Euler_Problem-024.b93

v0123456789
>"ddd"**1v
vp129p11-<
       >v>v     v     <  v    <   >  v   >\1-\vv    p14+1g14<
>21g:1+|>|>21g0\>:1-:#^_v>*\:#^_$:|  >31p 141p >31g41g*11g`!|
       $             >$1 ^ > p68*v>$1^   |-"x" g0:\<\1<g14  <
       @ >          #^0#<#v0#,+ #<$     v>    >1+\:|  1      
^p12-1g12p11-*-1g14g13g11 <^\"x"\-"0"g0:>#- #1 #$ #<  ^      



AB`      A>B

[1,1] Current Number (1m -> 0)
[2,1] Current Possibilities (9 -> 0)
[3,1] Fac(Current Possibilities)
[4,1] Current Digit Index (1 -> ..)

// Factorial function (stack -> stack)

          v     <  v    <  >  v
        0\>:1-:#^_$>*\:#^_:|  >
                           >$1^
// 1 million

"ddd"**

// Get X-th number and cross out (num on stack)

                      >\1-\v
         1\>\:0g "x"-#^_   >1+v>
           |:\                <
           >$1-:0g"0"-\"x"\0p  ^

// Solution: 2783915460

---------------------------------------

This is a really nice problem - and I kinda like the solution.

If we think about the ordered numbers we can see that the first few start with `0`, the next with `1` and so on ...  
We can also see that there are `8!` possible numbers that start with `0` (or any other digit).  
So if `1 * 8! <= (1000000 - 0)` the result starts with `0`. Otherwise if `2 * 8! <= (1000000 - 0)`, etc.  
Then after we got our first digit (`2`) we can similar calculate the second with `1 * 7! <= (1000000 - 2 * 8!)`, `2 * 7! <= (1000000 - 2 * 8!)` ...  
The last thing we have to be aware of is that this method yields us to the "n-th digit of the remaining digits". So if we got the 6th digit for our second calculation its in fact `7`, because we already used `2`.

The program now does exactly these calculations, you can see in the top row the already used digits (they are crossed out).

All in all this program pretty fast - they could really do another version of this problem with bigger numbers.