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Reduce allocations and refactoring

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This commit is contained in:
David Doblas Jiménez 2023-04-05 22:04:20 +02:00
parent 9983ee23d6
commit 28ba539204
2 changed files with 19 additions and 15 deletions
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18
src/Julia/Problems001-050/Problem012.jl
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@ -4,8 +4,9 @@ Created on 21 Jul 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 12 of Project Euler
https://projecteuler.net/problem=12 =#
Solution for Problem 012 of Project Euler
https://projecteuler.net/problem=12
=#
using BenchmarkTools
@ -16,7 +17,7 @@ function num_divisors(number::Int64)
end
function Problem12()
function Problem012()
#=
The sequence of triangle numbers is generated by adding the natural
numbers. So the 7th triangle number would be:
@ -38,9 +39,10 @@ function Problem12()
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred
divisors? =#
divisors?
=#
ans::Int64 = 0
ans = 0
for number in Iterators.countfrom(1)
ans += number
if num_divisors(ans) > 500
@ -50,7 +52,7 @@ function Problem12()
end
println("Time to evaluate Problem $(lpad(12, 3, "0")):")
@btime Problem12()
println("Took:")
@btime Problem012()
println("")
println("Result for Problem $(lpad(12, 3, "0")): ", Problem12())
println("Result for Problem $(lpad(12, 3, "0")): ", Problem012())

16
src/Julia/Problems001-050/Problem014.jl
View File

@ -4,8 +4,9 @@ Created on 24 Jul 2021
@author: David Doblas Jiménez
@email: daviddoji@pm.me
Solution for Problem 14 of Project Euler
https://projecteuler.net/problem=14 =#
Solution for Problem 014 of Project Euler
https://projecteuler.net/problem=14
=#
using BenchmarkTools
@ -18,7 +19,7 @@ function chain_length(n)
return length
end
function Problem14()
function Problem014()
#=
The following iterative sequence is defined for the set of positive
integers:
@ -37,7 +38,8 @@ function Problem14()
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million. =#
NOTE: Once the chain starts the terms are allowed to go above one million.
=#
ans = 0
limit = 1_000_000
@ -53,7 +55,7 @@ function Problem14()
end
println("Time to evaluate Problem $(lpad(14, 3, "0")):")
@btime Problem14()
println("Took:")
@btime Problem014()
println("")
println("Result for Problem $(lpad(14, 3, "0")): ", Problem14())
println("Result for Problem $(lpad(14, 3, "0")): ", Problem014())