Discussion @ 1000. A+B Problem @ Timus Online Judge

Alexander Prudaev can you solve this, without using "+"? :) [40] // Problem 1000. A+B Problem 27 Oct 2006 00:26

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

Krayev Alexey (PSU) Re: can you solve this, without using "+"? :) [34] // Problem 1000. A+B Problem 27 Oct 2006 02:58

-(-a-b)

Nurbek Re: can you solve this, without using "+"? :) [33] // Problem 1000. A+B Problem 21 Dec 2006 23:30

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

Edited by author 21.12.2006 23:31

KIRILL(ArcSTU) Re: can you solve this, without using "+"? :) [32] // Problem 1000. A+B Problem 22 Dec 2006 19:00

a--b
a------------------------b :)))

Ostap Korkuna (Lviv NU) I can do it without using + and - :-) [31] // Problem 1000. A+B Problem 23 Dec 2006 01:43

#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);
}

[SSAU_#617]snipious aka Pimenov Sergey Nikolaevich Of course!!! [30] // Problem 1000. A+B Problem 8 Jan 2007 17:42

var
a, b: integer;
begin
read(a, b);
asm
    mov ax, a;
    add ax, b;
    mov a, ax;
end;
writeln(a);
end.

it4.kp Or using bitwise operations only :) [29] // Problem 1000. A+B Problem 8 Jan 2007 22:13

#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;
}

Roma Labish[Lviv NU] Re: Or using bitwise operations only :) [19] // Problem 1000. A+B Problem 8 Jan 2007 22:16

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

Edited by author 08.01.2007 22:17

it4.kp Re: Or using bitwise operations only :) [18] // Problem 1000. A+B Problem 8 Jan 2007 22:26

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

RockBeat 119 ) [16] // Problem 1000. A+B Problem 8 Jan 2007 23:33

#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;
}

it4.kp 108 ) [15] // Problem 1000. A+B Problem 8 Jan 2007 23:48

#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

RockBeat 106 ) [14] // Problem 1000. A+B Problem 9 Jan 2007 00:30

#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

Slam [Tartu U] Re: 106 ) [13] // Problem 1000. A+B Problem 11 Jan 2007 07:41

RockBeat wrote 9 January 2007 00:30

#include<iostream.h>

main(){
    int x,y,i; <--- LOL :)
    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

KIRILL(ArcSTU) Re: Pascal forever!) [12] // Problem 1000. A+B Problem 15 Jan 2007 14:32

Assembler solutions are cheating, but about this one ;)

type pinteger = ^integer;
var a, b: integer;
p:procedure;
instr:array [1..11] of byte = ($8B, $45, $08, $8B, $55, $0C, $8B, $12, $01, $10, $C3);

procedure sum(a, b: pinteger); stdcall;
begin
p();
end;

begin
p:=@instr;
readln(a, b);
sum(@a, @b);
writeln(a);
end.

It's not forbidden by any rules
It may be done more simplify, but I had many troubles
to get AC with FreePascal on Timus

Lomir Re: Pascal forever!) // Problem 1000. A+B Problem 1 Apr 2007 23:26

Here is my solution without + :)
However it will be shorter in pascal...
#include <cstdio>
int main ()
{
    int a; int b;
    scanf("%d %d", &a, &b);
    __asm{
        mov eax, dword ptr a
        add eax, dword ptr b
        mov dword ptr a, eax
    }
    printf("%d\n", a);
    return 0;
}

Edited by author 01.04.2007 23:27

PSV Re: Pascal forever!) [3] // Problem 1000. A+B Problem 2 Apr 2007 02:40

KIRILL(ArcSTU) ----------> COOL

rumi Re: Pascal forever!) [2] // Problem 1000. A+B Problem 28 Apr 2007 00:37

begin
randomize;
write(300-199*random(2))
end.

Alias (Alexander Prudaev) Re: Pascal forever!) [1] // Problem 1000. A+B Problem 5 May 2007 23:10

#include <stdio.h>

int main()
{
    char m[1];
    int a,b;
    scanf("%d %d", &a, &b);
    printf("%d\n", &b[&m[a]]-m);
    return 0;
}

of course it must be &((&m[a])[b])-m;

Edited by author 05.05.2007 23:12

Alias (Alexander Prudaev) C++ forever // Problem 1000. A+B Problem 6 May 2007 14:34

#include <stdio.h>

int main()
{
    int a,b;
    scanf("%d %d", &a, &b);
    printf("%d\n", &b[&((char*)0)[a]]);
    return 0;
}

after compile :
mov    eax, DWORD PTR _a$[ebp]
add    eax, DWORD PTR _b$[ebp]

Mansurov Artur Re: Pascal forever!) [6] // Problem 1000. A+B Problem 16 Jul 2008 22:30

KIRILL(ArcSTU) ----------> COOL
+1

OpenGL Re: Pascal forever!) [5] // Problem 1000. A+B Problem 13 Oct 2008 19:06

program qwerty;
var a,b:integer;
begin
read(a,b);
if(a=b)then a:=a shl 1
else a:=(sqr(a)-sqr(b))div(a-b);
write(a);
end.

penartur Re: Pascal forever!) [4] // Problem 1000. A+B Problem 13 Oct 2008 23:18

2OpenGL: As it stated on previous page, you cannot use the '-' sign (otherwise, the task would have been extremely simple, you could write just "return a--b" or something like this).
--
C#:

using System;

namespace Task1000b {
    class Program {

        private static int Sum(int a, int b) {
            if(a == 0) {
                return b;
            } else if(b == 0) {
                return a;
            } else {
                return Sum(a^b, (a&b)<<1);
            }
        }

        static void Main(string[] args) {
            string[] input = Console.ReadLine().Split();
            Console.WriteLine(Sum(int.Parse(input[0]), int.Parse(input[1])));
        }
    }
}

Edited by author 13.10.2008 23:18

penartur Re: Pascal forever!) [3] // Problem 1000. A+B Problem 13 Oct 2008 23:55

Or, shortened:
using System;
namespace n {
    class c {
        static void Main() {
            string[] i = Console.ReadLine().Split();
            int a = int.Parse(i[0]);
            int b = int.Parse(i[1]);
            while((a&b)!=0) {
                a = a^b;
                b = (b&~a)<<1;
            }
            Console.WriteLine(a^b);
        }
    }
}

Anisimov Dmitry (Novosibirsk STU) Re: Pascal forever!) [2] // Problem 1000. A+B Problem 6 Nov 2008 01:18

#include <stdio.h>
#include <string.h>

char a[1024000];
char b[1024000];
int x,y;

int main() {
    scanf("%d %d",&x,&y);
    while(x>strlen(a)) strcat(a," ");
    while(y>strlen(b)) strcat(b," ");
    strcat(a,b);
    printf("%d\n",strlen(a));
    return 0;
}

Anisimov Dmitry (Novosibirsk STU) C [1] // Problem 1000. A+B Problem 6 Nov 2008 01:20

#include <stdio.h>

int x,y;

int main() {
      scanf("%d%d",&x,&y);
      fseek(stdin,x,SEEK_SET);
      fseek(stdin,y,SEEK_CUR);
    printf("%d\n",ftell(stdin));
    return 0;
}

Edited by author 06.11.2008 01:22

Edited by author 06.11.2008 01:23

Anisimov Dmitry (Novosibirsk STU) Re: C // Problem 1000. A+B Problem 6 Nov 2008 01:32

#include <cstdio>

int x,y;

int main() {
      scanf("%d%d",&x,&y);
    printf("%d\n",&(&((char*)0)[x])[y]);
    return 0;
}

Metallllllll No subject // Problem 1000. A+B Problem 10 Jan 2007 02:01

Edited by author 10.01.2007 02:09

[SSAU_#617]snipious_#1 aka Pimenov Sergey Nikolaevich Re: Or using bitwise operations only :) [7] // Problem 1000. A+B Problem 9 Jan 2007 00:35

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

KIRILL(ArcSTU) Re: Or using bitwise operations only :) [6] // Problem 1000. A+B Problem 9 Jan 2007 00:42

[SSAU_#617]snipious_#1 aka Pimenov Sergey Nikolaevich wrote 9 January 2007 00:35

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

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

[SSAU_#617]snipious_#1 aka Pimenov Sergey Nikolaevich Re: Or using bitwise operations only :) [5] // Problem 1000. A+B Problem 9 Jan 2007 00:45

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

Roma Labish[Lviv NU] Or using ':=' operator only :) [4] // Problem 1000. A+B Problem 10 Jan 2007 02:10

Without :

it4.kp wrote 8 January 2007 22:26

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.

KIRILL(ArcSTU) Re: Or using ':=' operator only :) [3] // Problem 1000. A+B Problem 10 Jan 2007 03:07

Tests are simple. You solution pass TL easily, but

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

Slam [Tartu U] Re: Or using ':=' operator only :) [2] // Problem 1000. A+B Problem 10 Jan 2007 03:08

inc() is using + itself so its forbidden

Roma Labish[Lviv NU] Re: Or using ':=' operator only :) // Problem 1000. A+B Problem 10 Jan 2007 15:49

So what? you can't see '+' in program => there isn't '+' :)

inc(b,a) -> I didn't know it, becouse I'm C++ programmer :)

And IV Re: Or using ':=' operator only :) // Problem 1000. A+B Problem 22 Jun 2007 17:16

rules on olimpiads: dont use assembler, but it works on timus. :)

program Project2;

{$APPTYPE CONSOLE}

uses
SysUtils;
var
a,b:integer;
begin
readln(a,b);
asm
mov eax,a;
add eax,b;
mov a,eax;
end;
writeln(a);
end.

Edited by author 22.06.2007 17:20

wangling219 Re: Or using bitwise operations only :) // Problem 1000. A+B Problem 22 Nov 2008 20:13

Great|!

KIRILL(ArcSTU) Re: can you solve this, without using "+"? :) [1] // Problem 1000. A+B Problem 9 Jan 2007 00:40

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.

blacksnow Re: can you solve this, without using "+"? :) // Problem 1000. A+B Problem 29 Apr 2009 19:51

WELL DONE!!!

mingyuangao Re: can you solve this, without using "+"? :) [1] // Problem 1000. A+B Problem 23 Nov 2008 17:21

For Pascal, it is easy.
Just use the procedure "inc".

Alias aka Alexander Prudaev Fast Fourier Transofmation Solution :) // Problem 1000. A+B Problem 20 Mar 2009 08:29

#include <complex>
typedef complex<double> com;

const double Pi = acos(-1.0);

void FastFourierTransformation(vector<com> &a, int z)
{
    if (a.size() == 1) return;
    vector<com> a0;
    vector<com> a1;
    for (int i = 0; i < a.size(); i++)
        if (i & 1)
            a1.push_back(a[i]);
        else
            a0.push_back(a[i]);
    FastFourierTransformation(a0, z);
    FastFourierTransformation(a1, z);
    com x = com(cos(z * 2 * Pi / a.size()), sin(z * 2 * Pi / a.size()));
    com xx(1);
    for (int i = 0; i < a.size() / 2; i++)
    {
        a[i] = a0[i] + a1[i] * xx;
        a[i + a.size() / 2] = a0[i] - a1[i] * xx;
        xx = xx * x;
    }
}

int main()
{
    int A, B;
    scanf("%d %d", &A, &B);
    vector<com> a(4, com(0));
    a[0] = com(A);
    a[1] = com(B);
    FastFourierTransformation(a, 1);
    int ans = (int)(a[0].real() + 0.5);
    printf("%d\n", ans);
    return 0;
}

Edited by author 20.03.2009 15:51

Net walker Re: can you solve this, without using "+"? :) // Problem 1000. A+B Problem 6 May 2009 17:25

yes I can...

Discussion of Problem 1000. A+B Problem