Edited by author 22.03.2011 02:51

My solution also says
1
Vasia

and for the BTW check test the same answer.
but i have wa 4.

please give me some other examples.
thanks.

I'm sorry, of course the answer will be
1
Vasya

To daftcoder [Yaroslavl SU]:
Your test has a mistake in the last line. It should be like this:
5
E 1:20.0
D 1:10.0
C 0:30.0
B 0:20.0
A 0:10.0

Edited by author 27.07.2011 20:48

It's obviously true.

why is wa 1? here the code of the my solution pleas chek:
uses
  SysUtils,math;

Type m = record
     name:string[21];
     min:integer;
     seconds:real;
     pos:integer;
end;
Var n,i,c:integer;
   tmp:string;
   mas:array[1..101]of m;
   tmpG1:m;
   j,Size,k:integer;
   TopName:array[1..101] of m;
   t:m;

function CompareStr1(const S1, S2: string): Integer; assembler;
asm
        PUSH    ESI
        PUSH    EDI
        MOV     ESI,EAX
        MOV     EDI,EDX
        OR      EAX,EAX
        JE      @@1
        MOV     EAX,[EAX-4]
@@1:    OR      EDX,EDX
        JE      @@2
        MOV     EDX,[EDX-4]
@@2:    MOV     ECX,EAX
        CMP     ECX,EDX
        JBE     @@3
        MOV     ECX,EDX
@@3:    CMP     ECX,ECX
        REPE    CMPSB
        JE      @@4
        MOVZX   EAX,BYTE PTR [ESI-1]
        MOVZX   EDX,BYTE PTR [EDI-1]
@@4:    SUB     EAX,EDX
        POP     EDI
        POP     ESI
end;
function timeCompare(m1:integer;S1:real;d1:integer; m2:integer; s2:real; d2:integer):boolean;
Var ch:boolean;
begin
if (m1+(floor(d1+s1) div 60))>(m2+(floor(d2+s1) div 60)) then
 begin
 ch:=true;
 end
 else if(m1+(floor(d1+s1) div 60))=(m2+(floor(d2+s1) div 60))
 then
   begin
   if  (s1+d1-(floor(s1+d1) div 60)*60)>(s2+d2-(floor(s2+d2) div 60)*60) then ch:=true
   else ch:=false;
   end
 else ch:=false;
result:=ch;
end;

begin
readln(n);
for i:=1 to n do
  begin
    readln(tmp);
    mas[i].name:=Copy(tmp,1,pos(' ',tmp)-1);
    Delete(tmp,1,pos(' ',tmp));
    Val(copy(tmp,1,pos(':',tmp)),mas[i].min,c);
    Delete(tmp,1,pos(':',tmp));
    Val(tmp,mas[i].seconds,c);
    mas[i].pos:=i-1;
  end;
tmpG1:=mas[1];
TopName[1]:=Mas[1];
j:=1;
for i:=2 to n do
 if timeCompare(TmpG1.min,TmpG1.seconds,TmpG1.pos*30, Mas[i].min,Mas[i].seconds,Mas[i].pos*30){Min>mas[i]} then
 begin
  for k:=j downto 1 do
   if timeCompare(TopName[k].min,TopName[k].seconds,TopName[k].pos*30,Mas[i].min,Mas[i].seconds,Mas[i].pos*30) then dec(j);
   inc(j);
   TopName[j]:=Mas[i];
 end
 else
   if timeCompare(TmpG1.min,TmpG1.seconds,0,Mas[i].min,Mas[i].seconds,0) then
   begin
    inc(j);
    TopName[j]:=Mas[i];
    tmpG1:=mas[i];
   end;

Size:=j;
for i:=1 to Size do
 for j:=1 to Size do
  begin
  if  CompareStr1(TopName[i].name,TopName[j].name)<0 then
    begin
    t:=TopName[i];
    TopName[i]:=TopName[j];
    TopName[j]:=t;
    end;
  end;
writeln(size);
for i:=1 to size do
 writeln(Topname[i].name);
end.

The problem states that contestants start each 30 seconds. Let's say that running times are:
- Runner1 30:00.0 (starting at 0:00.0, finishing at 30:00.0)
- Runner2 29:31.0 (starting at 0:30.0, finishing at 30:01.0)
This means that
- Runner1 has the best time after 30 minutes from the start of the race
- Runner2 has the best time, 1 second after Runner1
So they are both in the top.
However, if:
- Runner1 30:00.0 (starting at 0:00.0, finishing at 30:00.0)
- Runner2 29:29.0 (starting at 0:30.0, finishing at 29:59.0)
Then Runner2 finishes BEFORE Runner1, so Runner1 will never have a best time.

An explanation for the problem sample is a bit below (http://acm.timus.ru/forum/thread.aspx?id=26847&upd=634557023855807397). For me, it was easier to explain on this other example.

Hope this is clear, good luck!

P.S.: I see you already have AC, congratulations!

Edited by author 16.12.2011 19:35