Add Project Euler Problem 078 solution 01 (#5565)

* Create sol1.py

* updating DIRECTORY.md

* Create __init__.py

* Add docstring

* Reformat with black

* Fix flake8 issues

* Add EOL

* Fix formatting issues

* Add docstring

* Add func return type

* Change return type

* Remove test print statement

* Reformat code

* Fix return types

* Break loop

* Update doctest sol

* Update project_euler/problem_078/sol1.py

Co-authored-by: John Law <johnlaw.po@gmail.com>

* Added doctest and changed return type

* Add int()

* Fix flake8 issues

* Use argument instead of fixed constant

* Update sol1.py

* fix sol1.py

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>
Co-authored-by: John Law <johnlaw.po@gmail.com>

DIRECTORY.md

@@ -788,6 +788,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_076/sol1.py)
   * Problem 077
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_077/sol1.py)
+  * Problem 078
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_078/sol1.py)
   * Problem 080
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_080/sol1.py)
   * Problem 081

project_euler/problem_078/sol1.py

@@ -0,0 +1,55 @@
+"""
+Problem 78
+Url: https://projecteuler.net/problem=78
+Statement:
+Let p(n) represent the number of different ways in which n coins
+can be separated into piles. For example, five coins can be separated
+into piles in exactly seven different ways, so p(5)=7.
+
+            OOOOO
+            OOOO   O
+            OOO   OO
+            OOO   O   O
+            OO   OO   O
+            OO   O   O   O
+            O   O   O   O   O
+Find the least value of n for which p(n) is divisible by one million.
+"""
+
+import itertools
+
+
+def solution(number: int = 1000000) -> int:
+    """
+    >>> solution()
+    55374
+    """
+    partitions = [1]
+
+    for i in itertools.count(len(partitions)):
+        item = 0
+        for j in itertools.count(1):
+            sign = -1 if j % 2 == 0 else +1
+            index = (j * j * 3 - j) // 2
+            if index > i:
+                break
+            item += partitions[i - index] * sign
+            index += j
+            if index > i:
+                break
+            item += partitions[i - index] * sign
+            item %= number
+
+        if item == 0:
+            return i
+        partitions.append(item)
+
+    return 0
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    print(f"{solution() = }")