex.Matches(ss, @"a\d");
            for (int i = 0; i < mc.Count; i++)
            {
                list.Add(mc[i].Value.ToString());
                Console.WriteLine(mc[i].Value);
            }

            return (String[])list.ToArray(typeof(string));
        }

------解决方案--------------------
没必要吧。。。

MatchCollection 的内容远远不止一个string[]，做这种只能说华而不实，一定要做，自己写个方法，处理一下，单纯取$0的内容存到一个数组里，使用时调用这个方法也就是了

当然，对于一些简单的，可以用Regex.Split来做
C# code
string ss = "a1a2a3a4a5a6a7a8a9";
string[] result = Regex.Split(ss, @"(?<=a\d+)(?=a\d+)", RegexOptions.IgnoreCase);