C#，雷卡曼数（Recamán Number）的算法与源代码

1 雷卡曼数（Recamán Number）

雷卡曼数（Recamán Number），即Recaman序列被定义如下：

(1) a0=0;

(2) 如果am-1-m>0并且这个值在序列中不存在，则am=am-1-m;

(3) 否则am=am-1+m;

雷卡曼序列的前几个数的数值是：0,1,3,6,2,7,13,20,12,21,11,22,10,23,9,,,,

2 源程序

using System;

using System.Collections;

using System.Collections.Generic;

namespace Legalsoft.Truffer.Algorithm

{

public static partial class Number_Sequence

{

public static int Recaman_Number(int n)

{

int\[\] arr = new intn;

arr0 = 0;

for (int i = 1; i < n; i++)

{

int curr = arri - 1 - i;

for (int j = 0; j < i; j++)

{

if ((arrj == curr) || curr < 0)

{

curr = arri - 1 + i;

break;

}

}

arri = curr;

}

return arrn - 1;

}

public static int Recaman_Number_Second(int n)

{

if (n <= 0)

{

return 0;

}

HashSet<int> s = new HashSet<int>();

s.Add(0);

int prev = 0;

for (int i = 1; i < n; i++)

{

int curr = (prev - i);

if (curr < 0 || s.Contains(curr))

{

curr = (prev + i);

}

s.Add(curr);

prev = curr;

}

return prev;

}

}

}

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3 代码格式

using System;
using System.Collections;
using System.Collections.Generic;

namespace Legalsoft.Truffer.Algorithm
{
    public static partial class Number_Sequence
    {
        public static int Recaman_Number(int n)
        {
            int[] arr = new int[n];
            arr[0] = 0;
            for (int i = 1; i < n; i++)
            {
                int curr = arr[i - 1] - i;
                for (int j = 0; j < i; j++)
                {
                    if ((arr[j] == curr) || curr < 0)
                    {
                        curr = arr[i - 1] + i;
                        break;
                    }
                }
                arr[i] = curr;
            }
            return arr[n - 1];
        }

        public static int Recaman_Number_Second(int n)
        {
            if (n <= 0)
            {
                return 0;
            }
            HashSet<int> s = new HashSet<int>();
            s.Add(0);

            int prev = 0;
            for (int i = 1; i < n; i++)
            {
                int curr = (prev - i);
                if (curr < 0 || s.Contains(curr))
                {
                    curr = (prev + i);
                }
                s.Add(curr);
                prev = curr;
            }
            return prev;
        }
    }
}