2117. Abbreviating the Product of a Range

You are given two positive integers left and right with left <= right. Calculate the product of all integers in the inclusive range [left, right].

Since the product may be very large, you will abbreviate it following these steps:

• 1. Count all trailing zeros in the product and remove them. Let us denote this count as C.

• For example, there are 3 trailing zeros in 1000, and there are 0 trailing zeros in 546.

• 1. Denote the remaining number of digits in the product as d. If d > 10, then express the product as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf>denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it unchanged.

• For example, we express 1234567654321 as 12345…54321, but 1234567 is represented as 1234567.

• 1. Finally, represent the product as a string "<pre>...<suf>eC".

• For example, 12345678987600000 will be represented as “12345…89876e5”.

Return a string denoting the abbreviated product of all integers in the inclusive range [left, right].

class Solution {
public:
    string abbreviateProduct(int left, int right) {
        long long suf = 1, POW = 10000000000;
        double pre = 1.0;
        int zero = 0, remove = 0;
        for(int i = left; i <= right; i++) {
            suf *= i;
            pre *= i;
            while(pre >= 1.0) {
                pre /= 10.0;
                remove++;
            }
            while(!(suf % 10)) {
                suf /= 10;
                zero++;
            }
            if(suf >= POW) suf %= POW;
        }
        if(remove - zero <= 10) {
            return to_string((long) (pre * pow(10, remove - zero) + 0.5)) + 'e' + to_string(zero);
        }
        string ssuf = "0000" + to_string(suf);
        return to_string((long) (pre * pow(10, 5))) + "..." + ssuf.substr(ssuf.length() - 5) + 'e' + to_string(zero);
    }
};