What is this test?

1. #123ident;
2. #123'str';
3. #123#124;
4. #   (without following digits and not as part of {comment} and 'string')
5. 222.not_a_digit;
6. 222.222.222;
7. 222#123;
8. 222'str';
9. ident#123;
10. ident'str';

I didn't consider these expressions in my program and got WA4.

Edited by author 17.05.2009 16:56

type
  fu1=array['a'..'z'] of longint;
  su=record
       a:fu1;
       zf:integer;
       x:longint;
     end;
  kk1=array[1..1000] of su;
var
  a,b,c:kk1;
  d:su;
  i,j,k,l,x,y,n,m:longint;
  s:string;
  g:boolean;
  ch:char;
  function jun(g:boolean):integer;
    begin
      if g then exit(1) else exit(-1);
    end;
  procedure doit(var a:kk1;  var n:longint);
    var
      g:boolean;
      c:char;
      num:string;
      nu:longint;
    begin
      readln(s);
      num:='';
      for i:=1 to length(s) do
        if s[i]<>' ' then num:=num+s[i];
      s:='';
      for i:=1 to length(num) do
        begin
          if num[i] in[ '+','-']
            then begin
                   if i<>1 then
                   s:=s+' '+num[i]+' '
                   else s:=s+num[i];
                 end
            else begin
                   s:=s+num[i];
                 end;
        end;





      g:=true;
      if s[1] in ['+','-']
        then begin
               if s='-' then g:=false;
               delete(s,1,1);
             end;
      repeat
        i:=pos(' ',s);
        l:=length(s);
        if i=0 then i:=l+1;
        d.zf:=jun(g);
        d.x:=d.zf;
        fillchar(d.a,sizeof(d.a),0);
        if s[1] in ['0'..'9']
          then begin
                 num:='';
                 while (s[1] in ['0'..'9']) and (length(s)<>0) do
                   begin
                     num:=num+s[1];
                     delete(s,1,1);
                   end;
                 if s[1]<>' ' then  delete(s,1,1);
                 val(num,j,nu);
                 d.x:=j*d.zf;
                 i:=pos(' ',s);
                 l:=length(s);
                 if i=0 then i:=l+1;
               end;
          if d.x=0 then
            begin
              delete(s,1,i-1);
              g:=s[2]='+';
                 delete(s,1,3);
              continue;
            end;
        if s[1]=' ' then
          begin
            inc(n); a[n]:=d;
            g:=s[2]='+';
                 delete(s,1,3);
            continue;
          end;
        if length(s)=0 then begin inc(n); a[n]:=d; break; end;
        k:=1;
        j:=1;
        while j<=i do
          begin
            c:=s[j];
            k:=1;
            if (j>=i) then inc(j,2) else begin
              if s[j+1]='^'
                then begin
                       inc(j,2);
                       num:='';
                       while s[j] in ['0'..'9'] do
                         begin
                           num:=num+s[j];
                           inc(j);
                           if (j>i) or (j=l+1) then break;
                         end;
                       inc(j);
                       val(num,k,nu);
                     end
                     else inc(j,2);
                end;

            d.a[c]:=d.a[c]+k;
          end;
        delete(s,1,i-1);
        if i<>l+1
          then begin
                 g:=s[2]='+';
                 delete(s,1,3);
               end;
        inc(n);
        a[n]:=d;
      until i=l+1;
    end;
  function jia(a,b:su):su;
    var
      i:longint;
      c:char;
    begin
      a.x:=a.x*b.x;
      for c:='a' to 'z' do
        a.a[c]:=a.a[c]+b.a[c];
      exit(a);
    end;
  function ok(a,b:su):boolean;
    var
      c:char;
    begin
      for c:='a' to 'z' do
        begin
          if a.a[c]<>b.a[c]
            then exit(false);
        end;
      exit(true);
    end;
  function bijiao1(a,b:su):boolean;
    var
      c:char;
      i,x,y:longint;
    begin
      i:=2;
      x:=0; y:=0;
      for c:='a' to 'z' do
        begin
          x:=x+a.a[c];
          y:=y+b.a[c];
          if (i=2) then begin
            if a.a[c]>=b.a[c]
              then begin if a.a[c]>b.a[c] then  i:=1 end
               else i:=0;
            end;
        end;
      if (x<y) or ((i=0) and(x=y) ) then exit(true) else exit(false);
    end;
begin
 { fillchar(a,sizeof(a),0);
  fillchar(b,sizeof(b),0);
  fillchar(c,sizeof(c),0); }
  n:=0;
  m:=0;
  doit(a,n);
  doit(b,m);

  k:=0;
  for i:=1 to n do
    begin
      for j:=1 to m do
        begin
          d:=jia(a[i],b[j]);
          g:=true;
          for x:=1 to k do
            begin
              if ok(c[x],d)
                then begin
                       c[x].x:=c[x].x+d.x;
                       g:=false;
                       break;
                     end;
            end;
          if g then
           begin
             inc(k);
             c[k]:=d;
           end;
        end;
    end;
  for i:=1 to k-1 do
    for j:=i+1 to k do
      begin
        if bijiao1(c[i],c[j])
          then begin
                 d:=c[i];
                 c[i]:=c[j];
                 c[j]:=d;
               end;
      end;
    for i:=1 to k do
      begin
        if (c[i].x=0) then continue;
        if (c[i].x>0)
          then begin
                 if i<>1 then write('+');
                 if i<>1 then write(' ');
               end
          else begin
                 write('-');
                 if i<>1 then write(' ');
               end;
        c[i].x:=abs(c[i].x);
        g:=true;
        if c[i].x<>1 then write(c[i].x);
        for ch:='a' to 'z' do
          begin
            if c[i].a[ch]=0 then continue;
            if (c[i].x<>1) or (g=false)
              then write('*');
            g:=false;
            write(ch);
            if c[i].a[ch]<>1 then write('^',c[i].a[ch]);
          end;
        if g and (c[i].x=1) then write(1);
        if i<>k then
        write(' ');
      end;
end.


i think there's no problem with mine











Edited by author 17.05.2009 14:38

help~! >0<

please give me information of the test 2 of this problem.

Edited by author 09.11.2010 02:50

Edited by author 09.11.2010 02:49

Getting wrong answer. Can anybody post it, please!?

I found that it begins 0(not 1),and Accept.
..............................
..............................
..............................
http://cnxhk.wordpress.com.cn/2009/05/15/ural_1019/
This is my C code!

I solved this problem using:
Calendar calendar = new GregorianCalendar(year,month-1,1)
It helped me to get position of the first day of month and day's count at month.

Pay attention to:
1)Day's count in month.
2)Use trim() to remove  ending spaces.

How can N be 2000 digits long if N is less than 10^1000, which has only 1001 digit?

P.S. I didn't notice it was new year by server time, I am in America, here it's still December 31))) Happy new year)))

Edited by author 01.01.2008 01:13

Can anyone say something more than just disjoint sets or dfs?

Edited by author 07.05.2009 09:42

I got OLE in #10, and I also use an AC solution to check my code, but I can not find out any mistake, could somebody tell me what will lead to OLE in #10 instead of #1 ?

I do not understand one thing. In the task said that “inspector … took out from the table's pocket the LAST of the pocketed balls”. And about test 3-1-2 in the hint said that “Chichikov could pocket the balls in any order excepting the right one 1-2-3”. But how can we get the 1-3-2? Only if Chichikov pocketed balls 1, 3 and inspector took BOTH  bolls (in one moment) and then took the last ball - 2. What’s wrong?

ok, found it out ;)

Edited by author 13.05.2009 23:27

I am not sure from the problem statement, actually I don't get something. Suppose that we have this test case
5
1 20
1 0
2 0
3 20
3 0
2
1 5 10
3 2 25

Is answer = 2 for this test case?
4 buy debt from 5 to 1 and now 1 have 30 or it still has 20?
If it has 30, then it can buy debt from 2 to 3.

If this is wrong, can tell me what is the solution to this test case, please? Thanks.

Hello! My solution got accepted, but now as I thought it through it is wrong. Here are some tests that will fail me:
Test 1:
6 3
1 2
2 3
4 5
5 6
6 3
Answer:
First player wins flying to airport 6
Test 2:
15 12
1 2
2 3
4 5
5 6
6 3
3 11
7 8
8 9
9 10
10 11
11 12
13 12
14 13
15 14
Answer:
First player wins flying to airport 11

Edited by author 09.05.2009 13:47

I know the formula, but something is going wrong during the 3 test.