HACKTOBERFEST - Added solution to Euler 64. (TheAlgorithms#3706)

* Added solution to Euler 64.

Added Python solution to Project Euler Problem 64.
Added a folder problem_064.
Added __init__.py file.
Added sol1.py file.

* Update sol1.py

Made formatting changes as mentioned by pre-commit

* Update sol1.py

Minor changes to variable naming and function calling as mentioned by @ruppysuppy

* Update sol1.py

Changes to function call as mentioned by @cclauss

Author: sudoShikhar

Diff for: project_euler/problem_064/__init__.py

@@ -0,0 +1,77 @@
+"""
+Project Euler Problem 64: https://projecteuler.net/problem=64
+
+All square roots are periodic when written as continued fractions.
+For example, let us consider sqrt(23).
+It can be seen that the sequence is repeating.
+For conciseness, we use the notation sqrt(23)=[4;(1,3,1,8)],
+to indicate that the block (1,3,1,8) repeats indefinitely.
+Exactly four continued fractions, for N<=13, have an odd period.
+How many continued fractions for N<=10000 have an odd period?
+
+References:
+- https://en.wikipedia.org/wiki/Continued_fraction
+"""
+
+from math import floor, sqrt
+
+
+def continuous_fraction_period(n: int) -> int:
+    """
+    Returns the continued fraction period of a number n.
+
+    >>> continuous_fraction_period(2)
+    1
+    >>> continuous_fraction_period(5)
+    1
+    >>> continuous_fraction_period(7)
+    4
+    >>> continuous_fraction_period(11)
+    2
+    >>> continuous_fraction_period(13)
+    5
+    """
+    numerator = 0.0
+    denominator = 1.0
+    ROOT = int(sqrt(n))
+    integer_part = ROOT
+    period = 0
+    while integer_part != 2 * ROOT:
+        numerator = denominator * integer_part - numerator
+        denominator = (n - numerator ** 2) / denominator
+        integer_part = int((ROOT + numerator) / denominator)
+        period += 1
+    return period
+
+
+def solution(n: int = 10000) -> int:
+    """
+    Returns the count of numbers <= 10000 with odd periods.
+    This function calls continuous_fraction_period for numbers which are
+    not perfect squares.
+    This is checked in if sr - floor(sr) != 0 statement.
+    If an odd period is returned by continuous_fraction_period,
+    count_odd_periods is increased by 1.
+
+    >>> solution(2)
+    1
+    >>> solution(5)
+    2
+    >>> solution(7)
+    2
+    >>> solution(11)
+    3
+    >>> solution(13)
+    4
+    """
+    count_odd_periods = 0
+    for i in range(2, n + 1):
+        sr = sqrt(i)
+        if sr - floor(sr) != 0:
+            if continuous_fraction_period(i) % 2 == 1:
+                count_odd_periods += 1
+    return count_odd_periods
+
+
+if __name__ == "__main__":
+    print(f"{solution(int(input().strip()))}")