Translate numbers to French

Question

The French spell out numbers in a peculiar way.

• 1-16 are "normal"
• 17-19 are spelled out 10+7. 10+8, 10+9.
• 20-69 are "normal" (OK, OK! Not really, but they are in this challenge)
• 70-79 are 60+10, 60+11 ... 60+10+7, 60+10+8, 60+10+9.
• 80-99 are 4*20, 4*20+1, 4*20+2 ... 4*20+16, 4*20+10+7, 4*20+10+8, 4*20+10+9

Challenge:

Take a positive integer in the range [1,100], and output it the "French way". You must output it exactly as it's shown below, with * and +, so 97 is 4*20+10+7, not [4 20 10 7] or something else.

Test cases:

1  - 1
.. - ..
16 - 16
17 - 10+7
18 - 10+8
19 - 10+9
20 - 20
21 - 21
.. - ..
69 - 69
70 - 60+10
71 - 60+11
72 - 60+12
77 - 60+10+7
78 - 60+10+8
79 - 60+10+9
80 - 4*20
81 - 4*20+1
82 - 4*20+2
.. - ..
90 - 4*20+10
91 - 4*20+11
96 - 4*20+16
97 - 4*20+10+7
98 - 4*20+10+8
99 - 4*20+10+9
100 - 100

Answer 1

Excel, 153 149 Bytes

=IF(OR(A1<17,A1>99,AND(A1>19,A1<70)),A1,IF(A1<20,"10+"&A1-10,IF(A1<80,"60","4*20")&IF(A1=80,,IF(MOD(A1,20)>16,"+10+"&MOD(A1,20)-10,"+"&MOD(A1,20)))))

I'm sure this could be better, I struggled to find an efficient way to account for #80.

edit: Consolidated the 'Normal' cases better to save 4 bytes. #80 still sucks.

Can't find a specific answer on here, not sure the rules of code-golf tbh. Can I use multiple cells in Excel, and add the byte count of each?

ie. For an input in cell A1

A2: 11 Bytes

=MOD(A1,20)

A3 (result): 125 Bytes

=IF(OR(A1<17,A1>99,AND(A1>19,A1<70)),A1,IF(A1<20,"10+"&A1-10,IF(A1<80,"60","4*20")&IF(A1=80,,IF(A2>16,"+10+"&A2-10,"+"&A2))))

For a total of 136?

Answer 2

Retina, 52 48 bytes

4 bytes saved thanks to Neil

^7\B
60+1
^9\B
81
^8\B
4*20+
1(?=7|8|9)
10+
\+0

Try it online! or verify all inputs (provided by Neil)

Explanation

^7\B
60+1
^9\B
81
^8\B
4*20+

First we handle the translation of 70, 80, and 90. In these first 3 stages, a 7 at the start with another character following it is replaced by 60+1. Similarly, 9 is replaced by 81, and 8 by 4*20+1. The replacement of 9 is essentially changing it to "eighty-ten" and such, so that the 8 is then handled by the next replacement, which saves bytes over writing 4*20+1 twice.

1(?=7|8|9)
10+

This handles the cases of 17, 18, and 19, by replacing the 1 in each with 10+.

\+0

Finally, there should never be a +0 at the end, so delete it if it's there.

Answer 3

JavaScript (ES6), 73 71 bytes

f=n=>n>16&n<20?'10+'+n%10:n<70|n>99?n:n%20?f(n-n%20)+'+'+f(n%20):'4*20'

Bonus version that prints the numbers as they are actually spelled for an extra 2 bytes:

f=n=>n<17|n>99?n:n<20?'10+'+n%10:n%20?f(n-n%20)+'+'+f(n%20):n-80?n:'4*20'

Answer 4

Python 2, 98 bytes

f=lambda x:[`x`*(x%100<70),'10+'+`x-10`][16<x<20]or['60+'+f(x-60),'4*20'+x/81*('+'+f(x-80))][x/80]

Try it online!

Answer 5

R, 110 bytes

i=scan()
r=i%%10;paste0(ifelse(i>16&i<20,"10+",ifelse(i>69&i<80,"60+10+",ifelse(i>16&i<20,"4*20+",i-r/10))),r)

Answer 6

PHP, 99 bytes (I wanna be happy version)

a straight port of ETHproductions´ JS, 4 bytes golfed. Prints the numbers as asked for by the OP.

function f($n){return$n<17|$n>19?$n>60&$n<100?($m=$n%20)?f($n-$m)."+".f($m):'4*20':$n:"10+".$n%10;}

breakdown

function f($n){return
    $n<17|$n>19
        ?$n>69&$n<100
            ?($m=$n%20)
                ?f($n-$m)."+".f($m) # 70..79, 81..99
                :'4*20'             # 80
            :$n                     # 1..16, 20..69
        :"10+".$n%10                # 17..19
    ;
}

I wanna be right version, 114 98 bytes

new approach inspired by ETHproductions, prints the numbers as they are actually spelled out.

function f($n){return$n>16&$n<100?$n-80?($m=$n%($n>60?20:10))?f($n-$m)."+".f($m):$n-$m:'4*20':$n;}

try it online.

breakdown

function f($n){return
    $n>16&$n<100
        ?$n-80
            ?($m=$n%($n>60?20:10))
                ?f($n-$m)."+".f($m) # $n%$m>0
                :$n-$m              # 10,20,30,40,50,60
            :'4*20'                 # 80
        :$n                         # 1..16, 100
;}

Answer 7

Python 2, 130 108 bytes

22 bytes saved thanks to @mathjunkie

f=lambda x:79<x<100and('4*20'+('+'+f(x-80))*(x>80))or 69<x<100and'60+'+f(x-60)or 16<x<20and'10+'+`x-10`or`x`

Try it online!

Answer 8

Batch, 220 217 bytes

@set/pn=
@set s=
@if %n% gtr 99 goto g
@if %n% gtr 79 set s=+4*20&set/an-=80
@if %n% gtr 69 set s=+60&set/an-=60
@if %n% gtr 16 if %n% lss 20 set s=%s%+10&set/an-=10
:g
@if %n% gtr 0 set s=%s%+%n%
@echo %s:~1%

Takes input on STDIN. Generating and removing the leading + saves 1 byte over special-casing 80. Edit: Saved 3 bytes thanks to @ConorO'Brien.

Answer 9

Pyth, 61 56 bytes

L?}b}17 19++T\+eb|bk?}/QTr6T.s+?<Q80"60+""4*20+"y%Q20\+y

Test it online!

Thanks to Leaky Nun for a 5 byte improvement!

Explanation:

                     | Implicit: Q=eval(input())
L                    | Define y(b):
 ?}b}17 19           |  If b is in the inclusive range from 17 to 19:
          ++T\+eb    |   Return "10+" + (last digit of b)
                 |b  |  Else: if b!=0: return b
                   k |   Else: return an empty string (Leaves a trailing '+' for case 80)
_____________________|________________
?}/QTr6T                              | If 70<=Q<100:
          +                           |  Concatenate the next two expressions:
           ?<Q80                      |   If Q<80:
                "60+"                 |    Evaluate to "60+"
                     "4*20+"          |    Else: Evaluate to "4*20+"
                            y%Q20     |   y(Q%20)
        .s                       \+   |  Strip off trailing '+', if present (for case 80)
                                   y  | Else: return y(Q)
                                   (Q)| Trailing Q is implicitly added

Answer 10

Jelly, 55 bytes

⁹
’,ṃ60Ṁ€
⁹%80“4*20”,
Dj⁾0+µ¹e?“×ØŒ‘
%ȷ2:“FP‘S‘ŀḟ0Ç€j”+

Try it online! or see a test suite

No doubt there is a shorter way!

How?

+ - Link 1, helper for 1-69&100: number s=0, number n
⁹ - link's right argument, n

’,ṃ60Ṁ€ - Link 2, helper for 70-79: number s=1, number n
’       - decrement s -> 0
 ,      - pair -> [0,n]
  ṃ60   - base decompress (vectorises) using [1,2,...60]  effectively -> [60,[1,n%60]]
     Ṁ€ - maximum for €ach effectively -> [60,n%60]

⁹%80“4*20”, - Link 3, helper for 80-99: number s=2, number n
⁹           - link's right argument, n
 %80        - mod 80
    “4*20”  - literal ['4','*','2','0']
          , - pair -> [['4','*','2','0'],n]

Dj⁾0+µ¹e?“×ØŒ‘ - Link 4, reformat 17-19: element v (may be char list or number)
        ?      - if
       e       - v exists in
         “×ØŒ‘ - literal [17,18,19]
               - then:
D              -   convert to decimal list  e.g. [1,7]
  ⁾0+          -   literal ['0','+']
 j             -   join                          [1,'0','+',7]
     µ         - else:
      ¹        -   identity, v

%ȷ2:“FP‘S‘ŀḟ0Ç€j”+ - Main link: number n in [1,100]
 ȷ2                - literal 100
%                  - mod (convert 100 to 0)
    “FP‘           - literal [70,80]
   :               - integer division (vectorises)
        S          - sum (0 for 1-69&100; 1 for 70-79; 2 for 80-99)
         ‘         - increment (...1, 2 or 3)
          ŀ        - call link at index (1, 2 or 3) as a dyad(sum, n)
           ḟ0      - filter out zeros (remove 0 from 4*20+0)
             ǀ    - call the last link (4) as a monad for each
                ”+ - literal '+'
               j   - join

Answer 11

Python3, 127 bytes

m,r=map,range;l=[*r(1,17),*m("10+{}".format,(7,8,9))];f=[0,*l,*r(20,61),*m("60+{}".format,l),"4*20",*m("4*20+{}".format,l),100]

Each array element contains its representation:

for i in range(1,101):
    print(i, f[i])

The code does not actually create a function, just an array -- I don't know if that's allowed. Otherwise, I'd have to make this 139 bytes by adding f=[...].__getitem__.

Answer 12

Java 7, 97 96 109 bytes

String c(int i){return i>16&i<20?"10+"+(i-10):i%100<70?i+"":i<80?"60+"+c(i-60):"4*20"+(i<81?"":"+"+c(i-80));}

+13 bytes for bug-fixing case 80.. :(

Explanation:

String c(int i){      // Method with integer parameter and String return-type
  return i>16&i<20?   //  If 17..19:
    "10+"+(i-10)      //   Return "10+" + `i`-10
   :i%100<70?         //  Else if 1..16; 20..69; 100:
    i+""              //   Return `i`
   :i<80?             //  Else if 70..79:
    "60+"+c(i-60)     //   Return "60+" + recursive-call with `i`-60
   :                  //  Else (80..99):
    "4*20"+           //   Return "4*20" +
     (i<81?           //   If 80:
      ""              //    nothing
     :                //   Else (81..99)
      "+"+c(i-80));   //    recursive-call with `i`-80
}                     // End of method

Test code:

Try it here.

class M{
  static String c(int i){return i>16&i<20?"10+"+(i-10):i%100<70?i+"":i<80?"60+"+c(i-60):"4*20"+(i<81?"":"+"+c(i-80));}

  public static void main(String[] a){
    for (int i = 1; i <= 100; i++) {
      System.out.println(c(i));
    }
  }
}