Convert a fraction to a repeated decimal

Question

Almost the polar opposite if this challenge, and I suspect it will be slightly easier.

Your task is to take two integers in the format a/b (Forming a rational number) then output the number in decimal exactly.

For example, if you were to input 1/3, it would output:

0.33333333333333333

And would keep on printing 3s until the end of time, with an optional leading 0. (You could also print one character per line if and only if your language does not allow printing on the same line.)

The behaviour for x/0 will be undefined. For a number that looks like it doesn't repeat (Like, say 5/4) it actually does repeat. Either of the following two forms would be acceptable for 5/4:

1.25000000000000000
1.24999999999999999

(The same with whole numbers, 1.9999999 or 2.000000)

The fraction may not be in its simplest form, and a or b may be negative (Note -a/b = -(a/b), -a/-b = a/b, a/-b = -a/b, and -.6249999 is invalid, but -0.6249999 is acceptable, but you still can use.

Answer 1

CJam, 38 37 bytes

l'/%:i2*~*0<'-*o:Dmd\zo'.{oA*Dmd\z1}g

How it works

l     e# Read line from STDIN.            STACK '17/-13'
'/%   e# Split at '/'.                    STACK ['17' '-13']
:i    e# Cast each element to int.        STACK [17 -13]
2*~   e# Duplicate and dump the array.    STACK 17 -13 17 -13
*     e# Multiply.                        STACK 17 -13 -221
0<    e# Compare with zero.               STACK 17 -13 1
'-*o  e# Print '-' that many times.       STACK 17 -13
:D    e# Save the topmost integer in D.   STACK 17 -13
md    e# Perform modular division.        STACK -1 4
\z    e# Swap and take absolute value.    STACK 4 1
o'.   e# Print and push '.'.              STACK 4 '.'
{     e# do:
  o   e#   Print.                         STACK 4
  A*  e#   Multiply by 10.                STACK 40
  Dmd e#   Divide modulo D.               STACK -3 1
  \z  e#   Swap and take absolute value.  STACK 1 3
  o   e#   Print.                         STACK 1
1}g   e# while(1)

Answer 2

C, 108 79

Edit Modified to work with negative numbers.

Input from stdin. Old K&R style.

main(a,b){char*s="-%d.";scanf("%d/%d",&a,&b);for(a*b<0?(a<0?a=-a:(b=-b)):++s;printf(s,a/b);s="%d")a=a%b*10;}

Answer 3

Ruby, 83 69 102 91 89 bytes

->s{a,b=s.scan(/\d+/).map &:to_i
eval(s+?r)<0&&$><<?-
$><<a/b<<?.
loop{a=a%b*10
$><<a/b}}

Simple implementation of manual integer division based on the computer's integer division.

Thanks to @blutorange for the help in golfing.

Edit: Fixed the solution to include negative numbers.

Answer 4

Java, 177 176 170

s->{try{int x=new Integer(s.split("/")[0]),y=new Integer(s.split("/")[1]),z=1;for(;;x=x%y*10,Thread.sleep(999))System.out.print(x/y+(z-->0?".":""));}catch(Exception e){}}

The algorithm is straightforward; the tricky part was getting the printing to work. In the end, I had the computer sleep for a second between each step so it could print.

Expanded, runnable version

public class RepeatedDecimal {
    public static void main(String[] args) {
        java.util.function.Consumer<String> f = s -> {
                try {
                    int x = new Integer(s.split("/")[0]),
                        y = new Integer(s.split("/")[1]),
                        z = 1;
                    for (;; x = x % y * 10, Thread.sleep(999)) {
                        System.out.print(x / y + (z-- > 0 ? "." : ""));
                    }
                } catch (Exception e) { }
                };

        f.accept("5/7");
    }
}

Answer 5

R, 103 137 109 103

Bit happier with this now. Using scan with a separator save a lot of bytes. May still have some room for improvement. Replaced <- with =. Haven't always had the best of luck with this, but it worked this time.

cat(if(prod(i=scan(sep='/'))<0)'-',(n=(i=abs(i))[1])%/%(d=i[2]),'.',sep='');repeat cat((n=n%%d*10)%/%d)

Test runs

> cat(if(prod(i=scan(sep='/'))<0)'-',(n=(i=abs(i))[1])%/%(d=i[2]),'.',sep='');repeat cat((n=n%%d*10)%/%d)
1: -1/3
3: 
Read 2 items
-0.33333333333333333333...
> cat(if(prod(i=scan(sep='/'))<0)'-',(n=(i=abs(i))[1])%/%(d=i[2]),'.',sep='');repeat cat((n=n%%d*10)%/%d)
1: -5/-4
3: 
Read 2 items
1.250000000000000000000...

Answer 6

Python 3, 107 115 bytes

a,b=map(int,input().split("/"))
print(("%.1f"%(a/b))[:-1],end="")
b=abs(b)
while 1:a=abs(a)%b*10;print(a//b,end="")

Pretty straightforward:

• Input the numerator and denominator
• Output the the quotient with 1 digit after the decimal point, then take off the last digit (e.g. -1/3 -> -0.)
• Take absolute values*
• Loop:
  • Numerator is remainder after dividing out denominator
  • Multiply numerator by 10
  • Output integer quotient as next digit

* (Though the calculation for a was moved inside the loop to save a few bytes.)

Edit: Fixed bug with negative fractions > -1.

Answer 7

Python 2.7, 209 bytes

from sys import*;m,o=1,lambda x:stdout.write(str(x));a,b=[int(x)for x in argv[1].split('/')]
o(str(a*b)[0]);a,b=abs(a),abs(b);o('0'*(b>a))
while 1:
 while not((m*a)/b):o('0.'[m==1]);m*=10
 o((m*a)/b);a=(m*a)%b

edit:

Now outputs all characters on the same line, as asked.

edit2:

Now reads the fraction from commandline argument, as requested :)