Add type hints in problems 31-40

Python/p031.py

@@ -10,13 +10,16 @@
 #
 # How many different ways can £2 be made using any number of coins?
 
+from typing import List
+
 from projecteuler import timing
 
 
 # Simple recursive function that tries every combination.
-def count(coins, value, n, i):
+def count(coins: List[int], value: int, n: int, i: int) -> int:
     for j in range(i, 8):
         value = value + coins[j]
+
         if value == 200:
             return n + 1
 
@@ -30,7 +33,7 @@ def count(coins, value, n, i):
 
 
 @timing
-def p031():
+def p031() -> None:
     coins = [1, 2, 5, 10, 20, 50, 100, 200]
 
     n = count(coins, 0, 0, 0)

Python/p032.py

@@ -16,7 +16,7 @@ from projecteuler import is_pandigital, timing
 
 
 @timing
-def p032():
+def p032() -> None:
     n = 0
     # Initially I used a bigger array, but printing the resulting products
     # shows that 10 values are sufficient.
@@ -70,14 +70,14 @@ def p032():
     # Sort the found products to easily see if there are duplicates.
     products = np.sort(products[:n])
 
-    sum_ = products[0]
+    _sum = products[0]
 
     for i in range(1, n):
         if products[i] != products[i-1]:
-            sum_ = sum_ + products[i]
+            _sum = _sum + products[i]
 
     print('Project Euler, Problem 32')
-    print(f'Answer: {sum_}')
+    print(f'Answer: {_sum}')
 
 
 if __name__ == '__main__':

Python/p033.py

@@ -16,7 +16,7 @@ from projecteuler import timing
 
 
 @timing
-def p033():
+def p033() -> None:
     prod_n = 1
     prod_d = 1
 

Python/p034.py

@@ -14,9 +14,9 @@ from projecteuler import timing
 
 
 @timing
-def p034():
+def p034() -> None:
     a = 10
-    sum_ = 0
+    _sum = 0
     factorials = ones(10, int)
 
     # Pre-calculate factorials of each digit from 0 to 9.
@@ -34,12 +34,12 @@ def p034():
             sum_f = sum_f + factorials[digit]
 
         if a == sum_f:
-            sum_ = sum_ + a
+            _sum = _sum + a
 
         a = a + 1
 
     print('Project Euler, Problem 34')
-    print(f'Answer: {sum_}')
+    print(f'Answer: {_sum}')
 
 
 if __name__ == '__main__':

Python/p035.py

@@ -14,7 +14,7 @@ N = 1000000
 primes = sieve(N)
 
 
-def is_circular_prime(n):
+def is_circular_prime(n: int) -> bool:
     # If n is not prime, it's obviously not a circular prime.
     if primes[n] == 0:
         return False
@@ -31,6 +31,7 @@ def is_circular_prime(n):
     while tmp > 0:
         if tmp % 2 == 0:
             return False
+
         # Count the number of digits.
         count = count + 1
         tmp = tmp // 10
@@ -46,7 +47,7 @@ def is_circular_prime(n):
 
 
 @timing
-def p035():
+def p035() -> None:
     count = 13
 
     # Calculate all primes below one million, then check if they're circular.

Python/p036.py

@@ -10,20 +10,20 @@ from projecteuler import is_palindrome, timing
 
 
 @timing
-def p036():
+def p036() -> None:
     N = 1000000
 
-    sum_ = 0
+    _sum = 0
 
     # Brute force approach. For every number below 1 million,
     # check if they're palindrome in base 2 and 10 using the
     # function implemented in projecteuler.c.
     for i in range(1, N):
         if is_palindrome(i, 10) and is_palindrome(i, 2):
-            sum_ = sum_ + i
+            _sum = _sum + i
 
     print('Project Euler, Problem 36')
-    print(f'Answer: {sum_}')
+    print(f'Answer: {_sum}')
 
 
 if __name__ == '__main__':

Python/p037.py

@@ -10,7 +10,7 @@
 from projecteuler import is_prime, timing
 
 
-def is_tr_prime(n):
+def is_tr_prime(n: int) -> bool:
     # One-digit numbers and non-prime numbers are
     # not truncatable primes.
     if n < 11 or not is_prime(n):
@@ -23,6 +23,7 @@ def is_tr_prime(n):
     while tmp > 0:
         if not is_prime(tmp):
             return False
+
         tmp = tmp // 10
 
     # Starting from the last digit, check if it's prime, then add back one digit at a time on the left and check if it
@@ -33,6 +34,7 @@ def is_tr_prime(n):
     while tmp != n:
         if not is_prime(tmp):
             return False
+
         i = i * 10
         tmp = n % i
 
@@ -41,20 +43,20 @@ def is_tr_prime(n):
 
 
 @timing
-def p037():
+def p037() -> None:
     i = 0
     n = 1
-    sum_ = 0
+    _sum = 0
 
     # Check every number until 11 truncatable primes are found.
     while i < 11:
         if is_tr_prime(n):
-            sum_ = sum_ + n
+            _sum = _sum + n
             i = i + 1
         n = n + 1
 
     print('Project Euler, Problem 37')
-    print(f'Answer: {sum_}')
+    print(f'Answer: {_sum}')
 
 
 if __name__ == '__main__':

Python/p038.py

@@ -17,8 +17,8 @@ from projecteuler import is_pandigital, timing
 
 
 @timing
-def p038():
-    max_ = 0
+def p038() -> None:
+    _max = 0
 
     # A brute force approach is used, starting with 1 and multiplying
     # the number by 1, 2 etc., concatenating the results, checking if
@@ -35,8 +35,8 @@ def p038():
             n = n + tmp
             j = j + 1
 
-            if n > max_ and is_pandigital(n, 9):
-                max_ = n
+            if n > _max and is_pandigital(n, 9):
+                _max = n
             if i * j < 10:
                 n = n * 10
             elif i * j < 100:
@@ -52,7 +52,7 @@ def p038():
                 break
 
     print('Project Euler, Problem 38')
-    print(f'Answer: {max_}')
+    print(f'Answer: {_max}')
 
 
 if __name__ == '__main__':

Python/p039.py

@@ -12,8 +12,9 @@ from projecteuler import timing
 
 
 @timing
-def p039():
-    max_ = 0
+def p039() -> None:
+    _max = 0
+    res = 0
     savedc = zeros(1000, int)
 
     # Start with p=12 (the smallest pythagorean triplet is (3,4,5) and 3+4+5=12.
@@ -59,8 +60,8 @@ def p039():
 
         # If the current value is greater than the maximum,
         # save the new maximum and the value of p.
-        if count > max_:
-            max_ = count
+        if count > _max:
+            _max = count
             res = p
 
     print('Project Euler, Problem 39')

Python/p040.py

@@ -16,7 +16,7 @@ from projecteuler import timing
 
 
 @timing
-def p040():
+def p040() -> None:
     digits = zeros(1000005, int)
     i = 1
     value = 1