牛牛和罗马数字智力游戏

#include <unordered_map>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param num int整型 
     * @param limit int整型 
     * @return string字符串
     */
    string integerToRomanWithReverse(int num, int limit) {
        // write code here
        // 哈希表+数学
        unordered_map<int, string> um;
        um[1] = "I";
        um[4] = "IV";
        um[5] = "V";
        um[9] = "IX";
        um[10] = "X";
        um[40] = "XL";
        um[50] = "L";
        um[90] = "XC";
        um[100] = "C";
        um[400] = "CD";
        um[500] = "D";
        um[900] = "CM";
        um[1000] = "M";

        // 如何将阿拉伯数字转成罗马数字呢？
        string str = "";
        while(num)
        {
            if(um.count(num))
            {
                str += um[num];
                break;
            }

            if(num>1000)
            {
                str += um[1000];
                num -= 1000;
            }
            else if(num>900)
            {
                str += um[900];
                num -= 900;
            }
            else if(num>500)
            {
                str += um[500];
                num -= 500;
            }
            else if(num>400)
            {
                str += um[400];
                num -= 400;
            }
            else if(num>100)
            {
                str += um[100];
                num -= 100;
            }
            else if(num>90)
            {
                str += um[90];
                num -= 90;
            }
            else if(num>50)
            {
                str += um[50];
                num -= 50;
            }
            else if(num>40)
            {
                str += um[40];
                num -= 40;
            }
            else if(num>10)
            {
                str += um[10];
                num -= 10;
            }
            else if(num>9)
            {
                str += um[9];
                num -= 9;
            }
            else if(num>5)
            {
                str += um[5];
                num -= 5;
            }
            else if(num>4)
            {
                str += um[4];
                num -= 4;
            }
            else
            {
                str += um[1];
                num -= 1;
            }
        }

        // 反转字符串
        if((int)str.size()>=limit)
            reverse(str.begin(),str.end());
        
        return str;
    }
};