将 1 到 3999 之间的十进制数转换为罗马数字
原文:https://www . geesforgeks . org/converting-十进制数字-介于-1 到-3999 之间的数字-罗马数字/
给定一个数字，找到它对应的罗马数字。 例:
Input : 9
Output : IX
Input : 40
Output : XL
Input : 1904
Output : MCMIV
下面是罗马符号的列表，其中也包括减号:
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、十、百和千位。如果数字是 0，那么就没有对应的罗马数字符号。数字 4 和 9 的转换与其他数字略有不同，因为这些数字遵循减法符号。
将十进制数转换为罗马数字的算法 将给定的数字与 1000、900、500、400、100、90、50、40、10、9、5、4、1 的基值进行比较。刚好小于或等于给定数字的基础值将是初始基础值(最大基础值)。将该数除以其最大的基值，相应的基符号将重复商次，余数将成为未来除法和重复的数。该过程将重复进行，直到数字变为零。
演示上述算法的示例:
Convert 3549 to its Roman Numerals
输出:
MMMDXLIX
说明:
说明:
第一步
- 最初数字= 3549
- 自 3549 > = 1000；最大基础值最初为 1000。
- 除以 3549/1000。商= 3，余数=549。对应的符号 M 将重复三次。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于 0，所以我们再次调用函数。
第二步
- 现在，数字= 549
- 1000 > 549 >= 500 ;最大基础值将是 500。
- 除以 549/500。商= 1，余数=49。一旦&停止循环，相应的符号 D 将重复出现。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于 0，所以我们再次调用函数。
第三步
- 现在，数字= 49
- 50 > 49 >= 40 ;最大的基础值是 40。
- 平分 49/40。商= 1，余数= 9。一旦&停止循环，相应的符号 XL 将重复出现。
- 我们将结果值附加到第二个列表中。
- 现在余数不等于 0，所以我们再次调用函数。
第四步
- 现在，数字= 9
- 数字 9 出现在列表 ls 中，因此我们直接从字典字典中获取值，并将余数设置为 0 并停止循环。
- 余数= 0。对应的符号 IX 会重复一次，现在余数为 0，所以我们不再调用该函数。
第五步
- 最后，我们加入第二个列表值。
- 获得的输出 MMMDXLIX。
以下是上述算法的实现:
C++
// C++ Program to convert decimal number to
// roman numerals
#include <bits/stdc++.h>
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<<sym[i];
}
i--;
}
}
//Driver program
int main()
{
int number = 3549;
printRoman(number);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# Python3 program to convert
# decimal number to roman numerals
ls=[1000,900,500,400,100,90,50,40,10,9,5,4,1]
dict={1:"I",4:"IV",5:"V",9:"IX",10:"X",40:"XL",50:"L",90:"XC",100:"C",400:"CD",500:"D",900:"CM",1000:"M"}
ls2=[]
# Function to convert decimal to Roman Numerals
def func(no,res):
for i in range(0,len(ls)):
if no in ls:
res=dict[no]
rem=0
break
if ls[i]<no:
quo=no//ls[i]
rem=no%ls[i]
res=res+dict[ls[i]]*quo
break
ls2.append(res)
if rem==0:
pass
else:
func(rem,"")
# Driver code
if __name__ == "__main__":
func(3549, "")
print("".join(ls2))
# This code is contributed by
# VIKAS CHOUDHARY(vikaschoudhary344)
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
java 描述语言
<script>
// JavaScript Program to convert decimal number to
// roman numerals
// Function to convert decimal to Roman Numerals
function printRoman(number)
{
let num = [1,4,5,9,10,40,50,90,100,400,500,900,1000];
let sym = ["I","IV","V","IX","X","XL","L","XC","C","CD","D","CM","M"];
let i=12;
while(number>0)
{
let div = Math.floor(number/num[i]);
number = number%num[i];
while(div--)
{
document.write(sym[i]);
}
i--;
}
}
//Driver program
let number = 3549;
printRoman(number);
//This code is contributed by Manoj
</script>
Output
MMMDXLIX
参考文献:http://blog . function fun . net/2009/01/project-Euler-89-converting-to-and-from . html
另一种方法 1: : 在这种方法中我们首先要观察问题。问题陈述中给出的数字最多可为 4 位。解决这个问题的思路是:
- 把给定的数字在不同的地方分成数字，如一、二、百或千。
- 从千位开始打印相应的罗马值。例如，如果千位的数字是 3，那么打印罗马数字 3000。
- 重复第二步，直到我们到达某人的位置。
例 : 假设输入的数字是 3549。所以，从千之地开始，我们将开始印刷罗马版。在这种情况下，我们将按照下面给出的顺序打印: 第一个:相当于 3000 第二个:相当于 500 第三个:相当于 40 第四个:相当于 9 的罗马，因此，输出将是: MMMDXLIX
以下是上述想法的实现。
C++
// C++ Program for above approach
#include <bits/stdc++.h>
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 hundreds = c[(num % 1000) / 100];
string tens = x[(num % 100) / 10];
string ones = i[num % 10];
string ans = thousands + hundreds + tens + ones;
return ans;
}
// Driver program to test above function
int main()
{
int number = 3549;
cout << intToRoman(number);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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 hundreds = c[(num % 1000) / 100];
String tens = x[(num % 100) / 10];
String ones = i[num % 10];
String ans = thousands + hundreds + tens + ones;
return ans;
}
// Driver program to test above function
public static void main(String[] args)
{
int number = 3549;
System.out.println(intToRoman(number));
}
}
Python 3
# 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]
hundreds = c[(num % 1000) // 100]
tens = x[(num % 100) // 10]
ones = i[num % 10]
ans = (thousands + hundreds +
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 hundreds = c[(num % 1000) / 100];
String tens = x[(num % 100) / 10];
String ones = i[num % 10];
String ans = thousands + hundreds + tens + ones;
return ans;
}
// Driver program to test above function
public static void Main()
{
int number = 3549;
Console.WriteLine(intToRoman(number));
}
}
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP Program for above approach
// Function to calculate roman equivalent
function intToRoman($num)
{
// storing roman values of digits from 0-9
// when placed at different places
$m = array("", "M", "MM", "MMM");
$c = array("", "C", "CC", "CCC", "CD", "D",
"DC", "DCC", "DCCC", "CM");
$x = array("", "X", "XX", "XXX", "XL", "L",
"LX", "LXX", "LXXX", "XC");
$i = array("", "I", "II", "III", "IV", "V",
"VI", "VII", "VIII", "IX");
// Converting to roman
$thousands = $m[$num / 1000];
$hundreds = $c[($num % 1000) / 100];
$tens = $x[($num % 100) / 10];
$ones = $i[$num % 10];
$ans = $thousands . $hundreds . $tens . $ones;
return $ans;
}
// Driver Code
$number = 3549;
echo intToRoman($number);
// This code is contributed by Akanksha Rai
java 描述语言
<script>
// JavaScript Program for above approach
// Function to calculate roman equivalent
function intToRoman(num)
{
// storing roman values of digits from 0-9
// when placed at different places
let m = ["", "M", "MM", "MMM"];
let c = ["", "C", "CC", "CCC", "CD", "D",
"DC", "DCC", "DCCC", "CM"];
let x = ["", "X", "XX", "XXX", "XL", "L",
"LX", "LXX", "LXXX", "XC"];
let i = ["", "I", "II", "III", "IV", "V",
"VI", "VII", "VIII", "IX"];
// Converting to roman
let thousands = m[Math.floor(num/1000)];
let hundreds = c[Math.floor((num%1000)/100)];
let tens = x[Math.floor((num%100)/10)];
let ones = i[num%10];
let ans = thousands + hundreds + tens + ones;
return ans;
}
// Driver program to test above function
let number = 3549;
document.write(intToRoman(number));
//This code is contributed by Mayank Tyagi
</script>
Output
MMMDXLIX
感谢沙斯沃特·贾恩提供上述解决方案。
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:在 1234 中，主要有效数字是 1。同样，在 345 中，它是 3。 为了提取主有效数字，我们需要保持一个除数(让我们称之为 div)，比如 1000 代表 1234(因为 1234 / 1000 = 1)，100 代表 345 (345 / 100 = 3)。 另外，让我们维护一个名为 romannumber = { 1:“I”，5:“V”，10:“X”，50:“L”，100:“C”，500:“D”，1000:“M”}
算法如下:
如果主有效数字< = 3
- 罗马数字[div] * mainSignificantDigit
如果主有效数字== 4
- 罗马数字[div] +罗马数字[div * 5]
if 5 < =主有效数字< =8
- 罗马数字[div * 5] +(罗马数字[div] *(主数字-5))
如果主有效数字==3
- 罗马数字[div] +罗马数字[div*10]
例 : 假设输入的数字是 3649。
ITER 1
- 最初数字= 3649
- 主要有效数字是 3。Div = 1000。
- 所以，罗马数字[1000] * 3
- 给出:嗯
Iter 2
- 现在，数字= 649
- 主要有效数字是 6。Div = 100。
- 所以罗马数字[100*5] +(罗马数字[div] * ( 6-5))
- 给出:DC
Iter 3
- 现在，数字= 49
- 主要有效数字是 4。Div = 10。
- 所以罗马数字[10] +罗马数字[10 * 5]
- 给出:XL
Iter 4
- 现在，数字= 9
- 主要有效数字是 9。Div = 1。
- 所以罗马数字[1] 罗马数字[110]
- 给出:IX
最终结果通过杵所有以上:MMMDCXLIX
下面是上面想法的 Python 实现。
计算机编程语言
# 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)))
Output
Roman Numeral of Integer is:MMMDXLIX
感谢 Sriharsha Sammeta 提供上述解决方案。 本文由拉胡尔·阿格拉瓦尔供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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