Python | Use Log Value, explained
First we notice that the trailing zeroes should be abbreviated, so we can calculate the number of trailing zeroes by calculating how many 2 and 5 in the result 
First we notice that the trailing zeroes should be abbreviated, so we can calculate the number of trailing zeroes by calculating how many `2` and `5` in the result of prime factorization of the product, and the number of trailing zeroes is the minimum between the two values.\n\nNext we notice that we only need to calculate the first and the last 5 digits of the result. We can handle it separately.\n* How to calculate the last 5 digits?\nWell, it is easier than calculating first 5 digits. We only need to keep the remainder of the running product divided by `100000`.\n* How to calculate the first 5 digits?\nWe can calculate the cumulative sum of log value of the product, then convert it back to real numbers.\nFor example, `log(123456789) ~= 8.09151497717` and `pow(10, 4.09151497717) ~= 12345.6789`. We can find that if we only need to calculate the first 5 digits, the log value larger than 4 (i.e. `4.xx`) would be enough.\n\nThere is a small trick that we need to add a small epsilon (like `1e-11`) to handle the float precision.\n\n```python \nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        \n        twos, fives = 0, 0\n        tail, cum_log = 1, 0.0\n        for i in range(left, right + 1):\n            # eliminate all the 2\'s and 5\'s\n            while i % 2 == 0:\n                i //= 2\n                twos += 1\n            while i % 5 == 0:\n                i //= 5\n                fives += 1\n            # keep last 5 digits with division by 100000\n            tail = (tail * i) % 100000\n            # calculate cumulative log\n            cum_log += log(i, 10)\n        # the number of trailing zeros\n        zeros = min(twos, fives)\n        \n        # add the extra 2\'s or 5\'s into tail and cum_log\n        tail = (tail * pow(2, twos - zeros, 100000) * pow(5, fives - zeros, 100000)) % 100000\n        cum_log += log(2, 10) * (twos - zeros) + log(5, 10) * (fives - zeros)\n        \n        if cum_log >= 10:\n            # extract first 5 digits from cumulative log\n            head = str(pow(10, cum_log - floor(cum_log) + 1e-11 + 4))[:5]\n            tail = str(tail).zfill(5)\n            return f\'{head}...{tail}e{zeros}\'\n        else:\n            # use the cumulative log to calcualte the whole digits\n            whole = str(floor(pow(10, cum_log + 1e-11)))\n            return f\'{whole}e{zeros}\'\n```
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