C#用递归算法实现：一列数的规则如下: 1、1、2、3、5、8、13、21、34，求第30位数是多少

```/// <summary>
/// 一列数的规则如下: 1、1、2、3、5、8、13、21、34求第30位数是多少， 用递归算法实现。(C#语言)
/// </summary>
/// <param name="pos"></param>
/// <returns></returns>
public int GetNumberAtPos(int pos)
{
if(pos==0||pos==1)
{
return 1;
}
int res = GetNumberAtPos(pos - 1) + GetNumberAtPos(pos - 2);
return res;
}```

```using System;
using System.Collections;
using System.Collections.Generic;
using System.Text;

namespace Test
{
public class Class1
{
private ArrayList list = new ArrayList();

public Class1()
{
}

public Class1(int num)
: base()
{
int i;

for (i = 1; i <= num; i++)
{
}
}

private int Calculation(int num)
{
if (num == 1 || num == 2)
return 1;
else
return Convert.ToInt32(list[num - 2]) + Convert.ToInt32(list[num - 3]);
}

public int Calculation()
{
return Convert.ToInt32(list[list.Count - 1]);
}
}

public class test
{
public static void Main()
{
int j;
int num;
for (j = 1; j < 100; j++)
{
Console.WriteLine("你要计算第多少位:");
{
{
if (num < 1)
continue;
else
{
Class1 c1 = new Class1(num);
Console.WriteLine(c1.Calculation());
}
}
else
{
continue;
}
}
else
{
break;
}
}
}
}
}

```

```public long getNumber(int pos)
{
long one = 1;
long two = 1;
if (pos == 0 || pos == 1)
{
return 1;
}
int i = 3;
long sum = 1;
while (i <= pos)
{
sum = one + two;
one = two;
two = sum;
i++;
}
return sum;
}

```