We are given positive x and y. Let's go in the first loop.
Let's do some method refactoring for better understanding:
y0 = x*x+y;
x0 = x*x+y0;
y1 = sqrt(x0+(y0/labs(y0))*(-labs(y0)));
for (j = 1; j <= 2*y1; j++)
    x0 = x0-y1;
x = x0;
y = y1;
Let's go through the lines:
y0 = x*x + y
Next:
x0 = x*x + y0 = 2*x*x + y
As y0 > 0 (x, y are positive) => y0/labs(y0) = 1
So y1 = sqrt(x0+(y0/labs(y0))*(-labs(y0))) = sqrt(x0-labs(y0)) = sqrt(2*x*x + y - (x*x + y)) = sqrt(x*x) = x
Next 2 lines equals to this:
x0 = x0 - 2*y1*y1 = 2*x*x + y - 2*x*x = y
So, x=y and y=x. x and y are swapped. After that you need to count amount of swaps and print appropriate answer

See on one iteration of "for":
let be x and y on start:x1 and y1.
then after each change it'll be for x1 x2...x3; for y1 y2...y3;
so you have to find out x3 and y3 through x and y on start(x1 and y1);
Something like that x3=y1.
Good luck and sorry if my English is not enough good for explaining it)

The procedure given in the problem is the solution :D
my code:
long x,y;
cin>>x>>y;
P(x,y);
cout<<x<<" "<<y;

XD

hi all;
here's a bit simpler algorithm than in the question:

if(x > 0 && y > 0) {

    for(i = 1; i <= x + y; ++i) {

        y = x * x + y;
        x = x * x + y;
        y = sqrt((double)x - y);
        x -= (2 * y * y);
    }

}
could anyone prove it to me how this scary algorithm is the one in O(1),
i mean the one that tests reminders and then swap (if necessary).
thank you.

Edited by author 17.02.2013 15:05

Edited by author 17.02.2013 15:05

I got AC, but i dont understand why my code is correct,
in the other word in lines:
" y = sqrt(x+(y/labs(y))*(-labs(y)));"

sqrt(x+(y/labs(y)) > 0
(-labs(y))) < 0

but why when i compile this code y is positive !!!
sorry for my poor english.

Edited by author 15.11.2010 21:57

why is it wrong this code

if (x==0 || y==0)
 printf ("%lld %lld",x,y);
     else
       if (x % 2 != y % 2  )
       printf ("%lld %lld",y,x);
    else
       printf ("%lld %lld",x,y);

I think, the tests for this problem are not very good. In my code i forgot to check the case, where x or y are not positive, but i get AC. I think, that there must be a test case like that: -2 1.

(sorry my bad english)

What is solution for this problem?
I could not solve this problem up to the end

My not finished solution:

P(&x, &y)
{
s = x + y;
for (int i = 0; i < s; ++ i)
{
x0 = x;
y0 = y;
x = y0 - 2*x0*x0;
y = (int)(sqrt (2) * x0) // floor
}
}

but I don't know what to do further.

Tell me please.

i have this code:
<...>
    long x, y;
<...>
    if( (x + y) % 2 == 0 )
        printf("%ld %ld", x, y);
    else
        printf("%ld %ld", y, x);
<...>
and i don't check that x is > 0 and y > 0. Because "Input contains <...> output parameters of the function". so they are greater than 0! Can you explain me what is wrong?

#include <math.h>
#include <stdio.h>
void P(long x, long y)
{
     int i, j;
     if (x>0 && y>0)
     {
          for (i = 1; i <= x+y; i++)
          {
                y += x*x;
                x = x*x+y;
                y = sqrt(x+(y/labs(y))*(-labs(y)));
                for (j = 1; j <= 2*y; j++)
                     x = x-y;
          }
     }
     printf("%d %d",x,y);
}
int main()
{
    long t1,t2;
    scanf("%ld%ld",&t1,&t2);
    P(t1,t2);
    return 0;
}

This is easy this print "x y" or "y x":)




Edited by author 01.11.2007 02:17

Edited by author 01.11.2007 02:18

program Project1;


VAR
  x, y : longint;


begin

  readln(x, y);
  if ((x mod 2 = 0) or (y mod 2 = 0)) and (x > 0) and (y > 0) then
    writeln(y, ' ', x)
  else
    writeln(x, ' ', y);



end.

var x,y:longint;
 begin
   readln(x,y);
   if x=y then
     begin
       writeln(x,' ',y);halt;
     end;
  if ((x mod 2= 0)and(y mod 2 = 0))or((x mod 2<>0)and(y mod 2<>0))
then
    begin
      writeln(x,' ',y);halt;
    end;
  writeln(y,' ',x);
end.