[Python3] quasi brute-force
Please check out this commit for solutions of biweekly 68. \n\n\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        ans = 
Please check out this [commit](https://github.com/gaosanyong/leetcode/commit/8bba95f803d58a5e571fa13de6635c96f5d1c1ee) for solutions of biweekly 68. \n\n```\nclass Solution:\n    def abbreviateProduct(self, left: int, right: int) -> str:\n        ans = prefix = suffix = 1\n        trailing = 0 \n        flag = False \n        for x in range(left, right+1): \n            if not flag: \n                ans *= x\n                while ans % 10 == 0: ans //= 10 \n                if ans >= 1e10: flag = True \n            prefix *= x\n            suffix *= x\n            while prefix >= 1e12: prefix //= 10 \n            while suffix % 10 == 0: \n                trailing += 1\n                suffix //= 10 \n            if suffix >= 1e10: suffix %= 10_000_000_000\n        while prefix >= 100000: prefix //= 10 \n        suffix %= 100000\n        if flag: return f"{prefix}...{suffix:>05}e{trailing}"\n        return f"{ans}e{trailing}"\n```
['Python3']