<p>You are given two positive integers <code>left</code> and <code>right</code> with <code>left &lt;= right</code>. Calculate the <strong>product</strong> of all integers in the <strong>inclusive</strong> range <code>[left, right]</code>.</p>

<p>Since the product may be very large, you will <strong>abbreviate</strong> it following these steps:</p>

<ol>
	<li>Count all <strong>trailing</strong> zeros in the product and <strong>remove</strong> them. Let us denote this count as <code>C</code>.

	<ul>
		<li>For example, there are <code>3</code> trailing zeros in <code>1000</code>, and there are <code>0</code> trailing zeros in <code>546</code>.</li>
	</ul>
	</li>
	<li>Denote the remaining number of digits in the product as <code>d</code>. If <code>d &gt; 10</code>, then express the product as <code>&lt;pre&gt;...&lt;suf&gt;</code> where <code>&lt;pre&gt;</code> denotes the <strong>first</strong> <code>5</code> digits of the product, and <code>&lt;suf&gt;</code> denotes the <strong>last</strong> <code>5</code> digits of the product <strong>after</strong> removing all trailing zeros. If <code>d &lt;= 10</code>, we keep it unchanged.
	<ul>
		<li>For example, we express <code>1234567654321</code> as <code>12345...54321</code>, but <code>1234567</code> is represented as <code>1234567</code>.</li>
	</ul>
	</li>
	<li>Finally, represent the product as a <strong>string</strong> <code>&quot;&lt;pre&gt;...&lt;suf&gt;eC&quot;</code>.
	<ul>
		<li>For example, <code>12345678987600000</code> will be represented as <code>&quot;12345...89876e5&quot;</code>.</li>
	</ul>
	</li>
</ol>

<p>Return <em>a string denoting the <strong>abbreviated product</strong> of all integers in the <strong>inclusive</strong> range</em> <code>[left, right]</code>.</p>

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

<pre>
<strong>Input:</strong> left = 1, right = 4
<strong>Output:</strong> &quot;24e0&quot;
<strong>Explanation:</strong> The product is 1 &times; 2 &times; 3 &times; 4 = 24.
There are no trailing zeros, so 24 remains the same. The abbreviation will end with &quot;e0&quot;.
Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further.
Thus, the final representation is &quot;24e0&quot;.
</pre>

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

<pre>
<strong>Input:</strong> left = 2, right = 11
<strong>Output:</strong> &quot;399168e2&quot;
<strong>Explanation:</strong> The product is 39916800.
There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with &quot;e2&quot;.
The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further.
Hence, the abbreviated product is &quot;399168e2&quot;.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> left = 371, right = 375
<strong>Output:</strong> &quot;7219856259e3&quot;
<strong>Explanation:</strong> The product is 7219856259000.
</pre>

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

<ul>
	<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        # calc prod of l..r

        # abbreviate as follows:
        #      > count trailing zeros == C and remove them
        #      > let remainder be d
        #      > if d > 10 then report first and last five digits

        # then print either (<=10 digits)eC
        #       or (first5..last5)eC

        # last five pretty easy, mod 1e5
        # first five more interesting

        # if we have Ae10 + Be9 + Ce8 + ...
        # and we multiply by x
        # we get Axe10 + Bxe9 + ...
        # and the carry from D, E, F can roll up through A, B

        # solutions include:
        #      Python, Java: brute-force with arb. prec. ints
        #      C++: roll your own
        
        # is there a clever way to get the first several digits?
        # e.g. 1999 * 3 = 5997 via (2000-1)*3

        # maybe groups of 10?
        # if we know the product of a..b
        #    then a+10..b+10 would be
        #    (a+10)(a+1+10)..(b+10)
        # a..b + 10*(a+1..b + a,a+2..b + ... + a..b-1) + 100*(all ways to drop 2) + 1000*(all ways to drop 3) + ...
        # YIKES

        # so manual multiply is probably the way to go barring some other innovations

        # except manual multiply takes an order n*log10(n) digit number, does 3 mults on it, then sum
        # for four passes over n*log10(n) digits O(n) times, yikes
        # TOO EXPENSIVE

        # so we need a math trick for this
        #   last five digits are EZ

        # first five digits are hard
        # ... we're not expected to do the FFT multiply right? that would be very mean

        # is there a faster way to get just a specific digit? For example what's the first digit?
        # or can we get the number of digits?
        #    yes, from log10 of the numbers summed
        #
        # we can get log10 of the number, and the floating point might give us what we need

        twos = 0
        fives = 0
        for x in range(left, right+1):
            while x % 2 == 0:
                twos += 1
                x //= 2
            while x % 5 == 0:
                fives += 1
                x //= 5

        C = min(twos, fives)

        l10 = sum(log10(x) for x in range(left, right+1))
        digits = int(l10 + 1)
        
        if digits - C <= 10:
            # at most 10 digits besides trailing zeros: find the number via explicit multiply and divide-out-10s
            p = 1
            for x in range(left, right+1):
                p *= x
                while p % 10 == 0:
                    p //= 10

            return f"{p}e{C}"

        # FIXED: lost some essential precision when the product is huge, e.g. l10 is 5000.something
        # which means we dropped 3 digits from some of the values. So aggressively remove the integer
        # part of the log10 so we retain as many digits as possible
        l10 = 0
        for x in range(left, right+1):
            while x % 10 == 0: x //= 10
            l10 += log10(x)
            l10 -= int(l10)

        # separate calculation of first and last 5 digits,
        # potentially too many digits to divide out

        # for first five:
        #   log10(p) = log10(first*10**(digits-1) + second*10**(digits-2) + ...)
        #             digits-1 + log10(first + second/10 + third/100 + ...)
        #
        #     so 10**(l10-int(l10)) == first+second/10 + ... fifth/1e4
        #        (.................) * 1e4 is first*1e4 + .. + fifth   <- this is the number we want

        firstFive = int(10**l10*1e4)

        # fixed: last 5 are tricky
        # we know that the last five are from factors that are not 2 or 5
        # so what we need to do is get the digits other than 2 and 5
        # multiply in excess factors of 2 and 5 to get all of them
        # then mod 1e5
        lastFive = right
        for x in range(left, right):
            # get last five digits if we drop all powers of 2 and 5
            lastFive *= x
            while lastFive % 2 == 0: lastFive //= 2
            while lastFive % 5 == 0: lastFive //= 5

            lastFive %= 100_000

        # C handles 2**C and 5**C
        #    if we have excess 2s and 5s we need to add them in
        for _ in range(twos-C):
            lastFive *= 2
            lastFive %= 100_000
        for _ in range(fives-C):
            lastFive *= 5
            lastFive %= 100_000

        # FIXED: must print leading zeros in lastFive % 1e5
        return f"{firstFive}...{lastFive:05d}e{C}"
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