Reduce allocations and refactoring

This commit is contained in:
David Doblas Jiménez 2022-10-16 15:37:13 +02:00
parent 653a5cb17a
commit 7c0439ea30

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@ -1,4 +1,4 @@
#=
#=
Created on 28 Jul 2021
@author: David Doblas Jiménez
@ -9,39 +9,32 @@ https://projecteuler.net/problem=17 =#
using BenchmarkTools
function num2letters(num)
nums = Dict(
0 => "", 1 => "one", 2 => "two", 3 => "three", 4 => "four", 5 => "five",
6 => "six", 7 => "seven", 8 => "eight", 9 => "nine", 10 => "ten",
11 => "eleven", 12 => "twelve", 13 => "thirteen", 14 => "fourteen",
15 => "fifteen", 16 => "sixteen", 17 => "seventeen", 18 => "eighteen",
19 => "nineteen", 20 => "twenty", 30 => "thirty", 40 => "forty",
50 => "fifty", 60 => "sixty", 70 => "seventy", 80 => "eighty",
90 => "ninety", 100 => "hundred", 1000 => "thousand"
)
function num2letters(num, nums)
if num <= 20
return length(nums[num])
return nums[num]
elseif num < 100
tens, units = divrem(num, 10)
return (length(nums[Int(tens) * 10]) + num2letters(units))
return (nums[tens*10] + num2letters(units, nums))
elseif num < 1000
hundreds, rest = divrem(num, 100)
if rest != 0
return (num2letters(hundreds) + length(nums[100])
+ length("and") + num2letters(rest))
return (
num2letters(hundreds, nums) +
nums[100] +
length("and") +
num2letters(rest, nums)
)
else
return (num2letters(hundreds) + length(nums[100]))
return (num2letters(hundreds, nums) + nums[100])
end
else
thousands, rest = divrem(num, 1000)
return (num2letters(thousands) + length(nums[1000]))
return (num2letters(thousands, nums) + nums[1000])
end
end
function Problem17()
#=
#=
If the numbers 1 to 5 are written out in words: one, two, three, four,
five, then there are 3 + 3 + 5 + 4 + 4 = 19 letters used in total.
@ -53,11 +46,44 @@ function Problem17()
20 letters. The use of "and" when writing out numbers is in compliance
with British usage. =#
nums = Dict(
0 => 0,
1 => 3,
2 => 3,
3 => 5,
4 => 4,
5 => 4,
6 => 3,
7 => 5,
8 => 5,
9 => 4,
10 => 3,
11 => 6,
12 => 6,
13 => 8,
14 => 8,
15 => 7,
16 => 7,
17 => 9,
18 => 8,
19 => 8,
20 => 6,
30 => 6,
40 => 5,
50 => 5,
60 => 5,
70 => 7,
80 => 6,
90 => 6,
100 => 7,
1000 => 8,
)
n = 1000
letters = 0
for num in 1:n
letters += num2letters(num)
for num = 1:n
letters += num2letters(num, nums)
end
return letters
end