Показать все сообщения Спрятать все сообщенияcan you solve this, without using "+"? :) -(-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);
} Of course!!! [SSAU_#617]snipious aka Pimenov Sergey Nikolaevich 8 янв 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. #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 22:40 119 ) RockBeat 8 янв 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;
} 108 ) it4.kp 8 янв 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 00:14 106 ) RockBeat 9 янв 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 #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 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, $C3);
procedure sum(a, b: pinteger); stdcall;
begin
p();
end;
begin
p:=@instr;
readln(a, b);
sum(@a, @b);
writeln(a);
end.
It's not forbidden by any rules
It may be done more simplify, but I had many troubles
to get AC with FreePascal on Timus Here is my solution without + :)
However it will be shorter in pascal...
#include <cstdio>
int main ()
{
int a; int b;
scanf("%d %d", &a, &b);
__asm{
mov eax, dword ptr a
add eax, dword ptr b
mov dword ptr a, eax
}
printf("%d\n", a);
return 0;
}
Edited by author 01.04.2007 23:27 KIRILL(ArcSTU) ----------> COOL begin
randomize;
write(300-199*random(2))
end.
#include <stdio.h>
int main()
{
char m[1];
int a,b;
scanf("%d %d", &a, &b);
printf("%d\n", &b[&m[a]]-m);
return 0;
}
of course it must be &((&m[a])[b])-m;
Edited by author 05.05.2007 23:12 #include <stdio.h>
int main()
{
int a,b;
scanf("%d %d", &a, &b);
printf("%d\n", &b[&((char*)0)[a]]);
return 0;
}
after compile :
mov eax, DWORD PTR _a$[ebp]
add eax, DWORD PTR _b$[ebp] KIRILL(ArcSTU) ----------> COOL
+1 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. 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 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);
}
}
} #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;
} C Anisimov Dmitry (Novosibirsk STU) 6 ноя 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 Re: C Anisimov Dmitry (Novosibirsk STU) 6 ноя 2008 01:32 #include <cstdio>
int x,y;
int main() {
scanf("%d%d",&x,&y);
printf("%d\n",&(&((char*)0)[x])[y]);
return 0;
}
Edited by author 10.01.2007 02:09 But with ASM it'll run more quickly!!! But with ASM it'll run more quickly!!! 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;!!! Without : 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 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 So what? you can't see '+' in program => there isn't '+' :)
inc(b,a) -> I didn't know it, becouse I'm C++ programmer :) 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 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. For Pascal, it is easy.
Just use the procedure "inc". #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 |
