ให้เป็นจำนวนเต็มบวกอย่างเคร่งครัดคืนตัวเลขโรมันที่สั้นที่สุดโดยใช้กฎการบวกเท่านั้น เอาต์พุตจะต้องประกอบด้วยอักขระแต่ละตัวMDCLXVIในศูนย์หรือมากกว่านั้น จำนวน14ดังนั้นจึงต้องให้มากกว่าXIIIIXIV

ค่าตัวเลขของอักขระคือM= 1,000, D= 500, C= 100,L = 50, X= 10, V= 5, I= 1

ตัวอย่าง

3 → III

4 → IIII

9 → VIIII

42 → XXXXII

796 → DCCLXXXXVI

2017 → MMXVII

16807 → MMMMMMMMMMMMMMMMDCCCVII

ภาษาอังกฤษธรรมดา , 1059 1025 678 641 451 399 ไบต์

บันทึก 34 ไบต์โดยการลบกับดักข้อผิดพลาด จากนั้นบันทึกขนาด 384 ไบต์ด้วยการตีกอล์ฟ จากนั้นบันทึก 190 ไบต์ด้วยการรวมการดำเนินการหารเข้ากับการดำเนินการผนวก ("z") เป็นการดำเนินการใหม่ ("p") จากนั้นบันทึก 52 ไบต์โดยการเล่นกอล์ฟ

A s is a string.
To p a r remainder a s a x string a n number:
If the x is "", exit.
Divide the r by the n giving a q quotient and the r.
Fill a t s with the x's first's target given the q.
Append the t to the s.
To convert a r number to a s:
p the r the s "M" 1000.
p the r the s "D" 500.
p the r the s "C" 100.
p the r the s "L" 50.
p the r the s "X" 10.
p the r the s "V" 5.
p the r the s "I" 1.

นี่คือเวอร์ชันสุดท้ายของโค้ดสุดท้ายที่ไม่ถูกรวมกับกับดักข้อผิดพลาดสำหรับจำนวนลบ:

A roman numeral is a string.

To process a remainder given a roman numeral and a letter string is a number:
  If the letter is "", exit.
  Divide the remainder by the number giving a quotient and the remainder.
  Fill a temp string with the letter's first's target given the quotient.
  Append the temp string to the roman numeral.

To convert a number to a roman numeral:
  If the number is negative, exit.
  Put the number in a remainder.
  Process the remainder given the roman numeral and "M" is 1000.
  Process the remainder given the roman numeral and "D" is  500.
  Process the remainder given the roman numeral and "C" is  100.
  Process the remainder given the roman numeral and "L" is   50.
  Process the remainder given the roman numeral and "X" is   10.
  Process the remainder given the roman numeral and "V" is    5.
  Process the remainder given the roman numeral and "I" is    1.

APL (Dyalog) , 25 22 ไบต์

'MDCLXVI'/⍨(0,6⍴2 5)∘⊤

ลองออนไลน์!

เรติน่า , 57 42 ไบต์

แปลงเป็นเอกภาพจากนั้นโลภแทนที่กลุ่มของIs ด้วยค่าที่สูงขึ้นตามลำดับ

.*
$*I
I{5}
V
VV
X
X{5}
L
LL
C
C{5}
D
DD
M

ลองออนไลน์

_{บันทึก 15 ไบต์ขอบคุณ Martin}

Python 2 , 64 ไบต์

f=lambda n,d=5,r='IVXLCD':r and f(n/d,7-d,r[1:])+n%d*r[0]or'M'*n

ลองออนไลน์!

แทนที่จะสร้างเอาท์พุทสตริงจากจุดเริ่มต้นด้วยการโลภส่วนที่ใหญ่ที่สุดอย่างโลภสร้างสิ่งนี้จากปลาย ตัวอย่างเช่นจำนวนของIคือn%5จากนั้นจำนวนของVคือn/5%2และอื่น ๆ นี่คือการแปลงฐานแบบผสมที่มีอัตราส่วนต่อเนื่องสลับกัน 5 และ 2

นี่เทียบเท่าซ้ำ:

Python 2 , 68 ไบต์

n=input();s='';d=5
for c in'IVXLCD':s=n%d*c+s;n/=d;d^=7
print'M'*n+s

ลองออนไลน์!

M's จะต้องมีการจัดการแยกเพราะจำนวนของพวกเขาใด ๆ ที่อาจจะนำเสนอเป็นไม่มีหลักมีขนาดใหญ่ ดังนั้นหลังจากกำหนดค่าสถานที่อื่นแล้วค่าที่เหลือจะถูกแปลงเป็นMของ

สำหรับการเปรียบเทียบกลยุทธ์โลภ (69 ไบต์):

Python 2 , 69 ไบต์

f=lambda n,d=1000,r='MDCLXVI':r and n/d*r[0]+f(n%d,d/(d%3*3-1),r[1:])

ลองออนไลน์!

ค่าตัวเลขปัจจุบันdถูกหารด้วย 2 หรือ 5 เพื่อสร้างตัวเลขถัดไป คุณค่าของการd%3บอกกับเราว่าอันไหน: ถ้าd%3==1หารด้วย2; และถ้าd%3==2หารด้วย 5

Mathematica, 81 ไบต์

Table@@@Thread@{r=Characters@"MDCLXVI",#~NumberDecompose~FromRomanNumeral@r}<>""&

ใช้ค่าอย่างชัดเจนและได้รับตัวเลขที่สอดคล้องกันดูเหมือนว่าจะมีความยาวหนึ่งไบต์:

Table@@@Thread@{RomanNumeral[n={1000,500,100,50,10,5,1}],#~NumberDecompose~n}<>""&

Excel, 236 193 161 ไบต์

บันทึก 43 ไบต์ขอบคุณ @ BradC

ณ จุดนี้คำตอบนั้นเป็นของ @ BradCจริงๆ บันทึกอีก 32 ไบต์

=REPT("M",A1/1E3)&REPT("D",MOD(A1,1E3)/500)&REPT("C",MOD(A1,500)/100)&REPT("L",MOD(A1,100)/50)&REPT("X",MOD(A1,50)/10)&REPT("V",MOD(A1,10)/5)&REPT("I",MOD(A1,5))

จัดรูปแบบ:

=REPT("M",A1/1E3)
    &REPT("D",MOD(A1,1E3)/500)
    &REPT("C",MOD(A1,500)/100)
    &REPT("L",MOD(A1,100)/50)
    &REPT("X",MOD(A1,50)/10)
    &REPT("V",MOD(A1,10)/5)
    &REPT("I",MOD(A1,5))

Perl 5 , 66 ไบต์

65 ไบต์ของรหัส + -pธง

$s=1e3;for$@(MDCLXVI=~/./g){$\.=$@x($_/$s);$_%=$s;$s/=--$|?2:5}}{

ลองออนไลน์!

โดยไม่เปลี่ยนจำนวนไบต์MDCLXVI=~/./gสามารถถูกแทนที่ด้วยM,D,C,L,X,V,I; และ--$|?2:5$|--*3+2โดย

อีกต่อไป ( 99 ไบต์ ) มี:

$_=M x($_/1e3).D x($_%1e3/500).C x($_%500/100).L x($_%100/50).X x($_%50/10).V x($_%10/5).I x($_%5)

CJam , 35 28 ไบต์

-7 ไบต์ขอบคุณ Martin Ender

q~{5md\2md\}3*]W%"MDCLXVI".*

ลองออนไลน์!

คำอธิบาย

q~         e# Read and eval input (push the input as an integer).
{          e# Open a block:
 5md\      e#  Divmod the top value by 5, and bring the quotient to the top.
 2md\      e#  Divmod that by 2, and bring the quotient to the top.
}3*        e# Run this block 3 times.
]W%        e# Wrap the stack in an array and reverse it. Now we've performed the mixed-base
           e# conversion.
"MDCLXVI"  e# Push this string.
.*         e# Element-wise repetition of each character by the numbers in the other array.
           e# Implicitly join and print.

C #, 127 ไบต์

f=n=>n>999?"M"+f(n-1000):n>499?"D"+f(n-500):n>99?"C"+f(n-100):n>49?"L"+f(n-50):n>9?"X"+f(n-10):n>4?"V"+f(n-5):n>0?"I"+f(n-1):""

คำสั่งที่ประกอบไปด้วยรหัสยากอย่างหมดจดโดยใช้การเรียกซ้ำ

เวอร์ชันเต็ม / ฟอร์แมต:

using System;

class P
{
    static void Main()
    {
        Func<int, string> f = null;
        f = n => n > 999 ? "M" + f(n - 1000)
                         : n > 499 ? "D" + f(n - 500)
                                   : n > 99 ? "C" + f(n - 100)
                                            : n > 49 ? "L" + f(n - 50)
                                                     : n > 9 ? "X" + f(n - 10)
                                                             : n > 4 ? "V" + f(n - 5)
                                                                     : n > 0 ? "I" + f(n - 1)
                                                                             : "";

        Console.WriteLine(f(3));
        Console.WriteLine(f(4));
        Console.WriteLine(f(42));
        Console.WriteLine(f(796));
        Console.WriteLine(f(2017));
        Console.WriteLine(f(16807));

        Console.ReadLine();
    }
}

05AB1E , 29 26 25 ไบต์

¸5n3×Rvćy‰ì}"MDCLXVI"Ss×J

ลองออนไลน์!

คำอธิบาย

¸                           # wrap input in a list
 5n                         # push 5**2
   3×                       # repeat it 3 times
     Rv                     # for each digit y in its reverse
       ć                    # extract the head of the list 
                            # (div result of the previous iteration, initially input)
        y‰                  # divmod with y
          ì                 # prepend to the list
           }                # end loop
            "MDCLXVI"S      # push a list of roman numerals
                      s×    # repeat each a number of times corresponding to the result
                            # of the modulus operations
                        J   # join to string

JavaScript (ES6), 81 75 69 ไบต์

บันทึกแล้ว 6 ไบต์ขอบคุณ @Neil สำหรับคำตอบของ @ JörgHülsermann

บันทึกแล้ว 6 ไบต์ขอบคุณ @Shaggy

n=>'MDCLXVI'.replace(/./g,(c,i)=>c.repeat(n/a,n%=a,a/=i%2?5:‌​2),a=1e3)

กรณีทดสอบ:

f=
n=>'MDCLXVI'.replace(/./g,(c,i)=>c.repeat(n/a,n%=a,a/=i%2?5:‌​2),a=1e3)

const testCases = [3, 4, 42, 796, 2017, 16807]
testCases.forEach(testCase => console.log(f(testCase)))

///, 50 bytes

/1/I//IIIII/V//VV/X//XXXXX/L//LL/C//CCCCC/D//DD/M/

Try it online!

Takes input in unary, and I'm (ab)using the footer field on TIO for input so output is preceded by a newline.

Python 3, 100 97 96 94 93 91 90 bytes

• saved 4+2 bytes: use of def; array as default parameter reduced an indentation space; unwanted variable declaration removed
• @shooqie saved 1 byte a%= shorthand
• saved 2 bytes: rearranged and braces in (a//i) got removed
• @Wondercricket saved 1 byte: move the array from default parameter to within the function which removed [] at the cost of one indentation space, thus saving 1 byte.

def f(a):
 b=1000,500,100,50,10,5,1
 for i in b:print(end=a//i*'MDCLXVI'[b.index(i)]);a%=i

Try it online!

Cubix, 69 74 80 bytes

/.UI,..N&..0\0&/52"IVXLCDM"U,r%ws;rr3tu;pw..u;qrUosv!s.u\psq,!@Us(0;U

Try it online!

        / . U I
        , . . N
        & . . 0
        \ 0 & /
5 2 " I V X L C D M " U , r % w
s ; r r 3 t u ; p w . . u ; q r
U o s v ! s . u \ p s q , ! @ U
s ( 0 ; U . . . . . . . . . . .
        . . . .
        . . . .
        . . . .
        . . . .

Watch It Running

I have managed to compress it a bit more, but there is still some pesky no-ops, especially on the top face.

• 52"IVXLCDM"U put the necessary divisors and characters on the stack. The 5 and 2 will be used to reduce the div/mod value and the characters will be discarded after use.
• UIN0/&0\&,/U u-turns onto the top face and starts a long tour to get the input and push 1000 onto the stack. An initial divide is done and a u-turn onto the r of the next snippet. This was an area I was looking at to make some savings.
• ,r%ws;rr beginning of the divmod loop. integer divide, rotate the result away mod, then rearrange top of stack to be reduced input, current divisor and divide result.
• 3tus bring the current character to the top and swap it with the divide result.
• !vsoUs(0;U this is the print loop. while the div result is more than 0, swap with character output, swap back, decrement, push a 0 and drop it. On 0 redirect over the pop stack (remove divide result) and around the cube.
• \u;pwpsq,!@Urq;u with a bit of redirection, this removes the character from the stack, brings the 5 and 2 to the top, swaps them and pushes one back down. The remaining is used to reduce the divisor. Halt if it reduces to 0, otherwise push the 5 or 2 to the bottom and re-enter the loop.

Mathematica, 130 bytes

(f=#~NumberDecompose~{1000,500,100,50,10,5,1};""<>{Flatten@Table[Table[{"M","D","C","L","X","V","I"}[[i]],f[[i]]],{i,Length@f}]})&

Python 2, 109 90 bytes

lambda n,r=[1000,500,100,50,10,5,1]:''.join(n%a/b*c for a,b,c in zip([n+1]+r,r,'MDCLXVI'))

Try it online!

PHP, 70 bytes

for($r=1e3;$a=&$argn;)$a/$r<1?$r/=++$i%2?2:5:$a-=$r*print MDCLXVI[$i];

Try it online!

T-SQL, 164 bytes

SELECT REPLICATE('M',n/1000)+IIF(n%1000>499,'D','')
      +REPLICATE('C',n%500/100)+IIF(n%100>49,'L','')
      +REPLICATE('X',n%50/10)+IIF(n%10>4,'V','')
      +REPLICATE('I',n%5)
FROM t

Line breaks added for readability only.

This version is a lot longer (230 characters), but feels much more "SQL-like":

DECLARE @ INT,@r varchar(99)=''SELECT @=n FROM t
SELECT'I's,1v INTO m
INSERT m VALUES('V',5),('X',10),('L',50),('C',100),('D',500),('M',1000)
L:
    SELECT @-=v,@r+=s 
    FROM m WHERE v=(SELECT MAX(v)FROM m WHERE v<=@)
IF @>0GOTO L
SELECT @r

Makes a table m with all the char-value mappings, and then loops through finding the largest value <= the number, concatenating the matching character.

Japt, 34 bytes

"IVXLCD"£%(U/=Y=v *3+2Y)îXÃw i'MpU

Test it online!

"IVXLCD"£    %(U/=Y=v  *3+2Y )îXÃ w i'MpU
"IVXLCD"mXY{U%(U/=Y=Yv *3+2,Y)îX} w i'MpU : Ungolfed
                                          : Implicit: U = input number
"IVXLCD"mXY{                    }         : Map each char X and its index Y in this string to:
                  Y=Yv *3+2               :   Set Y to 5 for even indexes, 2 for odd.
               U/=                        :   Divide U by this amount.
            U%(            ,Y)            :   Modulate the old value of U by 5.
                              îX          :   Repeat the character that many times.
                                          : This returns e.g. "IIVCCCD" for 16807.
                                  w       : Reverse the entire string.
                                    i'MpU : Prepend U copies of 'M' (remember U is now the input / 1000).
                                          : Implicit: output result of last expression

JavaScript (ES6), 65 bytes

A recursive function.

f=(n,a=(i=0,1e3))=>n?a>n?f(n,a/=i++&1?5:2):'MDCLXVI'[i]+f(n-a):''

How?

The second recursive call f(n-a) really should be f(n-a,a). By omitting the 2nd parameter, a and i are reinitialized (to 1000 and 0 respectively) each time a new Roman digit is appended to the final result. This causes more recursion than needed but doesn't alter the outcome of the function and saves 2 bytes.

Test cases

f=(n,a=(i=0,1e3))=>n?a>n?f(n,a/=i++&1?5:2):'MDCLXVI'[i]+f(n-a):''

console.log(f(3))
console.log(f(4))
console.log(f(9))
console.log(f(42))
console.log(f(796))
console.log(f(2017))
console.log(f(16807))

J, 26 23 bytes

3 bytes saved thanks to Adám.

'MDCLXVI'#~(_,6$2 5)&#:

Try it online!

Similar to the APL answer basically the same thing.

'MDCLXVI'#~(_,6$2 5)&#:
           (       )&#:   mixed base conversion from decimal
              6$2 5       2 5 2 5 2 5
            _,            infinity 2 5 2 5 2 5
                          this gives us e.g. `0 0 0 0 1 0 4` for input `14`
'MDCLXVI'#~               shape each according to the number of times on the right
                          this is greedy roman numeral base conversion

Batch, 164 bytes

@set/pn=
@set s=
@for %%a in (1000.M 500.D 100.C 50.L 10.X 5.V 1.I)do @call:c %%~na %%~xa
@echo %s:.=%
@exit/b
:c
@if %n% geq %1 set s=%s%%2&set/an-=%1&goto c

Takes input on STDIN.

Oracle SQL, 456 bytes

select listagg((select listagg(l)within group(order by 1)from dual start with trunc((n-nvl(n-mod(n,p),0))/v)>0 connect by level<=trunc((n-nvl(n-mod(n,p),0))/v)))within group(order by v desc)from (select 2849n from dual)cross join(select 1000v,null p,'m'l from dual union select 500,1000,'d'from dual union select 100,500,'c'from dual union select 50,100,'l'from dual union select 10,50,'x'from dual union select 5,10,'v'from dual union select 1,5,'i'from dual)

Outputs:

mmdcccxxxxviiii

Please note the actual size of the line is 460bytes, because it includes the input number (2849).

Ungolfed:

select listagg(
            (select listagg(l, '') within group(order by 1) 
             from dual 
             start with trunc((n-nvl(p*trunc(n/p),0))/v) > 0 
             connect by level <= trunc((n-nvl(p*trunc(n/p),0))/v) )
        ) within group(order by v desc)
from (select 2348 n
    from dual
) cross join (
    select 1000v, null p, 'm' l from dual union 
    select 500, 1000, 'd' from dual union
    select 100, 500, 'c' from dual union
    select 50, 100, 'l' from dual union
    select 10, 50, 'x' from dual union
    select 5, 10, 'v' from dual union
    select 1, 5, 'i' from dual     
)

How it works: I calculate how many of each letter I need, by calculating the most I can get to with the higher value one (infinity for M), and then doing an integer division between the current letter's value and the result of that.

E.g. 2348, how many Cs do I need? trunc((2348-mod(2348,500))/100) = 3.

Then, I listagg that letter together 3 times (exploiting CONNECT BY to generate the 3 rows I need). Finally, I listagg everything together.

Kinda bulky, but most of it is the select from duals in the conversion table and I can't really do much about that...

Java (OpenJDK 8), 119 118 bytes

n->{String s="";for(int v[]={1,5,10,50,100,500,1000},i=7;i-->0;)for(;n>=v[i];n-=v[i])s+="IVXLCDM".charAt(i);return s;}

Try it online!

Saved a byte thanks to @TheLethalCoder

Charcoal, 61 50 46 bytes

ＮνＡ⁰χＷφ«Ｗ¬‹νφ«§MDCLXVIχＡ⁻νφν»Ａ⁺¹χχＡ÷φ⎇﹪χ²¦²¦⁵φ

Try it online!

Explanation:

Ｎν                   Take input as number and assign it to ν
Ａ⁰χ                  Let χ=0
Ｗφ«                  While φ>0 (φ has a predefined value of 1000)
    Ｗ¬‹νφ«               While v>=φ 
        §MDCLXVIχ             Take the char from string "MDCLXVI" at position χ
        Ａ⁻νφν»               Let ν=ν-φ
    Ａ⁺¹χχ                Increment χ
    Ａ÷φ⎇﹪χ²¦²¦⁵φ        If χ is odd, divide φ by 5, else divide φ by 2

• 4 bytes saved thanks to Neil, while I am still trying to figure out how to proceed with the second part of his comment.

C++, 272 Bytes

#include <cstdio>
#include <map>
std::map<int,char> m = {{1000,'M'},{500,'D'},{100,'C'},{50,'L'},{10,'X'},{5,'V'},{1,'I'}};
int main(){unsigned long x;scanf("%d",&x);for(auto i=m.rbegin();i!=m.rend();++i)while(x>=i->first){printf("%c", i->second);x=x-i->first;}return 0;}

C, 183 Bytes

#include <stdio.h>
int v[]={1000,500,100,50,10,5,1};
char*c="MDCLXVI";
int main(){int x;scanf("%d",&x);for(int i=0;i<sizeof v/sizeof(int);i++)for(;x>=v[i];x-=v[i])putc(c[i],stdout);}

Same algorithm as before, just using plain c arrays instead of an std::map, partially inspired by @xnor's answer and using a string to store the letters.

Common Lisp, 113 bytes

This is an anonymous function, returning the result as a string.

(lambda(value)(setf(values a b)(floor v 1000))(concatenate 'string(format()"~v,,,v<~>"a #\M)(format nil"~@:r"b)))

Ungolfed, with descriptive variable names and comments:

(defun format-roman (value)
  ;; Get "value integer-divided by 1000" and "value mod 1000"
  (setf (values n_thousands remainder) (floor value 1000))
  (concatenate 'string
               ;; Pad the empty string n_thousands times, using "M" as the 
               ;; padding character
               (format () "~v,,,v<~>" n_thousands #\M)
               ;; Format the remainder using "old-style" Roman numerals, i.e. 
               ;; numerals with "IIII" instead of "IV"
               (format nil "~@:r" remainder)))

CL has built-in Roman numeral formatter. Sadly it doesn't work for numbers larger than 3999.

Charcoal, 34 bytes

ＮςＡ²ξＦMDCLXVI«×ι÷ςφＡ﹪ςφςＡ÷φξφＡ÷χξξ

Originally based on @CarlosAlego's answer. A port of @xnor's Python solution is also 34 bytes:

ＮθＡ⁵ξＦIVXLCD«←×ι﹪θξＡ÷θξθＡ÷χξξ»←×Mθ

Edit: A port of @xnor's other Python solution turns out to be 33 bytes!

ＮθＦMDCLXVI«×ι÷θφＡ﹪θφθＡ÷φ⁺׳﹪φ³±¹φ

Try it online! Link is to verbose version of code. Note that I've used ⁺׳﹪φ³±¹ instead of ⁻׳﹪φ³¦¹ because the deverbosifier is currently failing to insert the separator.