var a:array[1..1000000] of int64;
    i,j:longint;
begin
{$IFNDEF ONLINE_JUDGE}
assign(input,'input.txt');
reset(input);
assign(output,'output.txt');
rewrite(output);
{$ENDIF}
i:=0;
while not seekeof do begin
inc(i);
read(a[i]);
end;
for j:=i downto 1 do
writeln(sqrt(a[j]):0:4);
{$IFNDEF ONLINE_JUDGE}
close(input);
close(output);
{$ENDIF}
readln;
end.

My decision:

1.) We consider polygon such what it is, we spend 2 straight linees in parallel OX, one passes through the lowermost point, another through the uppermost. We spend a third - a line of a break on middle between a theme to two. We look at fragments. If the area of the top fragment is more, then we lift a line of a break on 1/4 all intervals, else is lowered. Then shift on 1/8 etc. To put it briefly, binary search we find height of a horizontal that it divided into equal fragments.

2.) We turn a figure, concerning any fixed point on a small corner (about 0.01 radian)

3.) We carry out action number 1.


And so we shall consider all possible variants. Certainly sooner or later it is necessary to stop to rotate a figure. It at turn on 180 degrees from an initial arrangement of a figure.

My decision works for 0.078 seconds and 253 Kb.

What your thunk about it?


GOOD LUCK!

I am sorry for bad English

Help please

WA#5 help.
sorry for my poor english

Edited by author 25.03.2009 10:19

Sorry for the offtop, but still... To the situation nowadays, I can only praise my colleagues and myself for the problem statement that remains actual despite three years passed since the contest was held. Long life this problem! :)

"You should right a program that helps scientists get rid of troubleduons."
=>
"You should write a program that helps scientists get rid of troubleduons."

Is there any good method other than precalculating and using the values stored?

so sorry for taking an admins' expensive time, its my bug. i fix it & i got ac. TU

Edited by author 08.01.2009 12:27

#include <iostream>
using namespace std;
int main()
{ long long  n,a[65550],i,k=1,b=1;
cin>>n;
for(i=0; i<n; i++) { cin>>a[i];  k=1;  b=1;
while(k<=a[i]) { b+=k; k++; }
if(b==a[i]) cout<<1<<" ';
else cout<<0<<" "; }
return 0; }

In your example order of words in dictionary is same with order of outputing words, but in the problem's text there is no words about it.
Fix it , please.

Did anybody try to solve this problem via "splay trees" (self-organizing trees by Sleator)? Randomized trees and treaps easily accepted, but now I'am interested in splay trees and always TL :(

In test 17 point A is equal to point B. Is it correct?

In problem statement "a line segment with the ends A and B". I think that one point is _not_ a line segment.

excuse my english first.

i tried my program on a lot of examples of mine and it's ok.
but i got for 10 times WA on the test #4...why?

can anybody tell me what's about or give me some hint for this problem?

thx all

it seems that my program has no bug. it pass all test on board and all my tests.
I bicolored graph. then used DP to find answer but I got WA3.

***

I have found mistake. AC now!

this test helped me very much:
50 2
99 100
98 99

Edited by author 25.02.2007 19:13

1 2
2 3
3 1
1 10
in this test , i do not know why ?
what about the others vertex ? 4,5,6,7,8,9 ??

#include <iostream>
using namespace std;
int step(int, int);
int main()
{ int n,m,y,i,s=0,f=-1;
cin>>n>>m>>y;
for(i=0; i<m; i++)
if(step(i,n)%m==y) { cout<<i<<" "; s++; }
if(s==0) cout<<f;
return 0; }
int step(int x, int n)
{ int i,p=1;
for(i=0; i<n; i++) p*=x;
return p; }