Euler_Problem-063.b93

v XXXX   #######################################################################
  XX
                          v1\1p12::+1   \+g2<
0                 v9p03p02<         \g12:+1<^2\p22_v
1          >v     v              <  v-\"P"<       :$
>        :v1v>0p>9>:"0"\0p1+:"O"-|>:0g"0"-||!`g12<-$
          :p\3v p050p04"O"<p0\"1"<^+>#1 $#< :21g-|\.
          211p> 40g0g"0"-2 0g*50g+:55+%"0"v>$55+0v+@
          >^20|-8p04:-1g04 p05/+5 #5p0g04+<>5#$5#<^
            >^>30g1-:30p #^_9     ^








1+::21p1\120p30p

[2, 0]           (Pow)  i
[3, 0]           (Pow)  n
[4, 0]           (Pow)  pos
[5, 0]           (Pow)  carry

[9, 0] - [80, 0] (Pow) arr

[2, 1]           (GetNDigitPowerNumberCount)  n
[2, 2]           tmp


// Pow   (stack, stack) -> (array)
         (i    , n)     -> [2, 9] - [2,80]

        v              < 
21p20p>9>:"0"\0p1+:"O"-|
     vp320p22"O"<p0\"1"<
     >22g0g"0"-2 0g*23g+:55+%"0"v
     |-8p22:-1g2 2p32/+55p0g22+ <
     >21g1-:21p#^_@



// Len(arr)

   v-\"P"<
9>:0g"0"-|
 ^+>#1 $#<   @



// GetNDigitPowerNumberCount    (stack)->(stack)

        v9p03p02          \g12:+1<        
 >v     v              <  v-\"P"<         
v1v>0p>9>:"0"\0p1+:"O"-|>:0g"0"-||!`g12<>-
:p\3v p050p04"O"<p0\"1"<^+>#1 $#< :21g-|\ 
211p> 40g0g"0"-2 0g*50g+:55+%"0"v>$55+0v+ 
>^20|-8p04:-1g04 p05/+5 #5p0g04+<>5#$5#<^ 
  >^>30g1-:30p #^_9     ^                 


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

After all the previous programs, this one is surprisingly ... dense (the main code-block is 54x8).

The algorithm is quickly explained for each length n we calculate the numbers `1^n`, `2^n` ... until `9^n` and see which have a length of `n`.
(From `10^n` upwards the condition is impossible, because `10^n` has `(n+1)` digits).

The main problem is that the numbers exceed Int64. So we need to implement long multiplication ... again. (see problem 16, 20, 29, 56 and 57)