<p>There are two types of soup: <strong>type A</strong> and <strong>type B</strong>. Initially, we have <code>n</code> ml of each type of soup. There are four kinds of operations:</p>

<ol>
	<li>Serve <code>100</code> ml of <strong>soup A</strong> and <code>0</code> ml of <strong>soup B</strong>,</li>
	<li>Serve <code>75</code> ml of <strong>soup A</strong> and <code>25</code> ml of <strong>soup B</strong>,</li>
	<li>Serve <code>50</code> ml of <strong>soup A</strong> and <code>50</code> ml of <strong>soup B</strong>, and</li>
	<li>Serve <code>25</code> ml of <strong>soup A</strong> and <code>75</code> ml of <strong>soup B</strong>.</li>
</ol>

<p>When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability <code>0.25</code>. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.</p>

<p><strong>Note</strong> that we do not have an operation where all <code>100</code> ml&#39;s of <strong>soup B</strong> are used first.</p>

<p>Return <em>the probability that <strong>soup A</strong> will be empty first, plus half the probability that <strong>A</strong> and <strong>B</strong> become empty at the same time</em>. Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 50
<strong>Output:</strong> 0.62500
<strong>Explanation:</strong> If we choose the first two operations, A will become empty first.
For the third operation, A and B will become empty at the same time.
For the fourth operation, B will become empty first.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 100
<strong>Output:</strong> 0.71875
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def soupServings(self, n: int) -> float:
        # If n is very large, the probability converges to 1.
        if n > 4500:
            return 1

        # Possible serving options (in ml) for A and B
        SERVING_OPTIONS = [(100, 0), (75, 25), (50, 50), (25, 75)]

        # Dictionary to store the chance of (remaining A, remaining B) soup servings
        soup_probabilities = {(n, n): [1, 1]}
        # Queue for BFS
        q = deque()
        q.appendleft((n, n))

        total_numerator = 0
        total_denominator = 1

        while q:
            ml_A, ml_B = q.pop()  # Current amounts of soup A and B
            numerator, denominator = soup_probabilities.pop((ml_A, ml_B))

            # If either soup A or B is finished
            if min(ml_A, ml_B) <= 0:
                # Scale the total numerator and denominator for precision
                while total_denominator < denominator:
                    total_numerator <<= 2
                    total_denominator <<= 2

                # If soup A is finished, add to the numerator (probability of A finishing first)
                if ml_A <= 0:
                    total_numerator += numerator
                # If soup B is not finished, add to the numerator (since B finishing after A also counts)
                if ml_B > 0:
                    total_numerator += numerator
                continue

            # Process each serving option
            for serve_A, serve_B in SERVING_OPTIONS:
                remaining_soups = (ml_A - serve_A, ml_B - serve_B)
                # If this state is not already in the dictionary, add it to the queue and initialize its probability
                if remaining_soups not in soup_probabilities:
                    q.appendleft(remaining_soups)
                    soup_probabilities[remaining_soups] = [0, denominator << 2]
                # Update the numerator for this state
                soup_probabilities[remaining_soups][0] += numerator

        # Return the probability of soup A finishing first
        return total_numerator / (total_denominator << 1)
16800