Question

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.

Answer 1

#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 "";     } }

Answer 2

#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;
}