88 lines
2 KiB
Ruby
88 lines
2 KiB
Ruby
description=%{
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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.
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3
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7 4
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2 4 6
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8 5 9 3
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That is, 3 + 7 + 4 + 9 = 23.
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Find the maximum total from top to bottom of the triangle below:
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75
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95 64
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17 47 82
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18 35 87 10
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20 04 82 47 65
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19 01 23 75 03 34
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88 02 77 73 07 63 67
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99 65 04 28 06 16 70 92
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41 41 26 56 83 40 80 70 33
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41 48 72 33 47 32 37 16 94 29
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53 71 44 65 25 43 91 52 97 51 14
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70 11 33 28 77 73 17 78 39 68 17 57
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91 71 52 38 17 14 91 43 58 50 27 29 48
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63 66 04 68 89 53 67 30 73 16 69 87 40 31
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04 62 98 27 23 09 70 98 73 93 38 53 60 04 23
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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)
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}
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reflexion=%{
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use bottom to top method
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z
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x y chose max(x,y) path for z, new triangle = z + max(x,y)
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}
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data_test=%{3
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7 4
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2 4 6
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8 5 9 3}
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data=%{75
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95 64
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17 47 82
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18 35 87 10
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20 04 82 47 65
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19 01 23 75 03 34
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88 02 77 73 07 63 67
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99 65 04 28 06 16 70 92
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41 41 26 56 83 40 80 70 33
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41 48 72 33 47 32 37 16 94 29
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53 71 44 65 25 43 91 52 97 51 14
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70 11 33 28 77 73 17 78 39 68 17 57
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91 71 52 38 17 14 91 43 58 50 27 29 48
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63 66 04 68 89 53 67 30 73 16 69 87 40 31
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04 62 98 27 23 09 70 98 73 93 38 53 60 04 23}
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triangle=data.split("\n").collect! { |line| line.split(" ").collect! {|n| n.to_i} }.reverse!
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p triangle
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def max(x,y)
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return x if x>y
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return y
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end
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nbline=0
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triangle.each do |line|
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if nbline==0
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nbline +=1
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next
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end
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col=0
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line.each do |n|
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triangle[nbline][col] =
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n + max( triangle[nbline-1][col],
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triangle[nbline-1][col+1] )
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col +=1
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end
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nbline+=1
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puts '==='
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p triangle
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end
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p triangle
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puts triangle[-1][0]
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