I don't know why my program do WA!!! I tested it for hundreds test
and it works for them!!!
Has anybody tests for this problem?

var
 ok:array[1..10000]of longint;
 n,s,i,j,min:longint;
 nf,sf,max:longint;
begin
 fillchar(ok,sizeof(ok),0);
 readln(n,s);
 ok[s]:=1;max:=0;
 for i:=s to n-1 do
 begin
  if i*2>=n then min:=n else min:=i*2;
  for j:=i+1 to min do
   if (ok[i]+1>ok[j])and(j*100 mod i=0) then
    ok[j]:=ok[i]+1;
  if max<ok[i] then max:=ok[i];
 end;
 writeln(max);
end.

#include <stdio.h>
#include <stdlib.h>

#define ONLINE_CONTEST
void main()
{
    long n;
    long nr;
    int i, x, y, ant, first = 1;
#ifndef ONLINE_CONTEST
    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
#endif
    scanf("%ld", &n);
    nr  = 0;
    for(i = 0; i < n; i++)
    {
        scanf("%d%d", &x, &y);
        x = x + y;
        if(first)
        {
            first = 0;
            ant = x;
        }
        else if(x < 9)
        {
            printf("%d", ant);
            for(y = 0; y < nr;y++)
            {
                printf("9");
            }
            ant = x;
            nr=0;
        }
        else if(x == 9)
        {
            nr++;
        }
        else if(x > 9)
        {
            printf("%d", ant+1);
            for(y = 0; y < nr; y++)
            {
                printf("0");
            }
            ant = x % 10;
            nr = 0;
        }
    }
    if(!first)
        printf("%d", ant);
}

dear sir
i am not understandin the prob clearly what is the L1,L2,L3
in the prob

dear sir please help me out in the prob
sir i am not understanding the prob
what is mean by"ever ticket  has its 2N-digit number".
and what will be the first digit and what will be the last digit
and what is mean by "the sum of all digits in the number"
when we give 2 2
the answer is 4
how will u please explain me please

dear sir
please help me out here.
i have got some problem and plese do help me to  find it out its prob
if the length of each word is 4
1101
the answer is 1001 cos mod of 5 is =0;
in 11011 why not the answer 1001 cos it is also the muntiple of  N
and why not the answer of 1101 is 1111 cos bit 1 in the third
position has been changed to 0 after passing throught the noise.
and please sir let me know about the answers of the followring and
please explain why?
1)10011
2)110111
3)00110
4)1111
THE LENGTH OF ALL THESE IS 4

Can you correct my mistake please!!!!!!


var m               : Array[-10..20,-10..20] of 0..1;
    i,j,k,n,a,b     : INTEGER;
    Flag            : Boolean;
    e               : array[1..100] of record x,y : integer; end;
procedure work;
var z,l,index,beg,en : integer;
begin
index := 1; beg := 1; en := 1;
e[1].x:=a; e[1].y:=b;
m[a,b]:=0;
for j:=1 to n do
  begin
  for i:=beg to en do
     begin
     if m[e[i].x,e[i].y+1] = 1 then
        begin
        m[e[i].x,e[i].y+1] := 0;
        inc(index);
        e[index]:=e[i];
        inc(e[index].y);
        write('R');
        end;
     if m[e[i].x+1,e[i].y] = 1 then
        begin
        m[e[i].x+1,e[i].y] := 0;
        inc(index);
        e[index]:=e[i];
        inc(e[index].x);
        write('B');
        end;
     if m[e[i].x,e[i].y-1] = 1 then
        begin
        m[e[i].x,e[i].y-1] := 0;
        inc(index);
        e[index]:=e[i];
        dec(e[index].y);
        write('L');
        end;
     if m[e[i].x-1,e[i].y] = 1 then
        begin
        m[e[i].x-1,e[i].y] := 0;
        inc(index);
        e[index]:=e[i];
        dec(e[index].x);
        write('T');
        end;
     if i <> index then writeln(',') else writeln('.');
     end;
   beg := en + 1; en := index;
   end;
end;
begin
readln(n); for i:= 1 to n do begin readln(j,k); m[10-k+1,j]:=1; end;

{ Tursene na nai dolen lqv element }
Flag := False;
for i := 1 to 10 do
 if Flag then break else
  for j:= 10 downto 1 do
    if m[j,i] = 1 then begin Flag := True; a:=j; b:=i; Break end;
{ ... }

work;
end.

Hallo all,
there might be some people from Varna, Bulgaria here - as far as I
know - there should be some. I gonna be there (ok, not in Varna, at
Zlatni Piascy, of course) for the two weeks starting from August, 23.
It might be a good idea to meet somebody there - just for
conversation. I could also give some lecture - as an ACM-veteran and
contest organizer :-)

  The algorithms I found requires O(N^2) , which N = 1800 . But we
must calculate the large-number ( I use string) , so it takes about O
(N^3) !!! . Could anyone help me on this Problem  ?

var a:array[0..500]of string[85];
    N,i,j,k,num,p:integer;
procedure conv(i,j:integer;var num,p:integer);
var k:integer;
begin
{  if a[i]=a[j] then
  begin
    p:=length(a[i])+1;
    num:=0;
  end;}
  num:=0;p:=1;
  k:=0;
  for k:=1 to length(a[i]) do
  begin
    if a[i,k]<>a[j,k] then break
    else if a[i,k]=' ' then
    begin
      inc(num);
      p:=k+1;
    end;
  end;
  if (a[i,k]=a[j,k])and(k<>0)and((length(a[j])=length(a[i]))or(a
[j,k+1] = ' ')) then
  begin
      inc(num);
      p:=k+2;
  end;
end;

procedure sort(l,r: longint);
var i,j: longint;
    x,y: string[85];
begin
   i:=l;j:=r;x:=a[(l+r) div 2];
   repeat
     while a[i]<x do inc(i);
     while x<a[j] do dec(j);
     if not(i>j) then
       begin
          y:=a[i];
          a[i]:=a[j];
          a[j]:=y;
          inc(i);
          j:=j-1;
       end;
   until i>j;
   if l<j then sort(l,j);
   if i<r then sort(i,r);
end;

begin
{ assign(input,'disktree.in');
 reset(input);
 assign(output,'disktree.out');
 rewrite(output);}
 readln(n);
 for i:=1 to N do
 begin
   readln(a[i]);
   for j:=1 to length(a[i]) do if a[i,j] = '\' then a[i,j]:=' ';
 end;
 sort(1,N);
 for i:=1 to N do
 begin
   if a[i-1]=a[i] then continue;
   conv(i-1,i,num,p);
   for k:=1 to num do write(' ');
   for j:=p to length(a[i]) do
   begin
     if a[i,j]=' ' then
     begin
       inc(num);
       writeln;
       for k:=1 to num do write(' ');
     end
     else write(a[i,j]);
   end;
   writeln;
 end;
{ close(output);}
end.

just another(wrong) idea?

why do the Memory usage is always -8.
Even in such an experiment, I got Time Limit Exceeded and the
Memory usage was -8.

*********************************************
#include <iostream.h>

typedef struct node1
{int start;
 int end;
 char mark[10];
}node;

int main()
{
node ku[100];
int i;
int n;
short maxs;

cin>>maxs>>n;
while(n!=-1){
   for(i=1;i<=n;i++){
      cin>>ku[i].start>>ku[i].end;
      cin>>ku[i].mark;
   }
   cin>>n;
}
cout<<n;

return 1;
}
****************************************

help!

if the'\r' appears does it mean it start from the beginning of the line again?
i mean : "(*xxxxxx  '\r'*) "---> " *)"  isn't it?
if the above one is true, what will b the output for this one
(  '\r'(1+2)
i mean i've found space after '(' so it's Invalid now right?  \but when '\r'
appears it start again from the begginning of the line so ' (1+2) '
replace '(' so the now it's valid???????
** '\r'==carriege return

The reason is not Wrong answer as you can think. I use recursion in
my solution of this problem and get mem limit exceeded (it is stack
overflow); at all test cases, that I could make, my program works
successfull and I do not know what the matter.
Please - if you have any test cases - post it to this webboard or to
my email pflojd@yandex.ru.

const nmax=100;
type mas=array[1..nmax]of longint;
var aa:array[0..500]of mas;
    i:longint;
    s:longint;
    n:longint;
    m,j:longint;
    a,b,c:mas;
function dlina(a:mas):longint;
var i:longint;
begin
 i:=nmax;
 while a[i]=0 do dec(i);
 dlina:=i;
end;
procedure slog(a,b:mas;var c:mas);
var i,d1,d2,max:longint;
begin
 fillchar(c,sizeof(c),0);
 d1:=dlina(a);d2:=dlina(b);
 if d1>d2 then max:=d1 else max:=d2;
 for i:=1 to max do
  begin
   c[i]:=c[i]+a[i]+b[i];
   if c[i]>9 then
    begin
     c[i+1]:=c[i] div 10;
     c[i]:=c[i] mod 10;
    end;
  end;
end;
procedure umn(a:mas;var c:mas);
var i,j,p,z,d:longint;
begin
 d:=dlina(a);
 fillchar(c,sizeof(c),0);
 for i:=1 to d do
  begin
   p:=0;
   for j:=1 to d do
    begin
     z:=a[i]*a[j]+p+c[i+j-1];
     c[i+j-1]:=z mod 10;
     p:=z div 10;
    end;
   c[i+j]:=p;
  end;
end;
begin
 readln(n,s);
 if (s div 2>n*9)or(odd(n)) then
  begin
   writeln(0);
   halt;
  end;
 if n=1 then
  begin
   if s>9 then
    writeln(0)else writeln(1);
    halt;
  end;
 s:=s div 2;
 aa[0][1]:=1;
 for i:=1 to n do
  begin
  if s>9*i then m:=0 else m:=s;
   for j:=1 to m do
    begin
    slog(aa[j-1],aa[j],c);
    aa[j]:=c;
    end;
  end;
 umn(aa[m],c);
 for i:=dlina(c)downto 1 do
  write(c[i]);
 writeln;
end.

When I submit the solution for the problem I get a compilation error.
The error reads:

"You tried to solve problem 1017. Your solution on C++ was compiled
with the following errors:

143733
temp\143733(54) : error C2471: cannot update program
database 'e:\judge\vc60.pdb'"

can anyone tell me what is this about.

I am sure the problem is correctly solved and that it does no illegal
operations. Also it compiles perfectly under gcc for dos. I can't
test it with Visual C++ but I don't use anything other than ANSI so
it should be ok.

I attach the source code of my program, maybe you could indicate the
error. If the error is not with my source code, but with TIMUS, what
should I do?

#include <stdio.h>

#define MAX 501

int n;

double answer, d[MAX][MAX];

void solve()
{
    int i, j, k;

    // d[i][j] = numarul de moduri in care se pot aranja j
blocuri astfel
    // incat inaltzimea sa fie egazact j
    for (i = 1; i <= n; i++)
        for (j = 1; j <= n; j++)
            d[i][j] = 0;
    d[0][0] = 1;
    for (i = 1; i <= n; i++)
        for (j = 1; j <= n && i - j >= 0; j++)
        {
            d[i][j] = 0;
            for (k = 0; k < j; k++)
                d[i][j] += d[i - j][k];
        }
/*    for (i = 1; i <= n; i++)
    {
        for (j = 1; j <= n; j++)
            printf("%d ", d[i][j]);
        printf("\n");
    }*/
    for (k = 1; k < n; k++)
        answer += d[n][k];
}

int main()
{
/*#ifndef ONLINE_JUDGE
    freopen("staircases.in", "r", stdin);
    freopen("staircases.out", "w", stdout);
#endif*/

    scanf("%d", &n);
    solve();
    printf("%.0lf\n", answer);

/*#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
#endif*/

    return 0;
}

what will be the answer of
the following
and please let me know why
1) 1 2 -1 3
2) 1 2 0 4
3) -1 -2 3 0