Fix formatting for printing

src/Julia/Problem059.jl

@@ -0,0 +1,86 @@
+#=
+Created on 17 Oct 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for Problem 59 of Project Euler
+https://projecteuler.net/problem=59
+=#
+
+using BenchmarkTools
+using Combinatorics
+using DelimitedFiles
+
+function Problem59()
+    #=
+    Each character on a computer is assigned a unique code and the preferred
+    standard is ASCII (American Standard Code for Information Interchange).
+    For example, uppercase A = 65, asterisk (*) = 42, and lowercase k = 107.
+
+    A modern encryption method is to take a text file, convert the bytes to
+    ASCII, then XOR each byte with a given value, taken from a secret key.
+    The advantage with the XOR function is that using the same encryption key
+    on the cipher text, restores the plain text; for example, 65 XOR 42 = 107,
+    then 107 XOR 42 = 65.
+
+    For unbreakable encryption, the key is the same length as the plain text
+    message, and the key is made up of random bytes. The user would keep the
+    encrypted message and the encryption key in different locations, and
+    without both "halves", it is impossible to decrypt the message.
+
+    Unfortunately, this method is impractical for most users, so the modified
+    method is to use a password as a key. If the password is shorter than the
+    message, which is likely, the key is repeated cyclically throughout the
+    message. The balance for this method is using a sufficiently long password
+    key for security, but short enough to be memorable.
+
+    Your task has been made easy, as the encryption key consists of three
+    lower case characters. Using p059_cipher.txt, a file containing the
+    encrypted ASCII codes, and the knowledge that the plain text must contain
+    common English words, decrypt the message and find the sum of the ASCII
+    values in the original text.
+    =#
+
+    file = "/datos/Scripts/Gitea/Project_Euler/src/files/Problem59.txt"
+    data = readline(file)
+    data = parse.(Int, split(data, ","))
+    # encrypted = readdlm(file, ',', Int)
+
+    # print(encrypted)
+
+    # with open('../files/Problem59.txt', 'r') as f:
+    #     # encrypted = list(map(int, f.read().split(',')))
+    #     encrypted = [int(char) for char in f.read().split(',')]
+    # # print(encrypted)
+    # # print(test)
+    # plain_text = (length(encrypted) ÷ 3)
+    # println(plain_text)
+    # ascii_lowercase = join('a':'z')
+    # for key in combinations(ascii_lowercase, 3)
+    #     decrypted = ""
+    #     for (k, i) in zip(repeat(key, plain_text), encrypted)
+    #         decrypted += Char(round(BigInt(Int(k) ^ i)))
+    #     end
+
+    #     # assuming Euler will be in the text
+    #     if "Euler" in decrypted
+    #         return sum([Int(c) for c in decrypted])
+    #     end
+    # end
+    ascisum = 0
+    for shift=0:2, c='a':'z'
+        decrypted = data[1+shift:3:end] .⊻ Int(c)
+        if minimum(decrypted) == Int(' ') && maximum(decrypted) <= Int('z')
+            ascisum += sum(decrypted)
+        end
+    end
+
+    return ascisum
+end
+
+
+println("Time to evaluate Problem 59:")
+@btime Problem59()
+println("")
+println("Result for Problem 59: ", Problem59())