Added solution to Project Euler problem 301

Author: KumarUniverse

project_euler/problem_301/__init__.py

@@ -0,0 +1,74 @@
+"""
+Project Euler Problem 301: https://projecteuler.net/problem=301
+
+Problem Statement:
+Nim is a game played with heaps of stones, where two players take
+it in turn to remove any number of stones from any heap until no stones remain.
+
+We'll consider the three-heap normal-play version of
+Nim, which works as follows:
+- At the start of the game there are three heaps of stones.
+- On each player's turn, the player may remove any positive
+  number of stones from any single heap.
+- The first player unable to move (because no stones remain) loses.
+
+If (n1, n2, n3) indicates a Nim position consisting of heaps of size
+n1, n2, and n3, then there is a simple function, which you may look up
+or attempt to deduce for yourself, X(n1, n2, n3) that returns:
+- zero if, with perfect strategy, the player about to
+  move will eventually lose; or
+- non-zero if, with perfect strategy, the player about
+  to move will eventually win.
+
+For example X(1,2,3) = 0 because, no matter what the current player does,
+the opponent can respond with a move that leaves two heaps of equal size,
+at which point every move by the current player can be mirrored by the
+opponent until no stones remain; so the current player loses. To illustrate:
+- current player moves to (1,2,1)
+- opponent moves to (1,0,1)
+- current player moves to (0,0,1)
+- opponent moves to (0,0,0), and so wins.
+
+For how many positive integers n <= 2^30 does X(n,2n,3n) = 0?
+"""
+
+def X(n: int, n2: int, n3: int) -> int:
+    """
+    Returns:
+    - zero if, with perfect strategy, the player about to
+      move will eventually lose; or
+    - non-zero if, with perfect strategy, the player about
+      to move will eventually win.
+
+    >>> X(1)
+    0
+    >>> X(3)
+    12
+    >>> X(8)
+    0
+    >>> X(11)
+    60
+    >>> X(1000)
+    3968
+    """
+    return n ^ n2 ^ n3
+
+def solution(n: int = 2**30) -> int:
+    """
+    For a given integer n <= 2^30, returns how many Nim games are lost.
+    >>> solution(2)
+    2
+    >>> solution(10)
+    144
+    >>> solution(30)
+    2178309
+    """
+    lossCount = 0
+    for i in range(1,n+1):
+        if X(i,2*i,3*i) == 0:
+            lossCount += 1
+
+    return lossCount
+
+if __name__ == "__main__":
+    print(solution())