Adding without using a + or - sign

Question

There have been many "Do __ without __" challenges before, but I hope that this is one of the most challenging.

The Challenge

You are to write a program that takes two natural numbers (whole numbers > 0) from STDIN, and prints the sum of the two numbers to STDOUT. The challenge is that you must use as few + and - signs as possible. You are not allowed to use any sum-like or negation functions.

Examples

input

123
468

output

591

input

702
720

output

1422

Tie Breaker: If two programs have the same number of + and - characters, the winner is the person with fewer / * ( ) = . , and 0-9 characters.

Not Allowed: Languages in which the standard addition/subtraction and increment/decrement operators are symbols other than + or - are not allowed. This means that Whitespace the language is not allowed.

Answer 1

Perl (no +/-, no tie-breakers, 29 chars)

s!!xx!;s!x!$"x<>!eg;say y!!!c

As a bonus, you can make the code sum more than two numbers by adding more xs to the s!!xx!.

Alternatively, here are two 21-char solutions with 1 and 3 tie-breakers respectively

say length$"x<>.$"x<>

say log exp(<>)*exp<>

Note: These solutions use the say function, available since Perl 5.10.0 with the -E command line switch or with use 5.010. See the edit history of this answer for versions that work on older perls.

---

How does the solution with no tie-breakers work?

• s!!xx! is a regexp replacement operator, operating by default on the $_ variable, which replaces the empty string with the string xx. (Usually / is used as the regexp delimiter in Perl, but really almost any character can be used. I chose ! since it's not a tie-breaker.) This is just a fancy way of prepending "xx" to $_ — or, since $_ starts out empty (undefined, actually), it's really a way to write $_ = "xx" without using the equals sign (and with one character less, too).

• s!x!$"x<>!eg is another regexp replacement, this time replacing each x in $_ with the value of the expression $" x <>. (The g switch specifies global replacement, e specifies that the replacement is to be evaluated as Perl code instead of being used as a literal string.) $" is a special variable whose default value happens to be a single space; using it instead of " " saves one char. (Any other variable known to have a one-character value, such as $& or $/, would work equally well here, except that using $/ would cost me a tie-breaker.)

  The <> line input operator, in scalar context, reads one line from standard input and returns it. The x before it is the Perl string repetition operator, and is really the core of this solution: it returns its left operand (a single space character) repeated the number of times given by its right operand (the line we just read as input).

• y!!!c is just an obscure way to (ab)use the transliteration operator to count the characters in a string ($_ by default, again). I could've just written say length, but the obfuscated version is one character shorter. :)

Answer 2

R (24 characters)

length(sequence(scan()))

What this does:

• scan reads input from STDIN (or a file)
• sequence generates integer sequences starting from 1 and concatenates the sequences. For example, sequence(c(2, 3)) results in the vector 1 2 1 2 3
• length calculates the number of elements in the concatenated vector

Example 1:

> length(sequence(scan()))
1: 123
2: 468
3:
Read 2 items
[1] 591

Example 2:

> length(sequence(scan()))
1: 702
2: 720
3:
Read 2 items
[1] 1422

Answer 3

D

main(){
    int a,b;
    readf("%d %d",&a,&b);
    while(b){a^=b;b=((a^b)&b)<<1;}
    write(a);
}

bit twiddling for the win

as a bonus the compiled code doesn't contain a add operation (can't speak for the readf call though)

Answer 4

Python 2, 43 bytes

print len('%s%s'%(input()*'?',input()*'!'))

Answer 5

GolfScript

No +/- or tie-breakers:

# Twiddle some bits to get a few small integers
[]!~abs:two~abs:three!:zero~:minusone;

~[two base minusone%\two base minusone%]zip
zero:c;
{
    # Stack holds [x y] or [x] with implicit y is zero
    # Half adder using x y c: want to end up with sum on stack and carry back in c
    [~c zero]three<zero$
    ~^^\
    $~;:c;;
}%
[~c]minusone%two base

Much simpler version with two tie-breaker characters, using the same list-concatenation trick that other people are using:

~[\]{,~}%,

I'm assuming that GolfScript isn't disqualified for having ) as an increment operator, as I'm not actually using it.

Answer 6

C (32bit only)

int main(int ac, char *av) {
    scanf("%d\n%d", &ac, &av);
    return printf("%d\n", &av[ac]);
}

Pointer arithmetic is just as good.
How does it match the requirements?
* No + or -
* No /, =, ., 0-9
* Only 3 pairs of parenthesis, which seems to me minimal (you need main, scanf, printf).
* One * (the pointer approach requires it).
* Four , (could save one by defining normal variables, not ac,av)

Answer 7

C++ 0 +/-, 3 tie-breakers

#include <vector>
#include <iostream>

#define WAX (
#define WANE )
#define SPOT .

int main WAX WANE {
    unsigned x;
    unsigned y;
    std::cin >> x >> y;
    std::vector<int> v WAX x WANE;
    std::vector<int> u WAX y WANE;
    for WAX auto n : u WANE {
        v SPOT push_back WAX n WANE;
    }
    std::cout << v SPOT size WAX WANE;
}

Answer 8

Haskell, 0+2

import Monad
main = join $ fmap print $ fmap length $ fmap f $ fmap lines getContents
f x = join $ flip replicate [] `fmap` fmap read x

This uses no + or - characters, and only two = from the set of tie breaker characters, one of which is mandatory for binding main. The sum is done by concatenating lists of the appropriate lengths.

Answer 9

EDIT This was posted BEFORE the rules changed to disallow sum...

The R language: No calls to + or -... And 9 tie-breaker characters!

sum(as.numeric(readLines(n=2)))

Example:

> sum(as.numeric(readLines(n=2)))
123
456
[1] 579

The [1] 579 is the answer 579 (the [1] is to keep track of where in the result vector your are since in R all values are vectors - in this case of length 1)

Note that R has + operators just like most languages - it just so happens that it has sum too that sums up a bunch of vectors.

In this case, readLines returns a string vector of length 2. I then coerce it to numeric (doubles) and sum it up...

Just to show some other features of R:

> 11:20 # Generate a sequence
 [1] 11 12 13 14 15 16 17 18 19 20

> sum(1:10, 101:110, pi)
[1] 1113.142

Answer 10

The R language

New rules, new answer, same language. No calls to + or -

UPDATE Using scan, it drops to 11 tie-breaker characters (and 27 characters in all).

as.numeric(scan())%*%c(1,1)

Original: 13 tie-breaker characters!

as.numeric(readLines(n=2)) %*% c(1,1)

Example:

> as.numeric(readLines(n=2)) %*% c(1,1)
123
456
     [,1]
[1,]  579

This time the result is achieved by matrix multiplication. The answer is displayed as a 1x1 matrix.

Answer 11

Haskell, 0 +/-, 6 2 tie-breakers (=)

(does not use the string/list concatenation trick)

main = interact f
f x = show $ log $ product $ map exp $ map read $ lines x

Answer 12

Seed, 3904 3846 11 bytes, 0 +/-, 10 tie breakers

4 141745954

Answer 13

Javascript, 56

p=prompt;alert(Array(~~p()).concat(Array(~~p())).length)

Thanks to @JiminP on the ~~ tip! I'm going for least bytes, so the 1 byte saving on the p=prompt; is still worth it. I understand your argument about tie-breaker chars, but to be honest wouldn't you rather the least bytes :-p

Version, 69

i=parseInt;p=prompt;alert(Array(i(p())).concat(Array(i(p()))).length)

Thanks to some feedback from @Ilmari and @JiminP, I've shaved 13 bytes off my original solution.

Originally, 82

i=parseInt;p=prompt;a=Array(i(p()));a.push.apply(a, Array(i(p())));alert(a.length)

Answer 14

C

b[500];i;j;main(){scanf("%d%d",&i,&j);printf("%d\n",sprintf(b,"%*s%*s",i,"",j,""));

Answer 15

APL (no +/-, no tie breakers, 8 or 10 characters)

This entry is similar to the other ones that concatenate sequences generated from the input and find the length... but it's in APL, which can appear confusing even for a small problem like this. I used Dyalog APL, which offers a free educational license.

Code:

⍴⊃⍪⌿⍳¨⎕⎕

From right to left:

• Each quote-quad ( ⎕ ) requests input from the user and evaluates it.
• The each operator ( ¨ ) applies the index generator function ( ⍳ ) to each of the items in the array to its right.
• This flattens the resulting array of arrays into one array. The input array is reduced to a flat list by the reduction operator ( / ), which folds the array using the concatenation function ( , ). For the sake of this challenge, the one-dimensional reduction operator ( ⌿ ) is used, along with the concatenation operator along the first axis ( ⍪ ).
• As a result of using the reduction operator, the array is enclosed, which is like placing it in the bag; all we see on the outside is a bag, not its contents. The disclose operator ( ⊃ ) gives us the contents of the enclosed array (the bag).
• Finally, the shape-of function ( ⍴ ) gives us the lengths of the dimensions of an array. In this case, we have a one-dimensional array, so we obtain the number of items in the array, which is our result.

If we need to explicitly output the result, we can do so like this:

⎕←⍴⊃⍪⌿⍳¨⎕⎕

Comparable Python code, with corresponding APL symbols above:

import operator

⎕←     ⍴    ⌿        ⍪                 ¨              ⍳                ⎕        ⎕         ⊃
print (len (reduce (operator.__add__, [map (lambda n: range (1, n+1), [input(), input()])][0])))

I'd like to know if there's a shorter version possible in APL - another, simpler version I came up with that has more tie breakers (although still at 8 characters) is: ⍴(⍳⎕),⍳⎕.

Answer 16

I didn't see anyone do it the Electrical Engineering way, so here's my take (in ruby):

def please_sum(a, b)
    return (a&b !=0)? please_sum( ((a&b)<<1) , a^b ):a^b
end

It is a little bit ugly, but it gets the job done. The two values are compared by a bitwise AND. If they don't have any bits in common, there is no "carry" into the next binary column, so the addition can be completed by bitwise XORing them. If there is a carry, you have to add the carry to the bitwise XOR. Here's a little ruby script I used to make sure my digital logic was not too rusty:

100.times do
    a=rand 10
    b=rand 10
    c=please_sum(a,b)
    puts "#{a}+#{b}=#{c}"
    end

Cheers!

Answer 17

Shell, 52

read a
read b
(seq 1 $a;seq 1 $b)|wc|awk '{print$1}'

This is basically the same answer I gave for another problem.

Answer 18

Javascript (17 tie-breaker characters)

eval('걢갽거걲걯걭거건갨걡갽거걲걯걭거건갨걹갽걦걵걮걣건걩걯걮갨걡갩걻걣갽걮걥걷갠걕걩걮건갸걁걲걲걡걹갨걡갩갻걦걯걲갨걩갠걩걮갠걣갩걩걦갨걩갽갽걾걾걩갩걸갮거걵걳걨갨갱갩걽갬걸갽걛걝갩갩갻걹갨걡갩갻걹갨걢갩갻걡걬걥걲건갨걸갮걬걥걮걧건걨갩갻'['split']('')['map'](function(_){return String['fromCharCode'](_['charCodeAt'](~~[])^0xac00)})['join'](''))

:P ("Obfuscated" to reduce number of tie-breaker characters. Internally, it's b=prompt(a=prompt(y=function(a){c=new Uint8Array(a);for(i in c)if(i==~~i)x.push(1)},x=[]));y(a);y(b);alert(x.length); .)

Answer 19

C

a,b;A(int a,int b){return a&b?A(a^b,(a&b)<<1):a^b;}
main(){scanf("%d%d",&a,&b);printf("%d\n",A(a,b));}

Answer 20

C#

It's not the shortest by any stretch:

private static int getIntFromBitArray(BitArray bitArray)
{
    int[] array = new int[1];
    bitArray.CopyTo(array, 0);
    return array[0];
}

private static BitArray getBitArrayFromInt32(Int32 a)
{
    byte[] bytes = BitConverter.GetBytes(a);
    return new BitArray(bytes);
}

static void Main(string[] args)
{
    BitArray first = getBitArrayFromInt32(int.Parse(Console.ReadLine()));
    BitArray second = getBitArrayFromInt32(int.Parse(Console.ReadLine()));
    BitArray result = new BitArray(32);

    bool carry = false;
    for (int i = 0; i < result.Length; i++)
    {
        if (first[i] && second[i] && carry)
        {
            result[i] = true;
        }
        else if (first[i] && second[i])
        {
            result[i] = false;
            carry = true;
        }
        else if (carry && (first[i] || second[i]))
        {
            result[i] = false;
            carry = true;
        }
        else
        {
            result[i] = carry || first[i] || second[i];
            carry = false;
        }
    }
    Console.WriteLine(getIntFromBitArray(result));
}

Answer 21

J, 15 7 chars, 1 tie breaker, incomplete program

This is my J attempt. It is not a full program, because I have not yet figured out how to write one. Just put that line in a script to get the function p that can be used for adding an arbitrary amount of numbers. It is a monad and takes a list of numbers to add (such as p 1 2 3 4):

p=:#@#~

The idea is very simple. The function is written in tacit aka pointless style. Here is a pointed definition:

p=:3 :'##~y'

Read from right to left. In the tacit version, @ composes the parts of the function. (like a ∘ in mathematics [(f∘g)(x) = f(g(x)])

• y is the parameter of p.
• ~ makes a verb reflexive. For some verb m, m~ a is equal to a m a.
• # (copy, a#b): Each element in a is replicated i times, where i is the element at the same index as the current element of a of b. Thus, #~ replicates an item n n times.
• # (count, #b): Counts the number of elements in b.

Conclusion: J is awsome and less readable than Perl (that makes it even more awsome)

Edits

• 15 -> 7 using # instead of i.. Yeah! Less chars than golfscript.

More of a program

This one queries for input, but it still isn't a full program: (13 chars, 3 breakers)

##~".1!:1<#a:

Answer 22

C#,

Program works on 1 line; separated on multiple lines to avoid horizontal scrolling.

using C=System.Console;
class A{
static void Main(){
int a,b,x,y;
a=int.Parse(C.ReadLine());
b=int.Parse(C.ReadLine());
do{x=a&b;y=a^b;a=x<<1;b=y;}while(x>0);
C.WriteLine(y);
}}

Answer 23

Keg (SBCS on Keg wiki)

Basically a port of the R answer.

¿¿Ï_"Ï_!.

Explanation

¿¿#        Take 2 integer inputs
  Ï_"Ï_#   Generate 2 arrays the length of the integer inputs
       !.# Output the length of the stack

Answer 24

PowerShell, 27 42 bytes, 0 +-, 4 1 secondary

^{Thanks to mazzy for saving a + and 4 secondaries}

$args|%{[int[]]$_*$_}|measure|select count

Try it online! or Pretty Table for an extra 3 bytes

-Or- adding four secondaries to save 19 bytes:

32 23 bytes, 1 0 +-, 12 5 secondaries

^{-9 bytes thanks to mazzy}

($args|%{,$_*$_}).count

Try it online!

For each argument, we push n array elements (consisting of [n] but that's not important) to the pipeline which are grouped by the parens and then counted.

Answer 25

05AB1E, 2 4 bytes, 0 +/-

F>

Try it online!

Apologies if I misunderstood this challenge, but I was surprised there was no 05AB1E answer. Shortest answer in this language I could come up with that doesn't use + or the built in sum function.

Explanation:

 F   #Loop A many times
  >  #Increment B
     #(Implicit output)

-2 Bytes thanks to Grimy.

Answer 26

Python 3

Without relying on hidden summations in other functions.

h=input;x,y=int(h()),int(h())
while y:c=x&y;x^=y;y=c<<1
print(x)

Try it online!

Answer 27

Clojure (44 chars)

(pr(#(count(concat(%)(%)))#(repeat(read)0)))

Edit: fixed to print on STDOUT instead of just returning sum.

Answer 28

Scala

• score:
  • +- : 0
  • (). : 5+5+3=13

Code:

(List.fill (readInt) (1) ::: List.fill (readInt) (2)).size

• List.fill (4)(7) produces List (7, 7, 7, 7)
• a ::: b concatenates 2 Lists into one
• The rest should be obvious

Answer 29

K, 11

{#,[!x;!y]}

Same concatenation trick as the R solution. Reading right to left: Enumerate the two input variables, concatenate and then count.

Answer 30

R, 44 bytes

nchar(paste(strrep(" ",scan()),collapse=""))

Try it online!

Converts the two inputs to unary and "sums" via concatenation. Check out the other R answers:

• Using sequence
• Using dot product

Answer 31

C, score 0 with 8 tie-breakers

f(a,b) { return b ? f(a^b, a<<1 & b<<1) : a; }
 ^ ^ ^               ^   ^    ^      ^^

I've highlighted the tie-break chars.

Negative numbers are assumed to be 2s-complement.

Answer 32

Javascript (ES6) - 8 Tie-breaker characters

add: {
    let a = prompt``|false;
    let b = prompt``|false;
    while (a & b) {
        a = a ^ b;
        b = [b & ~a][false|false] << true
    }
    alert(a | b);
}

• 2 * (
• 2 * )
• 4 * =

Answer 33

Gol><>, 12 bytes

IITM:{P}?trh

Thanks to JoKing for reminding me there was a Teleport Pad!!

1st version, 25 bytes

II!/M:{P}?\rh
   \     :/

Try it online!

This was alot easier than I thought it would be!!! I will be golfing this alot more!!!

Answer 34

Runic Enchantments, (no +/-, 1 tie-breaker, 11 bytes)

'V2,k!$wi|;

Try it online!

Not allowed to have + characters? Fine, I'll create them myself with math and reflection. V has the byte value 86, which gets divided by 2 (the , and only tie-breaker character) converted to a character, and finally, reflectively written into the program under the instruction pointer. It then reads input, hits a reflector (|), reads input again, then reaches the space originally containing a w and adds them together before printing. The IP then performs an illegal action at , and is silently terminated.

Is this cheating? Probably. But the challenge says "don't use + characters" and the source code contains 0 of them.

Alternate method, (no +/-, 3 tie-breakers, 16 bytes)

"ab"1,i*}i*qul$;

Try it online!

Makes two strings, one of length x and one of length y, concatenates them together, then computes the length of the string. Does this count as a "sum-like function"? Perhaps. + on two strings does do the same thing.

Answer 35

APL (Dyalog Unicode), 4 bytes^SBCS, 0 +/-, 0 tie breakers

Can sum any number of non-negative integers.

≢⎕⌿#

Try it online!

# reference to the root object

⎕/ replicate by the number(s) from STDIN
^{for each number, we get that many copies of the corresponding element on the right, but since there is only one, all numbers get paired up with that one, forming a single list of N₁+N₂+…+N_n references}

≢ tally

Output is implicitly to STDOUT.

Answer 36

Befunge-98 (FBBI), 4 bytes, one +.

&&+.

Try it online!

Answer 37

MathGolf, 3 bytes, 0 +/-

α▓∞

Try it online!

While perhaps violating the spirit of the "no sum" rule, this solution wraps the inputs into an array, calculates the average of the values in that array, and duplicates the result. Of course, the avg operator uses a sum internally, but its behavior is inherently different compared to the sum operator Σ.

The output is printed as a float, with .0 at the end. If this is not acceptable, just add i to the end to convert it back into an integer.

Explanation

α     wrap last two elements in array
 ▓    get average of list
  ∞   pop a, push 2*a

Answer 38

Befunge-98 (PyFunge), 0 +/- characters, 0 tiebreaker characters, 47 bytes

brqzzzzzzzzzzzzzzzzzzzpyadgyacpyappyacybyabybay

Try it online!

Made all out letters, because why not?

More readable variant that does pretty much the same thing:

b<@                               &&pyapyabybay

Abuses the y GetSysInfo command, which can be used to retrieve the instruction pointer's current X coordinate. This is used to push the ASCII codes for + (add) and . (output), which later get put onto the field.

Answer 39

Ruby -na0, no +/-, no tiebreakers, 60 bytes

$F<<?a until~%r[
]&&$F::join[%r[a{#$`}a{#$'}]]
p$F::count ?a

Try it online!

Actually harder than I thought, I'm almost certainly missing a trick.

Reads in the input, then adds a characters to the argument array until the array, when joined into a string, matches a regular expression that uses the two input numbers as quantifiers, then counts how many it added.

Syntax tricks include using %r[] instead of // for a regexp, and :: instead . for a method call.

Answer 40

Zsh, 26 bytes, 7 tiebreakers

Positional parameters are pretty punishing

x=({1..$1} {1..$2})
<<<$#x

Try it online!

---

62 bytes, 0 tiebreakers

read a b
{repeat $a;printf x;repeat $b;printf x}|read x
<<<$#x

Try it online!

Answer 41

C++, () only for main

#include <iostream>

int main()
{
    unsigned int a, b;
    std::cin >> a;
    std::cin >> b;


    unsigned int p = a ^ b;
    unsigned int g = a & b;

    unsigned int gg {g | p & g << 1};
    unsigned int pp {p & p << 1};

    unsigned int ggg {gg | pp & gg << 2};
    unsigned int ppp {pp | pp << 2};

    unsigned int gggg {ggg | ppp & ggg << 4};
    unsigned int pppp {ppp & ppp << 4};

    unsigned int ggggg {gggg | pppp & gggg << 8};
    unsigned int ppppp {pppp | pppp << 8};

    unsigned int gggggg {ggggg | ppppp & ggggg << 16};

    unsigned int result {a ^ b ^ gggggg << 1};
    std::cout << result;
}

Assumes 32-bit unsigned ints. Very ugly code to remove tiebreak characters. Emulates a Kogge-Stone adder.

Answer 42

Octave/MATLAB, 37 bytes

f=@()e^(input(''));disp(log(f()*f()))

Try it online!

Answer 43

Python 3

import numpy as n;i=lambda:n.e**int(input());print(int(n.log(i()*i())))

Try it online!

Answer 44

Ruby, 0 +/-, 0 tie-breakers, 39 bytes

And fairly unreadable.

p"#{"%#{"%ss%%%ss"%$F}"%%w[o o]}"::size

Try it online!

Answer 45

Gol><>, 8 bytes

IRmIRmlh

Minus one byte!, I figured out that I didn't need to take both inputs first

Try it online!

Previous answer, 9 bytes

IIRfrRflh

Explanation below

II        take both inputs
  Rm      Pop and repeat that number of times, pushing -1
    r     Reverse stack (to get the first num on top)
     Rm   Pop and repeat that number of times, pushing -1
       lh Push length of stack, and output as a number halting

Basically what I did was count it all out in unary, it would only take a few more bytes to expand it to be able to handle more than two numbers

Try it online!

Answer 46

JavaScript (ES6), 48 bytes, 12 tie-breakers

a=>b=>"0".repeat(a).concat("0".repeat(b)).length

Answer 47

C#, 12 tiebreakers (), 3 tiebreakers .

using System;
using System.Data;

class Program {
  static string ReadLine { get { return Console.ReadLine(); } }
  static void Main(string[] args) {
    Console.Out.Write(
     new DataTable().Compute(string.Join("\u002b", 
     ReadLine,
     ReadLine), "")
    );
  }
}

Answer 48

JavaScript 69

My attempt, using bitwise operators

for(x=(z=prompt().split(" "))[0],y=z[1];y;)x^=y,y=(y&x^y)<<1;alert(x)

Answer 49

D

main(){
    int a,b;
    readf("%d %d",&a,&b);
    write((new int[a]~new int[b]).length);
}

this time using array lengths

Answer 50

Python

i=input
print eval('%d\x2b%d'%(i(),i()))

37 chars with a little bit of cheating :p

Answer 51

A very yucky C version

Has no +, -, / or any digits. One use of * for a pointer type. The others (()=,) are needed a lot in the language itself.

#include <malloc.h>
#include <string.h>
#include <stdio.h>
typedef unsigned char *ptr;

char
  y;

const size_t
  one = sizeof y;

unsigned char
  zero='a'^'a',
  a [] = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
  z [] = "xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx",
  x [] = "xxxxxxxxxxxxxxxxx";

size_t plus_one (size_t i)
{
  ptr b=&a[one];
  memset (a,'x',sizeof a);
  b[i]=zero;
  return strlen(a);
}

size_t minus_one (size_t i)
{
  ptr b=&a[one];
  memset (a,'x',sizeof a);
  a[i]=zero;
  return strlen(b);
}

size_t minus_zero (size_t i)
{
  size_t p;
  for (p = zero ; z[p] ; p = plus_one (p))
  {
    i = minus_one(i);
  }
  return i;
}

unsigned char to_zero (unsigned char i)
{
  size_t p;
  for (p = one>>one ; x[p] ; p = plus_one (p))
  {
    i = (unsigned char) minus_one(i);
  }
  return i;
}

main (int argc, ptr argv [])
{
  ptr
    a = argv [argc>>one],
    b = argv [argc>>one<<one];

  size_t
    len_a = strlen (a),
    len_b = strlen (b),
    len = len_a > len_b ? len_a : len_b,
    o = plus_one(len),
    i,
    c=one>>one;

  ptr
    res=(ptr)malloc(plus_one(o));

  res[o]=zero;

  for (i = zero ; i < len ; i=plus_one(i))
  {
    size_t va;
    unsigned char r;

    va = (len_a > zero ? minus_zero (a[minus_one(len_a)]) <<one<<one<<one<<one : zero) | (len_b > zero ? minus_zero (b[minus_one (len_b)]) : zero);
    r = (c?"bcdefghijAuhjsadcdefghijABsadlkfdefghijABCvsxwerefghijABCDDNIWjsfghijABCDEOQJdccghijABCDEFIndDjlhijABCDEFGCnjndwijABCDEFGHsadXSkjABCDEFGHICSjshdABCDEFGHIJcbaCBA":"abcdefghijaswhHDbcdefghijAQWuciucdefghijABCOasdpdefghijABCCnjaskefghijABCDCIUdasfghijABCDEuoiDSAghijABCDEFCsalkjhijABCDEFGcapCPcijABCDEFGHCWEoerjABCDEFGHIkjIjIj") [va];
    c=r<'a';
    r&=~(one<<one<<one<<one<<one<<one);
    o=minus_one(o);
    res[o]=to_zero(r);

    if (len_a) len_a = minus_one (len_a);
    if (len_b) len_b = minus_one (len_b);
  }
  if(c) 
  {
    o=minus_one(o);
    res[o]=to_zero('B');
  }
  printf("%s\n",&res[o]);
  free (res);
}

Answer 52

Common Lisp

Always surprising for such a verbose language.

((lambda(n m)(princ(length(append(make-list n)(make-list m)))))(read)(read))

Answer 53

VBA - No +/-, 7 tie-breakers

+,-,/,*,,,.,0-9 = 0
( = 3 (3 total pairs)
) = 3
= = 1

Function a(b)
    For Each c In b
        a = a & Space(c)
    Next
    MsgBox Len(a)
End Function

Takes an array of natural numbers as an argument. This will actually add more than just 2 values, if provided, as well as return the value of a single argument.

Older versions

---

+,-,/,*,=,.,0-9 = 0
( = 4 (4 total pairs)
) = 4
, = 1

Function a(b, c)
MsgBox Len(Space(b) & Space(c))
End Function

+,-,/,*,=,.,0-9 = 0
( = 4 (4 total pairs)
) = 4
, = 3

Sub a(b,c)
MsgBox Len(String(b," ") & String(c," "))
End Sub

---

+,-,/,*,. = 0
= = 4
( = 2 (2 total pairs)
) = 2
0-9 = 2
, = 1

Sub d(e,f)
For g=1 To e:h=h & " ":Next
For g=1 To f:h=h & " ":Next
MsgBox Len(h)
End Sub

Answer 54

C #, 26 tiebreakers

+/- : 0
=   : 2
.   : 6 
()  : 18

using System;
using System.Linq;
class P
{
    static void Main()
    {
        Func<int[]> f=()=>new int[int.Parse(Console.ReadLine())];
        Console.Write(f().Concat(f()).Count());
    }
}

Answer 55

PHP, 22 chars

echo array_sum($argv);

Documentation: array_sum() and $argv
Usage: php -r 'echo array_sum($argv);' 5 6 will output 11.

Answer 56

APL (10, 3 tie-breakers)

⎕←⍴(⍳⎕),⍳⎕

Answer 57

Python 3 - 159 characters | ().,*/= count: 39

A little long, but I felt like doing it HDL-style.

a,b,c,d=int(input()),int(input()),0,0;l=max(a,b,key=int.bit_length)
for i in range(len(bin(l<<1)[2:])):
 e=a>>i&1;f=b>>i&1;c|=(e^f^d)<<i;d=e&f|d&(e^f)
print(c)

Answer 58

C# 141 characters 27 tiebreakers

+/- : 0
=   : 0
.   : 9 
()  : 18


Console.WriteLine(Enumerable.Range(1,int.Parse(Console.ReadLine()))
.Concat(Enumerable.Range(1,int.Parse(Console.ReadLine())))
.Count());

Answer 59

Python 42

a,b=input()
print eval("~-"*a+"~-"*b+"0")

Seven tie breakers, but those are string operations