Code cleanup and fixes using Black and linters

Author: vegadodo

Diff for: p001.py

@@ -1,15 +1,18 @@
-# Multiples of 3 and 5
-# Problem 1
-# https://projecteuler.net/problem=1
+"""
+Problem 1.
 
-sum = 0
+Multiples of 3 and 5
+https://projecteuler.net/problem=1
+"""
 
 # Pretty self-exlanatory code.
 # Check if given number is dividable by 3 or 5.
 # Rinse and repeat for 1 to 1000.
 
+answer = 0
+
 for i in range(1000):
     if i % 3 == 0 or i % 5 == 0:
-        sum += i
+        answer += i
 
-print(sum)
+print(answer)

Diff for: p002.py

@@ -1,28 +1,29 @@
-# Even Fibonacci numbers
-# Problem 2
-# https://projecteuler.net/problem=2
+"""
+Problem 2.
+
+Even Fibonacci numbers
+https://projecteuler.net/problem=2
+"""
 
 # Basic example of using a recursion.
 
+
 def fibonacci(n):
-    """
-    Find the nth fibonacci number.
-    """
+    """Return the nth fibonacci number."""
     if n == 1:
         return 1
-    elif n == 2:
+    if n == 2:
         return 2
-    else:
-        # Follow the definition.
-        return fibonacci(n - 1) + fibonacci(n - 2)
+    # Follow the definition.
+    return fibonacci(n - 1) + fibonacci(n - 2)
 
 
-sum = 0
+answer = 0
 n = 1
 
-while (fibonacci(n) < 4000000):
+while fibonacci(n) < 4000000:
     if fibonacci(n) % 2 == 0:
-        sum += fibonacci(n)
+        answer += fibonacci(n)
     n += 1
 
-print(sum)
+print(answer)

Diff for: p003.py

@@ -1,11 +1,13 @@
-# Largest prime factor
-# Problem 3
-# https://projecteuler.net/problem=3
+"""
+Problem 3.
+
+Largest prime factor
+https://projecteuler.net/problem=3
+"""
+
 
 def is_prime(n):
-    """
-    Find if the given number n is prime.
-    """
+    """Determine whether the given integer n is prime."""
     result = True
     for i in range(n - 1, 2, -1):
         if n % i == 0:
@@ -25,7 +27,7 @@ def is_prime(n):
 # However, we will not use sqrt() function!
 # We can just compare the quotient with the divisor.
 
-while (600851475143 // current_number >= current_number):
+while 600851475143 // current_number >= current_number:
     # We first check if current_number can divide 600851475143.
     # If 600851475143 is divisible
     # then we check if current_number is a prime, with the help of is_prime(n).

Diff for: p004.py

@@ -1,12 +1,13 @@
-# Largest palindrome product
-# Problem 4
-# https://projecteuler.net/problem=4
+"""
+Problem 4.
+
+Largest palindrome product
+https://projecteuler.net/problem=4
+"""
 
-def is_palindrome(n):
-    """
-    Find if n is palindrome.
-    """
 
+def is_palindrome(n):
+    """Determine whether n is palindrome."""
     # Convert the given number into list.
     number_original = list(map(int, str(n)))
 
@@ -15,14 +16,11 @@ def is_palindrome(n):
     # Then, reverse the list.
     number_reversed.reverse()
 
-    return (number_original == number_reversed)
+    return number_original == number_reversed
 
 
 def is_prod_of_two_3_digit_num(n):
-    """
-    Find if n is the product of 3-digit numbers.
-    """
-
+    """Determine whether n is the product of 3-digit numbers."""
     result = False
 
     for i in range(100, 1000):

Diff for: p005.py

@@ -1,29 +1,32 @@
-# Smallest multiple
-# Problem 5
-# https://projecteuler.net/problem=5
+"""
+Problem 5.
+
+Smallest multiple
+https://projecteuler.net/problem=5
+"""
 
-# For integer a, b, it holds that
-# a * b = LCM(a, b) * GCD(a, b).
-# Starting from range(1, 21), we replace two head elements x, y to LCM(x, y).
 
 def gcd(a, b):
     """
     Find the greatest common divisor of two integers a, b.
-    This uses Euclid's algorithm.
+
+    This uses Euclid's algorithm. For integer a, b, it holds that
+    a * b = LCM(a, b) * GCD(a, b).
+    Starting from range(1, 21), we replace two head elements x, y to LCM(x, y).
     """
     dividend = max(a, b)
     divisor = min(a, b)
 
     if divisor == 0:
         return dividend
-    else:
-        return gcd(dividend % divisor, divisor)
+
+    return gcd(dividend % divisor, divisor)
 
 
-lcm_list = range(1, 21)
+lcm_list = [*range(1, 21)]
 
 while len(lcm_list) != 1:
-    lcm_head = (lcm_list[0] * lcm_list[1]) / gcd(lcm_list[0], lcm_list[1])
+    lcm_head = (lcm_list[0] * lcm_list[1]) // gcd(lcm_list[0], lcm_list[1])
     del lcm_list[0:2]
     lcm_list.insert(0, lcm_head)