我对加密非常新，我非常感谢你对这个谜题的投入，因为我被卡住了。
谜题：
<?php> blinker
defined( '_JEXEC' ) or die( 'Restricted access' );
终端：〜一个梦幻般的实习机会......
-bash：解决这个难题如果你感兴趣那么[email protected] = [email protected]将带你到171225.13916.co.za
终端：〜答案是什么$
请注意我不会输入答案，我对如何解决这个问题非常感兴趣。
答案是job.mip.co.za取字母并将数字与每个字母相关联。
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
拿单词work：w = 23; o = 15; r = 18; k = 11如果你用fun这个词来做这个就行不通。那是因为乐趣是“与工作相反”。所以：f = 6但你必须从字母表的末尾开始计数到6。所以你会得到f = 21。你= 21，但你必须从字母表的末尾开始计算，你会得到u = 6。 n将是13。
现在你取数字171225，你也可以做有趣的事情。你会得到一份工作。至于数字13916你做同样的工作，你会得到单词mip。现在输入job.mip.co.za作为答案，你将被重定向到http://job.mip.co.za/
而已
if [email protected] = [email protected]
Then it is safe to assume that we working with a number and alphabet
correlation where a number equals an alphabet
Mathematically speaking 1 = A, and 2 = B .... and the are only 26 alphabets meaning we can't go higher than the number 26.
The first set of numbers are 5, and "fun" has three Alphabets where the only viable combination is
[21][6][13] = [U][f][M]
[2][16][13] = [A][P][M]
but none of them equals FUN
The second set of number are 8, and "work" has four Alphabets where the only viable combination is
[23][15][18][11] = [W][O][R][K]
Which give us the hint that the first set of numbers do in fact equals the word 'Fun'
we just have to look at the numbers differently.
how can 'F' be equal to [21]
and 'U' be equal to [6]
and 'N' be equal to [13]
or
how can 'F' be equal to [2]
and 'U' be equal to [16]
and 'N' be equal to [13]
we discovered counting from the right to the left side leaves us at undesired position
now lets try counting from the left side to right and see if it leave us at the desired position
[21] from left side to right side leave me at position [F]
[6] from left side to right side leave me at position [U]
[13] from left side to right side leave me at position [N]
[2] from left side to right side leave me at position [Y]
[6] from left side to right side leave me at position [U]
[13] from left side to right side leave me at position [N]
thus the first combination equals my desired value.
now using that same method it's safe to repeat on the next set of numbers.
171225.13916.co.za
[17] from left side to right side leave me at position [J]
[12] from left side to right side leave me at position [O]
[25] from left side to right side leave me at position [B]
and second set of numbers read from right side to left side
[13][9][16] = [M][I][P]
[13][9][1][6] = [M][I][A][F]
So that leaves me with only two possibilities.
'job.mip.co.za' or 'job.miaf.co.za' only of them is takes me where I need to be... my new home. Hi :)
job.米跑.co.咋
171225.13916.co.咋
M = 16; I = 9，P = 16
l = 17; o = 12; b = 25，即lob
取mip：m = 16; i = 9; p = 16，l = 12从字母表的末尾开始计数到12，得到l = 15。 o-15并从字母表的结尾开始计算得到o = 12和b = 25
job.米跑.co.咋 171225.13916.co.咋
我把mip取为M = 13 I = 9 P = 16
因此，对于Job J = 10并且从字母表的末尾开始计数到10，我得到j = 17，对于o = 15，从最后一个字母表的末尾开始计数到15，得到12，对于b = 2，从最后开始计数到2最后一个字母表我得到25。