Задание.Для записи римскими цифрами используются символы I, V, X, L, С, D, М, обозначающие соответственно числа 1, 5, 10, 50, 100, 500, 1000. Составить программу, которая запись любого данного числа п (п ≤ 3999) арабскими цифрами переводила бы в запись римскими цифрам. На С++.

string RomeZifr[13] = { "I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M" };
int ArabZifr[13] = { 1,4,5,9,10,40,50,90,100,400,500,900,1000 };
int ara,i=13;//iдля цикла
string S=" ";//s=для вывода числа на роме
cout << "Ведите число<=3999";
cin >> ara;
while (ara > 0) {
while (ArabZifr[i] > ara)
i -= 1;//уменьшаем
S = S + RomeZifr[i];
ara -= ArabZifr[i];
}
cout << "Число=" << S;
Может можно по другому?У меня она не работает.

#include<iostream> #include<string> using std::string; string func(int num, string, string, string); int main(int argc, char argv[]) {     system("chcp 1251>0");     std::cout << "Введите число не большее 3999: ";       int arabien_num;       if (!(std::cin >> arabien_num) || (arabien_num > 3999)) {         std::cerr << "Невозможно инициализировать число!" << std::endl;         exit(1);     }       std::string rome_num;       rome_num += func(arabien_num / 1000, "", "", "M");     arabien_num %= 1000;     rome_num += func(arabien_num / 100, "M", "D", "C");     arabien_num %= 100;     rome_num += func(arabien_num / 10,  "C", "L", "X");     arabien_num %= 10;     rome_num += func(arabien_num ,   "X", "V", "I");     std::cout << rome_num << std::endl;     return 0; } string func(int num, string hi, string re, string lo) {     switch(num)     {         case 9:             return lo + hi;         case 8:             return re + lo + lo + lo;         case 7:             return re + lo + lo;         case 6:             return re + lo;         case 5:             return re;         case 4:             return lo + re;         case 3:             return lo + lo + lo;         case 2:             return lo + lo;         case 1:             return lo;         case 0:             return "";     } }

#include <iostream>
#include <string>

using namespace std;

string solve(int num, int divd)
{
    string result = "";
    int count = 0;
    while(num)
    {
        count = num / divd;
        if(count)
        {
            num %= divd;
            for (int i = 0; i < count; i++)
                switch(divd)
                {
                    case 1000:
                    {
                        result+="M";
                        break;
                    }
                    case 900:
                    {
                        result+="CM";
                        break;
                    }
                    case 800:
                    {
                        result+="DCC";
                        break;
                    }
                    case 700:
                    {
                        result+="DCC";
                        break;
                    }
                    case 600:
                    {
                        result+="DC";
                        break;
                    }
                    case 500:
                    {
                        result+="D";
                        break;
                    }
                    case 400:
                    {
                        result+="CD";
                        break;
                    }
                    case 300:
                    {
                        result+="CCC";
                        break;
                    }
                    case 200:
                    {
                        result+="CC";
                        break;
                    }
                    case 100:
                    {
                        result+="C";
                        break;
                    }
                    case 90:
                    {
                        result+="XC";
                        break;
                    }
                    case 80:
                    {
                        result+="LXXX";
                        break;
                    }
                    case 70:
                    {
                        result+="LXX";
                        break;
                    }
                    case 60:
                    {
                        result+="LX";
                        break;
                    }
                    case 50:
                    {
                        result+="L";
                        break;
                    }
                    case 40:
                    {
                        result+="XL";
                        break;
                    }
                    case 30:
                    {
                        result+="XXX";
                        break;
                    }
                    case 20:
                    {
                        result+="XX";
                        break;
                    }
                    case 10:
                    {
                        result+="X";
                        break;
                    }
                    case 9:
                    {
                        result+="IX";
                        break;
                    }
                    case 8:
                    {
                        result+="VIII";
                        break;
                    }
                    case 7:
                    {
                        result+="VII";
                        break;
                    }
                    case 6:
                    {
                        result+="VI";
                        break;
                    }
                    case 5:
                    {
                        result+="V";
                        break;
                    }
                    case 4:
                    {
                        result+="IV";
                        break;
                    }
                    case 3:
                    {
                        result+="III";
                        break;
                    }
                    case 2:
                    {
                        result+="II";
                        break;
                    }
                    case 1:
                    {
                        result+="I";
                        break;
                    }
                }
        }
        if(divd <= 1000 && divd > 100) divd-=100;
        else if(divd <= 100 && divd > 10) divd-=10;
        else divd-=1;
    }
    return result;
}

int main()
{
    int num,count;
    cin >> num;
    if(num <= 3999) cout << solve(num,1000);
    else cout << "Incorrect input.";
    return 0;
}