can you solve this, without using "+"? :)

-(-a-b)

-(-a-b)
BRAVO !!! :)

Edited by author 21.12.2006 23:31

a--b
a------------------------b  :)))

#include <stdio.h>
#include <math.h>

void main()
{
double a, b;
scanf("%lf%lf", &a, &b);
double c = log(exp(a)*exp(b));
printf("%.0lf", c);
}

#include <iostream>

using namespace std;

int sum(int a, int b){
    return (!a||!b ? a|b : sum((a&b)<<1,a^b));
}

int main(){
    int a, b;
    cin >> a >> b;
    cout << sum(a,b);
    return 0;
}

You stole my Idea :) I just think about it, but you write it faster then me :).

Edited by author 08.01.2007 22:17

Ok, let's make problem more interesting...
What is the shortest program to add A and B?

Rules:
1. Whitespace chars are ignored.
2. You cannot use +, -, * and / operations.
3. Assembler is "off the table".
4. Java's BigInteger too
5. any non-integer functions (like log() and exp()) are forbidden

My very raw variant's length is 129:

#include <iostream>

int s(int a, int b){
return (!a||!b ? a|b : s((a&b)<<1,a^b));
}

int main(){
int a, b;
std::cin >> a >> b;
std::cout << s(a,b);
return 0;
}

Edited by author 08.01.2007 22:40

#include<iostream.h>

#define _(x,y)    for(i=1;x&i;y^=i,i<<=1);y^=i;

main(){
    int x,y,i;
    for(cin>>x>>y;y;){_(x,x)_(~y,y)}
    cout<<x;
}

#include <iostream.h>

int s(int a, int b){
return !a||!b ? a|b : s((a&b)<<1,a^b);
}

main(){
int a, b;
cin >> a >> b;
cout << s(a,b);
}

Edited by author 09.01.2007 00:14

#include<iostream.h>

main(){
    int x,y,i;
    for(cin>>x>>y,x=~x;x^~0;x^=y^=x^=y)
        for(y^=i=1;y&i;y^=i<<=1);
    cout<<~y;
}

Edited by author 09.01.2007 01:03

But with ASM it'll run more quickly!!!

I've solved it on pascal some time ago, but not post
It's not short of course :)


var
  i,a,b,s,c:integer;
begin
  read(a,b);
  for i:=0 to 31 do
  begin
    s:=s or (a and 1 xor b and 1 xor c) shl i;
    c:=(c and (a and 1 xor b and 1)) or (a and 1 and b and 1);
    a:=a shr 1;
    b:=b shr 1
  end;
 write(s)
end.

Compiler generates better assembler code
for a:=a+b than your one

And what is it??? Try to use 10^10 my code and c:=a+b;!!!

Edited by author 10.01.2007 02:09

Without :and without bitwise operation:) :

63 symbols:

var a,b,i:integer;
begin
read(a,b);
for i:=1 to a do inc(b);
write(b)
end.

Tests are simple. You solution pass TL easily, but

for i:=1 to a do inc(b); equal to inc(b,a);

inc() is using + itself so its forbidden