使用C#进行数学概率 - 如何在特定尝试中获得3个不同字符的枚举值

Question

嗨,我昨天已经问了这个问题而且因为不被理解而被关闭了,所以我会尝试更具体,因为这次我可以.所以情况就是这样.

1.我有3个字符:字符数组中的A,B,C如下:

char[] characters = {'A', 'B', 'C'};

2.在枚举这3个字符的所有可能匹配时,我得到例如以下内容:

Text    = A
tries   = 1
indexes = characters[0] 
------------
Text    = B
tries   = 2
indexes = characters[1] 
------------
Text    = C
tries   = 3
indexes = characters[2] 
------------
Text    = AA
tries   = 4
indexes = characters[0] characters[0] 
------------
Text    = AB
tries   = 5
indexes = characters[0] characters[1] 
------------
Text    = AC
tries   = 6
indexes = characters[0] characters[2] 
------------
Text    = BA
tries   = 7
indexes = characters[1] characters[0] 
------------
Text    = BB
tries   = 8
indexes = characters[1] characters[1] 
------------
Text    = BC
tries   = 9
indexes = characters[1] characters[2] 
------------
Text    = CA
tries   = 10
indexes = characters[2] characters[0] 
------------
Text    = CB
tries   = 11
indexes = characters[2] characters[1] 
------------
Text    = CC
tries   = 12
indexes = characters[2] characters[2] 

3.现在给出一定的尝试次数,我们可以在这个特定的尝试中得到索引的数量吗？例如,在尝试号码10的意思下,尝试文本是CA(如上所示),因为索引是字符[2]和字符[0],那么是否有一个数学方程式来知道这些索引数字的尝试次数？

谢谢

Answer 1

我们可以在这个特定的尝试中获得索引的数量

对于任何给定数量的字母,您只需计算基数为3的数字.因此,只需将十进制转换为base-3,然后再返回.

为了确定较少的字符串长度有多少种可能性:有3 ^ n种不同的长度字符串n使用3个字母.