add commas to Numbers without String manipulation

Question

This challenge consist in adding commas to numbers every 3 spaces like:

123456789 => 123,456,789

Specifications:

• The possible inputs will be 1,000 <= input <= 100,000,000
• You can ONLY use Mathematical Operators
• (*) You can NOT do string, array, index or character manipulation
• You can NOT use any kind of built-in functions that add these commas
• The zeros should remain, ex 10000 => 10,000
• You need to print/alert/log the output (so "the output is at x var" is not allowed)

(*) the only exception is when forming/printing the output (but not in a manner that suggest cheating)

Examples

10100 => 10,100
120406 => 120,406
999999999 => 999,999,999

---

A basic idea on how to get numbers legally:

n = 123456;
m = roundDown(n / 10);//even if ends with .9
d = n - m;//we get the last number

Answer 1

You weren't very specific on the output format...

Haskell, 56

reverse.takeWhile(>0).map(`mod`1000).iterate(`div`1000)

example:

GHCi> reverse.takeWhile(>0).map(`mod`1000).iterate(`div`1000)$93842390862398
[93,842,390,862,398]

Answer 2

GolfScript, 13 37 chars

0:|;~{.10%|):|3%''','if@10/.9>}do]-1%

Look, Ma, no arrays!

Well, almost. The pieces of the output are collected on the stack in right-to-left order, so the final ]-1% is needed to reverse them. But I assume that's OK, as it counts as "forming/printing the output".

Here's an ungolfed version with comments:

0 :| ;         # initialize the variable | to the value 0
~              # eval the input, turning it from a string into a number

{              # start of loop
  .            # duplicate the number on top of the stack; we'll need it later
  10 %         # take the remainder modulo 10 and leave it on the stack

  | ) :| 3 %   # increment | by one and take the remainder modulo 3 of the result
  '' ',' if    # if the remainder is 0, push a comma onto the stack, else an empty string

  @            # pull the value we saved at the start of the loop to top of the stack
  10 /         # divide it by ten, rounding down
  . 9 >        # check if the result is greater than 9...
} do           # ...and repeat the loop if so

]              # finally, collect the digits and commas off the stack into an array...
-1 %           # ...and reverse it for output

---

If it weren't for a few pesky details, the following 13-char solution would be pretty hard to beat:

~1000base','*

However, it fails on two counts: it doesn't correctly preserve leading zeros in digit groups, and the built-in base conversion operator base might be considered "array manipulation", since it takes a number and returns an array of digits.

---

Ps. Here's an older 49-char arrayless solution:

~{.999>}{:^1000%:&9&>99&>+{0}*','^1000/}while]-1%

Answer 3

awk, 47 chars

awk '{for(_=NF-2;1<_;_-=3)$_=","$_}NR' FS= OFS=

10,100
120,406
999,999,999

Answer 4

Python 2, 86 chars

I figured this warranted a different enough answer. The first answer is for 3.0 only because of the print statement with end=, and this answer because division automatically floors instead of converts to float, and the use of backticks

b=input()
k=0
x=''
while b>0:
 x=`b%10`+x;b/=10;k+=1
 if k%3==0and b>0:x=','+x
print x

Only string manipulation is building the output

edit: by adding 5 characters int(), this program can accept any integer, even above 1 billion. The previous program inserts Ls (python longs) if the number is too large, but is still within the constraints of the challenge. Here is the program that works with any integer:

b=input()
k=0
x=''
while b>0:
 x=`int(b%10)`+x;b/=10;k+=1
 if k%3==0and b>0:x=','+x
print x

-

879612587634598623589762354009
879,612,587,634,598,623,589,762,354,009

Answer 5

AWK, no strings, 84 chars:

{for(p=$0;.1<p/=10;++i);for(;i-->0;printf i&&i%3==0?"%d,":"%d",$0/10**i%10);print""}

Test runs:

% awk -f commas.awk <<< 1230
1,230
% awk -f commas.awk <<< 123012
123,012
% awk -f commas.awk <<< 12301 
12,301
% awk -f commas.awk <<< 12301000
12,301,000

Answer 6

C, 136

i,j,k;main(m,n){scanf("%d",&n);for(m=n;n^0;i++,n/=10);k=i%3;while(j++<i)printf("%d%c",(int)(m/pow(10,i-j))%10,j%3==k?',':0);putchar(8);}

Answer 7

Python, 82 chars

n=input()
x=''
while n>999:x=','+`n/100%10`+`n/10%10`+`n%10`+x;n/=1000
print `n`+x

I can get it shorter (67 chars) with

while n>999:x=',%03d'%(n%1000)+x;n/=1000

Not sure if that's legal.

Answer 8

befunge, 62

&# <v%3-\3%3y+8a$j*d`0:/a\%a:
1`j@>:3%\3/!3*j',,\'0+,1+a8+y

well this code is array free, since befunge itself is array free.

• the first line simply reads a number and push every digit into stack.
• the second line reads those digits and prints then once at a time. and put commas if needed.

note that if you are trying to run it using rcfunge, you need run using "-Y" option.

Answer 9

Scala 65:

def f(n:Int){if(n<1000)print(n)else{f(n/1000)
print("."+n%1000)}}

Testcode:

val l=List(1,12,123,1234,12345,123456,1234567)
l.foreach(x=>{f(x);println})

Answer 10

Python 3, 165 chars

from math import*
b=int(input())
j=10**8
k=l=0
while j>0:
 z=int(floor(b/j));b-=z*j;k+=1;j=floor(j/10)
 if not l:l=z!=0
 if l:print(z,end=','if k%3==0and j>0 else'')

No strings except for '' and ',', no arrays/lists/tuples/dicts, or indexing

edit: nixxed some characters

Answer 11

Python 2.6.5, 65 94 chars

n=input()
s=''
while n:
    s=','+str(x)+s if x else','+'000'+s
    n/=1000
print s[1:]

Just used the string to store the output.

*Now preserves zeros

Answer 12

C, 80 chars

Using recursion within printf to reverse the printing order (recursion handles the least significant digits first, but printing is done when returning, so they're printed last).

n;
p(x) {
    x=n%1000;
    n/=1000;
    printf(n?p(),",%03d":"%d",x);
}
main(){
    p(scanf("%d",&n));
}

Answer 13

JavaScript 104, 99, 90, 81

Edit: I just noticed I also have this missing zeros bug when any group of contains 000 ,00X or 0X0. If you see this message, the code below isn't fixed yet

Code:

for(n=0,s="",i=m=1e3,r=0;i<n*m;i*=m)r=~~((n%i-r)/i*m),s=r+(i-m?",":"")+s;alert(s)

Note: change the value of n for different inputs