#=
By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method! ;o)
=#
a = Array[
[75],
[95, 64],
[17, 47, 82],
[18, 35, 87, 10],
[20, 04, 82, 47, 65],
[19, 01, 23, 75, 03, 34],
[88, 02, 77, 73, 07, 63, 67],
[99, 65, 04, 28, 06, 16, 70, 92],
[41, 41, 26, 56, 83, 40, 80, 70, 33],
[41, 48, 72, 33, 47, 32, 37, 16, 94, 29],
[53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14],
[70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57],
[91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48],
[63, 66, 04, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
[04, 62, 98, 27, 23, 09, 70, 98, 73, 93, 38, 53, 60, 04, 23]]

function calc(arr)
  result = Array[]
  push!(result, arr[length(arr)])
  for row in reverse(1:length(arr)-1)
    nextrow = Int64[]
    for column in 1:length(arr[row])
      lowleft = last(result)[column]
      lowright = last(result)[column+1]
      center = arr[row][column]
      push!(nextrow, max(center+lowleft, center+lowright))
    end
    push!(result, nextrow)
  end
  last(result)[1]
end
@time println(calc(a))

