Quine multiple times

Question

Your task is make program that do following:

1. You should take number. (Positive, negative, fraction is possible input)
2. If it is negative, you reverse the quine. and negate that number (Become positive)
3. Then you repeat <integer part of the input number> times and print first <floor(fraction part of the input number*length)> from your source program. If it is integer, then the fraction part is zero.

-10% bonus if your program isn't palindrome.

Example

If your program is "ABCDEFG", then

1.

5
ABCDEFGABCDEFGABCDEFGABCDEFGABCDEFG

Explanation

ABCDEFG five times

2.

-2
GFEDCBAGFEDCBA

Explanation

GFEDCBA (reversed ABCDEFG) 2 times

3.

7.5
ABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABCDEFGABC

Explanation

ABCDEFG 7 times followed by ABC (first 3(floor(0.5*7)=floor(3.5)=3) letter on ABCDEFG)

4.

-0.3
GF

Explanation

GFEDCBA (reversed ABCDEFG) 0 times followed by GF (first 2(floor(0.3*7)=floor(2.1)=2) letter on GFEDCBA(reversed ABCDEFG))

5.

0
<empty>

Explanation:

<empty> here means that your program doesn't output. It is ABCDEFG zero times that is defined as empty string.

Answer 1

CJam, 28 26 bytes * 0.9 = 23.4

Thanks to Sp3000 for saving 2 bytes.

{`"_~"+rd_z26*,\g1|@%f=}_~

Test it here.

Explanation

{`"_~"+ e# Generalised quine framework. Leaves the source code on the stack.
  rd    e# Read input and parse as double.
  _z    e# Copy and get absolute value.
  26*   e# Multiply by 26, the length of the quine to get length N of the output.
  ,     e# Get a range [0 1 ... floor(N-1)].
  \g    e# Swap with other copy and computer signum to determine direction of string.
  1|    e# This might be zero though, so take bitwise OR with 1 to avoid an error.
  @%    e# Pull up source string. Reverse it if the signum was -1 (no-op otherwise).
  f=    e# The range we pushed earlier corresponds to the (cyclic) indices of the source
        e# which make up the desired result, so we map each index to the corresponding
        e# character to get the final result.
}_~

Answer 2

Vitsy, 34 * 0.9 = 30.6 Bytes

Thanks to @Sp3000 for pointing out a flaw in my code!

Woo. My Physics teacher reminded me I had power functions to help me with this. Go figure.

'r(;Vd3*V2^12/^DvV/1+(rvl1-*\[DO{]
'                                   Start recording as a string - this grabs everything and pushes it to the stack as a string.
 r                                  Reverse the stack order.
  (;                                If the top item is zero, exit the program.
    V                               Grab the input as a final global variable.
     d3*                            Push the character ' to the stack.
        V2^12/^                     Get the absolute value of the input value.
               Dv                   Duplicate and save in a temp variable.
                 V                  Push the global variable to the stack.
                  /                 Divide the top two items - this gets -1 or 1 depending on the polarity of the input.
                   1+(              If it's -1, do the next instruction. Otherwise, don't.
                      r             Reverse the stack
                       v            Push the temporary variable to the stack.
                        l1-*        Multiply by the length of the stack minus 1.
                            \[   ]  Repeat everything in brackets top item of the stack times.
                              DO{   Duplicate an item, pop it off the stack and output it, then move one item over in the stack.

Answer 3

Perl, 104 bytes - 10% = 93.6

perl -i-0.3 -e '$_=q{$_="\$_=q{$_};eval";$_=reverse if$^I<0;$n=abs$^I;print+($_ x$n).substr$_,0,y///c*($n-int$n)};eval'

102 bytes + 2 bytes for -i - 10% for not being a palindrome. Input is passed as the argument to -i (e.g. -0.3 above).

How it works

This solution is based on the following quine:

$_=q{print"\$_=q{$_};eval"};eval

This works as follows. First, set $_ to the string:

print"\$_=q{$_};eval"

Next, call eval, which works on $_ by default. This calls print with one argument, a string literal:

"\$_=q{$_};eval"

Since this string is double-quoted, variables are interpolated. After interpolating $_, the value of the string is:

\$_=q{print"\$_=q{$_};eval"};eval

When printed, this outputs:

$_=q{print"\$_=q{$_};eval"};eval

which is the source code of the program itself.

The nice thing about this quine is that you can embed arbitrary code inside the string to be eval'd.

---

Here's a breakdown of the full solution:

perl -i-0.3 -e'
    $_=q{                     # string to be eval'd
        $_="\$_=q{$_};eval";  # append beginning and end of quine so the
                              #  entire thing can be reversed if necessary
        $_=reverse if $^I<0;  # reverse if input < 0
        $n=abs $^I;           # set $n to absolute value of input
        print                 # print
            +($_ x $n)        # $_ repeated $n times
            .                 # concatenated with
            substr $_,        # substring of $_
                   0,         # starting at the beginning
                   y///c      # having length x, where x is length of $_
                   *          # multiplied by
                   ($n-int$n) # fractional part of $n
    };
    eval                      # eval $_
'

Answer 4

Python 2, 193 bytes - 10% = 173.7

x=input();y=int(x);_='x=input();y=int(x);_=%r;_=(_%%_)[::y/abs(y)];x,y=abs(x),abs(y);_=_*y+_[:int(y*(x-y)*193)];print _';_=(_%_)[::y/abs(y)];x,y=abs(x),abs(y);_=_*y+_[:int(y*(x-y)*193)];print _

Errors out on 0, but, ignoring STDERR, you still get empty output.

Answer 5

Mathematica, 139 - 10% = 125.1 bytes

StringJoin[Table[s = If[#1 > 0, #1 & , StringReverse][ToString[#0, InputForm]], {Abs[Floor[#1]]}], StringTake[s, Floor[Mod[#1, 1]*139]]] & 

Note the trailing space. The whitespace, standard notation, etc. are the result of the ToString[#0, InputForm].

Answer 6

Haskell, 158 * 0.9 = 142.2 bytes

c i|i<0=reverse|1<2=id;f i=putStr$take(floor$abs i*158)$cycle$c i$s++show s;s="c i|i<0=reverse|1<2=id;f i=putStr$take(floor$abs i*158)$cycle$c i$s++show s;s="

A quine function.

*Main> f (-0.3)
"=s;s wohs++s$i c$elcyc$)851*i sba$roolf(ekat$r

*Main> f 1.1
c i|i<0=reverse|1<2=id;f i=putStr$take(floor$abs i*158)$cycle$c i$s++show s;s="c i|i<0=reverse|1<2=id;f i=putStr$take(floor$abs i*158)$cycle$c i$s++show s;s="c i|i<0=reverse

*Main> f 0
              <-- empty