将介于1到3999之间的十进制数转换为罗马数字

给定一个数字，找到其对应的罗马数字。
例子：

Input : 9
Output : IX

Input : 40
Output : XL

Input :  1904
Output : MCMIV

以下是罗马符号的列表，其中还包括减号：

SYMBOL       VALUE
I             1
IV            4
V             5
IX            9
X             10
XL            40
L             50
XC            90
C             100
CD            400
D             500
CM            900 
M             1000       

想法是分别转换给定数字的单位，数十，数百和数千个位置。如果数字为0，则没有相应的罗马数字符号。数字4和9的转换与其他数字稍有不同，因为这些数字采用减法表示法。

将十进制数字转换为罗马数字的算法
将给定数字与基数按顺序进行比较，顺序为1000、900、500、400、100、90、50、40、10、9、5、4、1。正好小于或等于给定数字的基数将是初始基值(最大基值)。将数字除以其最大基值，对应的基符号将被重复商次，然后剩余的将成为将来除法和重复的数字。将重复此过程直到该数字变为零。

演示上述算法的示例：

Convert 3549 to its Roman Numerals

输出：

MMMDXLIX

解释：

解释：

步骤1

• 初始编号= 3549
• 由于3549> = 1000;最初的最大基值将是1000。
• 除以3549/1000。商= 3，余数= 549。对应的符号M将重复三次。

第2步

• 现在，数字= 549
• 1000> 549> = 500;最大基值将是500。
• 除以549/500。商= 1，余数= 49。对应的符号D将重复一次。

第三步

• 现在，数字= 49
• 50> 49> = 40;最大基值为40。
• 除以49/40。商= 1，余数=9。相应的符号XL将重复一次。

第四步

• 现在，数字= 9
• 10> 9> = 9;最大基值为9。
• 除以9/9。商= 1，余数=0。相应的符号IX将重复一次。

第5步

• 最终数字变为0，算法在此处停止。
• 获得的输出为MMMDXLIX。

以下是上述算法的实现：

C++

// C++ Program to convert decimal number to
// roman numerals
#include 
using namespace std;
 
// Function to convert decimal to Roman Numerals
int printRoman(int number)
{
    int num[] = {1,4,5,9,10,40,50,90,100,400,500,900,1000};
    string sym[] = {"I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"};
    int i=12;   
    while(number>0)
    {
      int div = number/num[i];
      number = number%num[i];
      while(div--)
      {
        cout<

---

Java

// Java Program to convert decimal number to
// roman numerals
class GFG {
 
// To add corresponding base symbols in the array
// to handle cases that follow subtractive notation.
// Base symbols are added index 'i'.
    static int sub_digit(char num1, char num2, int i, char[] c) {
        c[i++] = num1;
        c[i++] = num2;
        return i;
    }
 
// To add symbol 'ch' n times after index i in c[]
    static int digit(char ch, int n, int i, char[] c) {
        for (int j = 0; j < n; j++) {
            c[i++] = ch;
        }
        return i;
    }
 
// Function to convert decimal to Roman Numerals
    static void printRoman(int number) {
        char c[] = new char[10001];
        int i = 0;
 
        // If number entered is not valid
        if (number <= 0) {
            System.out.printf("Invalid number");
            return;
        }
 
        // TO convert decimal number to roman numerals
        while (number != 0) {
            // If base value of number is greater than 1000
            if (number >= 1000) {
                // Add 'M' number/1000 times after index i
                i = digit('M', number / 1000, i, c);
                number = number % 1000;
            } // If base value of number is greater than or
            // equal to 500
            else if (number >= 500) {
                // To add base symbol to the character array
                if (number < 900) {
                    // Add 'D' number/1000 times after index i
                    i = digit('D', number / 500, i, c);
                    number = number % 500;
                } // To handle subtractive notation in case of number
                // having digit as 9 and adding corresponding base
                // symbol
                else {
                    // Add C and M after index i/.
                    i = sub_digit('C', 'M', i, c);
                    number = number % 100;
                }
            } // If base value of number is greater than or equal to 100
            else if (number >= 100) {
                // To add base symbol to the character array
                if (number < 400) {
                    i = digit('C', number / 100, i, c);
                    number = number % 100;
                } // To handle subtractive notation in case of number
                // having digit as 4 and adding corresponding base
                // symbol
                else {
                    i = sub_digit('C', 'D', i, c);
                    number = number % 100;
                }
            } // If base value of number is greater than or equal to 50
            else if (number >= 50) {
                // To add base symbol to the character array
                if (number < 90) {
                    i = digit('L', number / 50, i, c);
                    number = number % 50;
                } // To handle subtractive notation in case of number
                // having digit as 9 and adding corresponding base
                // symbol
                else {
                    i = sub_digit('X', 'C', i, c);
                    number = number % 10;
                }
            } // If base value of number is greater than or equal to 10
            else if (number >= 10) {
                // To add base symbol to the character array
                if (number < 40) {
                    i = digit('X', number / 10, i, c);
                    number = number % 10;
                } // To handle subtractive notation in case of
                // number having digit as 4 and adding
                // corresponding base symbol
                else {
                    i = sub_digit('X', 'L', i, c);
                    number = number % 10;
                }
            } // If base value of number is greater than or equal to 5
            else if (number >= 5) {
                if (number < 9) {
                    i = digit('V', number / 5, i, c);
                    number = number % 5;
                } // To handle subtractive notation in case of number
                // having digit as 9 and adding corresponding base
                // symbol
                else {
                    i = sub_digit('I', 'X', i, c);
                    number = 0;
                }
            } // If base value of number is greater than or equal to 1
            else if (number >= 1) {
                if (number < 4) {
                    i = digit('I', number, i, c);
                    number = 0;
                } // To handle subtractive notation in case of
                // number having digit as 4 and adding corresponding
                // base symbol
                else {
                    i = sub_digit('I', 'V', i, c);
                    number = 0;
                }
            }
        }
 
        // Printing equivalent Roman Numeral
        System.out.printf("Roman numeral is: ");
        for (int j = 0; j < i; j++) {
            System.out.printf("%c", c[j]);
        }
    }
 
//Driver program
    public static void main(String[] args) {
        int number = 3549;
 
        printRoman(number);
    }
}
// This code is contributed by PrinciRaj1992

Python3

# Python3 program to convert
# decimal number to roman numerals
 
# Function to convert decimal to Roman Numerals
def printRoman(number):
    num = [1, 4, 5, 9, 10, 40, 50, 90,
           100, 400, 500, 900, 1000]
    sym = ["I", "IV", "V", "IX", "X", "XL",
           "L", "XC", "C", "CD", "D", "CM", "M"]
    i = 12
    while number:
        div = number // num[i]
        number %= num[i]
 
        while div:
            print(sym[i], end = "")
            div -= 1
        i -= 1
 
# Driver code
if __name__ == "__main__":
    number = 3549
    print("Roman numeral is:", end = " ")
    printRoman(number)
     
# This code is contributed by
# sanjeev2552

C#

// C# Program to convert decimal number
// to roman numerals
using System;
class GFG
{
 
// To add corresponding base symbols in the
// array to handle cases which follow subtractive
// notation. Base symbols are added index 'i'.
static int sub_digit(char num1, char num2,
                         int i, char[] c)
{
    c[i++] = num1;
    c[i++] = num2;
    return i;
}
 
// To add symbol 'ch' n times after index i in c[]
static int digit(char ch, int n, int i, char[] c)
{
    for (int j = 0; j < n; j++)
    {
        c[i++] = ch;
    }
    return i;
}
 
// Function to convert decimal to Roman Numerals
static void printRoman(int number)
{
    char[] c = new char[10001];
    int i = 0;
 
    // If number entered is not valid
    if (number <= 0)
    {
        Console.WriteLine("Invalid number");
        return;
    }
 
    // TO convert decimal number to
    // roman numerals
    while (number != 0)
    {
        // If base value of number is
        // greater than 1000
        if (number >= 1000)
        {
            // Add 'M' number/1000 times after index i
            i = digit('M', number / 1000, i, c);
            number = number % 1000;
        }
         
        // If base value of number is greater
        // than or equal to 500
        else if (number >= 500)
        {
            // To add base symbol to the character array
            if (number < 900)
            {
                 
                // Add 'D' number/1000 times after index i
                i = digit('D', number / 500, i, c);
                number = number % 500;
            }
             
            // To handle subtractive notation in case
            // of number having digit as 9 and adding
            // corresponding base symbol
            else
            {
                 
                // Add C and M after index i/.
                i = sub_digit('C', 'M', i, c);
                number = number % 100;
            }
        }
         
        // If base value of number is greater
        // than or equal to 100
        else if (number >= 100)
        {
            // To add base symbol to the character array
            if (number < 400)
            {
                i = digit('C', number / 100, i, c);
                number = number % 100;
            }
             
            // To handle subtractive notation in case
            // of number having digit as 4 and adding
            // corresponding base symbol
            else
            {
                i = sub_digit('C', 'D', i, c);
                number = number % 100;
            }
        }
         
        // If base value of number is greater
        // than or equal to 50
        else if (number >= 50)
        {
             
            // To add base symbol to the character array
            if (number < 90)
            {
                i = digit('L', number / 50, i, c);
                number = number % 50;
            }
             
            // To handle subtractive notation in case
            // of number having digit as 9 and adding
            // corresponding base symbol
            else
            {
                i = sub_digit('X', 'C', i, c);
                number = number % 10;
            }
        }
         
        // If base value of number is greater
        // than or equal to 10
        else if (number >= 10)
        {
             
            // To add base symbol to the character array
            if (number < 40)
            {
                i = digit('X', number / 10, i, c);
                number = number % 10;
            }
             
            // To handle subtractive notation in case of
            // number having digit as 4 and adding
            // corresponding base symbol
            else
            {
                i = sub_digit('X', 'L', i, c);
                number = number % 10;
            }
        }
         
        // If base value of number is greater
        // than or equal to 5
        else if (number >= 5)
        {
            if (number < 9)
            {
                i = digit('V', number / 5, i, c);
                number = number % 5;
            }
             
            // To handle subtractive notation in case
            // of number having digit as 9 and adding
            // corresponding base symbol
            else
            {
                i = sub_digit('I', 'X', i, c);
                number = 0;
            }
        }
         
        // If base value of number is greater
        // than or equal to 1
        else if (number >= 1)
        {
            if (number < 4)
            {
                i = digit('I', number, i, c);
                number = 0;
            }
             
            // To handle subtractive notation in
            // case of number having digit as 4
            // and adding corresponding base symbol
            else
            {
                i = sub_digit('I', 'V', i, c);
                number = 0;
            }
        }
    }
 
    // Printing equivalent Roman Numeral
    Console.WriteLine("Roman numeral is: ");
    for (int j = 0; j < i; j++)
    {
        Console.Write("{0}", c[j]);
    }
}
 
// Driver Code
public static void Main()
{
    int number = 3549;
 
    printRoman(number);
}
}
 
// This code is contributed by Rajput-Ji

---

输出：

Roman numeral is: MMMDXLIX

参考资料：http://blog.functionalfun.net/2009/01/project-euler-89-converting-to-and-from.html

另一种方法1 ：
在这种方法中，我们必须首先观察该问题。问题陈述中给出的数字最多可以为4位数字。解决此问题的想法是：

1. 将给定数字在一个，两个，几百或几千个不同的地方分成数字。
2. 从千位开始打印相应的罗马值。例如，如果千位的数字为3，则打印罗马等值3000。
3. 重复第二步，直到到达一个位置。

范例：
假设输入数字为3549。因此，从千位开始，我们将开始打印罗马等值字。在这种情况下，我们将按照以下顺序打印：
第一：相当于3000的罗马字母
第二名：相当于罗马的500
第三名：相当于罗马的40
第四名：相当于罗马的9
因此，输出为： MMMDXLIX

以下是上述想法的实现。

C++

// C++ Program for above approach
#include 
using namespace std;
 
// Function to calculate roman equivalent
string intToRoman(int num)
{  
     // storing roman values of digits from 0-9
     // when placed at different places
     string m[] = {"", "M", "MM", "MMM"};
     string c[] = {"", "C", "CC", "CCC", "CD", "D",
                        "DC", "DCC", "DCCC", "CM"};
     string x[] = {"", "X", "XX", "XXX", "XL", "L",
                        "LX", "LXX", "LXXX", "XC"};
     string i[] = {"", "I", "II", "III", "IV", "V",
                        "VI", "VII", "VIII", "IX"};
         
     // Converting to roman
     string thousands = m[num/1000];
     string hundereds = c[(num%1000)/100];
     string tens =  x[(num%100)/10];
     string ones = i[num%10];
         
     string ans = thousands + hundereds + tens + ones;
         
     return ans;
}
 
// Driver program to test above function
int main()
{
      int number = 3549;
      cout << intToRoman(number);
      return 0;
}

Java

// Java Program for above approach
 
class GFG
{
    // Function to calculate roman equivalent
    static String intToRoman(int num)
    {
        // storing roman values of digits from 0-9
        // when placed at different places
        String m[] = {"", "M", "MM", "MMM"};
        String c[] = {"", "C", "CC", "CCC", "CD", "D",
                            "DC", "DCC", "DCCC", "CM"};
        String x[] = {"", "X", "XX", "XXX", "XL", "L",
                            "LX", "LXX", "LXXX", "XC"};
        String i[] = {"", "I", "II", "III", "IV", "V",
                            "VI", "VII", "VIII", "IX"};
             
        // Converting to roman
        String thousands = m[num/1000];
        String hundereds = c[(num%1000)/100];
        String tens = x[(num%100)/10];
        String ones = i[num%10];
             
        String ans = thousands + hundereds + tens + ones;
             
        return ans;
    }
     
    // Driver program to test above function
    public static void main(String []args)
    {
        int number = 3549;
        System.out.println(intToRoman(number));
         
    }
 
}

Python3

# Python3 program for above approach
  
# Function to calculate roman equivalent
def intToRoman(num):
 
     # Storing roman values of digits from 0-9
     # when placed at different places
     m = [ "", "M", "MM", "MMM" ]
     c = [ "", "C", "CC", "CCC", "CD", "D",
           "DC", "DCC", "DCCC", "CM "]
     x = [ "", "X", "XX", "XXX", "XL", "L",
           "LX", "LXX", "LXXX", "XC" ]
     i = [ "", "I", "II", "III", "IV", "V",
           "VI", "VII", "VIII", "IX"]
          
     # Converting to roman
     thousands = m[num // 1000]
     hundereds = c[(num % 1000) // 100]
     tens =  x[(num % 100) // 10]
     ones = i[num % 10]
          
     ans = (thousands + hundereds +
                 tens + ones)
          
     return ans;
 
# Driver code
if __name__=="__main__":
     
    number = 3549
     
    print(intToRoman(number))
 
# This code is contributed by rutvik_56

C＃

// C# Program for above approach
 
using System;
class GFG
{
    // Function to calculate roman equivalent
    static String intToRoman(int num)
    {
        // storing roman values of digits from 0-9
        // when placed at different places
        String []m = {"", "M", "MM", "MMM"};
        String []c = {"", "C", "CC", "CCC", "CD", "D",
                            "DC", "DCC", "DCCC", "CM"};
        String []x = {"", "X", "XX", "XXX", "XL", "L",
                            "LX", "LXX", "LXXX", "XC"};
        String []i = {"", "I", "II", "III", "IV", "V",
                            "VI", "VII", "VIII", "IX"};
             
        // Converting to roman
        String thousands = m[num/1000];
        String hundereds = c[(num%1000)/100];
        String tens = x[(num%100)/10];
        String ones = i[num%10];
             
        String ans = thousands + hundereds + tens + ones;
             
        return ans;
    }
     
    // Driver program to test above function
    public static void Main()
    {
        int number = 3549;
        Console.WriteLine(intToRoman(number));
         
    }
 
}

输出：

MMMDXLIX

感谢Shashwat Jain提供了上述解决方案。

另一种方法2 ：：
在这种方法中，我们考虑数字中的主要有效数字。例如：在1234中，主要有效数字是1。在345中，主要有效数字是3。
为了提取主要有效数字，我们需要维持一个除数(让我们称之为div)，例如1234(因为1234/1000 = 1)和100等于345(345/100 = 3)。
另外，让我们维护一个名为romanNumeral = {1：’I’，5：’V’，10：’X’，50：’L’，100：’C’，500：’D’，1000：’M ‘}

以下是算法：

如果主要有效数字<= 3

• romanNumeral [div] * mainSignificantDigit

如果主要有效数字== 4

• romanNumeral [div] + romanNumeral [div * 5]

如果5 <=主要有效数字<= 8

• romanNumeral [div * 5] +(romanNumeral [div] *(mainSignificantDigit-5))

如果主要有效数字== 3

• romanNumeral [div] + romanNumeral [div * 10]

范例：
假设输入数字为3649。

Iter 1

• 初始编号= 3649
• 主要有效数字为3。Div= 1000。
• 因此，romanNumeral [1000] * 3
• 给：MMM

Iter 2

• 现在，数字= 649
• 主有效数字为6。Div = 100。
• 因此romanNumeral [100 * 5] +(romanNumeral [div] *(6-5))
• 给出：DC

Iter 3

• 现在，数字= 49
• 主要有效数字为4。Div = 10。
• 所以romanNumeral [10] + romanNumeral [10 * 5]
• 给：XL

Iter 4

• 现在，数字= 9
• 主要有效数字为9。Div = 1。
• 因此romanNumeral [1] * romanNumeral [1 * 10]
• 给出：IX

通过合并以上所有内容而获得最终结果：MMMDCXLIX

以下是上述想法的Python实现。

Python

# Python 3 program to convert Decimal
# number to Roman numbers.
import math
 
def integerToRoman(A):
    romansDict = \
        {
            1: "I",
            5: "V",
            10: "X",
            50: "L",
            100: "C",
            500: "D",
            1000: "M",
            5000: "G",
            10000: "H"
        }
 
    div = 1
    while A >= div:
        div *= 10
 
    div /= 10
 
    res = ""
 
    while A:
 
        # main significant digit extracted
        # into lastNum
        lastNum = int(A / div)
 
        if lastNum <= 3:
            res += (romansDict[div] * lastNum)
        elif lastNum == 4:
            res += (romansDict[div] +
                          romansDict[div * 5])
        elif 5 <= lastNum <= 8:
            res += (romansDict[div * 5] +
            (romansDict[div] * (lastNum - 5)))
        elif lastNum == 9:
            res += (romansDict[div] +
                         romansDict[div * 10])
 
        A = math.floor(A % div)
        div /= 10
         
    return res
 
# Driver code
print("Roman Numeral of Integer is:"
                   + str(integerToRoman(3549)))

输出：

Roman Numeral of Integer is MMMDXLIX