Got AC 0.001 sec from the first effort :) I saw many requests for clarification here, but it's easier to describe grammar in terms of behavior.

Go sequentially through input data.

1. If you meet '{' - everything until closest corresponding '}' is a single comment, no matter what's inside. '}' corresponds to '{' if all {,} characters between them form valid bracket sequence.

2. If you meet '//' - everything until '\n' is a single comment, no matter what's inside.

3. If you meet ' - everything until next ' is a string, no matter what's inside.

3. If you meet '#' followed by a digit - everything until first non-digit is a string.

4. If you meet a digit, then everything until first non-digit is a number. If first non-digit character is '.' AND it is followed by another digit, then this post-dot digit sequence is appended to the number as well as the dot itself. I.e. "1.23" is a single number, but "123.a" is number '123' followed by '.', followed by identifier 'a'.

5. If you meet underscore or letter, everything until first non-digit-non-underscore-non-letter is an identifier. It may be a keyword.

6. Output all other characters as is.

Edited by author 17.08.2008 01:02

<PRE>
and a line
</PRE>

replace "and a line" with it's translation (и строку) :-)

Please help!!!Tell me what is wrong in my algo???Help to orgonise input!Here is my code:
#include <iostream.h>
int main()
{
    const int N=50000;
    int x[N],y[N];
    bool s;
    int i,k;
    int a,b;
    cin>>a;
    for(i=0;i<a;i++)
    {
        cin>>x[i];
    }
    cin>>b;
    for(i=0;i<b;i++)
    {
        cin>>y[i];
    }
    for(i=0;i<a;i++)
    {
        for(k=0;k<b;k++)
        {
            if(x[i]+y[k]==10000)
            {
                s=true;
            }
            if(x[i]+y[i]!=10000)
            {
                s=false;
            }
        }
    }
    if(s=true)
    {
        cout<<"YES"<<endl;
    }
    if(s=false)
    {
        cout<<"NO"<<endl;
    }
    return 0;

}

Help somebody!!
Why my code has WA6..or GIVE SOME TESTS, please))

#include<iostream>
#include<vector>
#include<cmath>
using namespace std;

vector<int>easy_numbers(0);

int next_number(int numb_of_numb)
{
        int step=easy_numbers.size()-1;
        int curr_number=easy_numbers[step];
        while(step+1<numb_of_numb)
        {
                while(1)
                {
                        curr_number++;
                        int sq_root=sqrt(static_cast<double>(curr_number))+1;
                        int buf_step;
                        buf_step=0;
                        while(easy_numbers[buf_step]<sq_root)
                                if (curr_number%easy_numbers[buf_step++]!=0) continue;
                                else goto g1;
                        break;
                        g1:
                }
                easy_numbers.push_back(curr_number);
                step++;
        }
        return curr_number;
}

int main()
{
        int K;
        int L=-1;
        cin >> K;
        if (K==4) {cout << 3; exit(0);}
        easy_numbers.push_back(3);
        int buf=0;
        while((easy_numbers[buf]*easy_numbers[buf])<=K)
        next_number(++buf);
        for (int i=0;i<easy_numbers.size();i++)
        {
                if (K%easy_numbers[i]==0) {L=easy_numbers[i]-1; break;}

        }
        if (L==-1 && K%2==0) {L=K/2-1; cout << L;
        exit(0);}
        if (L==-1) {L=K-1; cout << L;
        exit(0);}
        cout << L;
        return 0;
}

Please add test

aa

I know AC solution that answers to this test 'a'.

We can use Matrix-Tree theoremd(生成树计数) to solve it!
O(N^3)

have a try.
{
while not seekeof
begin

end;
}

Does the order in which you print the word matters?

I cannot pass thru 5th test.

What's the trick on Test #11?..

Help me ...

I have writen the program if many similar ways. I don't know why I got ac with this program:
const u : set of char = ['A'..'Z','a'..'z'];
var
  s   : string;
  i   : integer;
  ch  : char;
begin
  s := '';
  while not (eof) do
    begin
      read(ch);
      if ch in u then s := ch+s
                 else begin
                        write(s);
                        write(ch);
                        s := '';
                      end;
    end;
  if ch in u then write(s);
end.

two rivers are considered to be connected iff they share at least one vertex (no matter inner vertex or end vertex)

example 1:
2
2
-1 0
1 0
2
0 1
0 -1
not connected, because their intersection (0, 0) is not a vertex of river 1 or river 2

example 2:
2
2
-1 0
1 0
3
0 1
0 0
0 -1
not connected, because (0, 0) is not a vertex of River 1

example 3:
2
3
0 1
0 0
0 -1
3
-1 0
0 0
1 0
connected, because (0, 0) appears on both rivers

After 3 test use a/b=>(a/nod(a,b))/(b/nod(a,b))

Hello.

I have an array of numbers from 1 to N. The order is random. I have to answer a lot(M) of questions like: how many numbers from interval [x, y] bigger than k. I tried to build two-dimensional interval tree: each edge of interval tree has an internal tree. But that tree was too large.

1 <= N, M <= 100000
Memory limit = 64 Mb

Also I tried to build Fenwick tree. But it also needs too much memory: O(N^2).

Please, can you help me to compress an interval tree or Fenwick tree? Or, maybe, there is another way to solve this problem?

Thank you.

I still have WA on test 18 after many efforts.
Does anyone know some tricks about this test?
My program gives right answers on all my tests.
Can anyone give me more tests?

Shouldn't the answer to the sample be 0.3660?
I've solved this problem assuming, that pyramid lies on triangular edge. It seems to me that it is wrong suggestion, nevertheless it got AC. It seems to me, that pyramid should be firstly rotated to lay on one of the triangular face - then the answer is 0.3660. Am I wrong?

Edited by author 09.11.2009 02:17

from the statements of the problem ,I don't know how to end my input .Is there anyone can help me?
Thank you very much

N=5 Q=2
N=6 Q=3
N=7 Q=4
N=8 Q=5
N=11 Q=11
N=17 Q=36

Right?

var
  n:byte;
procedure a(now,step:byte);
  begin
    if now=step then
      write('sin(',now,')')
    else
      begin
        write('sin(',now);
        if now mod 2=1 then write('-') else write('+');
        a(now+1,step);
        write(')');
      end;
  end;
procedure s(step:byte);
  begin
    if step=1 then
      begin
        write('(');
        a(1,step);
        write('+',n-step+1,')');
      end
    else
      begin
        if step<>n then
          write('(');
        s(step-1);
        a(1,step);
        write('+',n-step+1);
        if step<>n then
          write(')');
      end;
  end;
begin
  readln(n);
  s(n);
  writeln;
end.

Somebody can respond why incorrectly? On turbo pascal all tests passes.


var a:array[1..1000] of integer;
    b:array[1..1000] of integer;
    i,j,n,m,r: integer; s1,s2,s: string;
begin
read(n);
for i:=1 to n do readln(a[i]);
for i:=1 to n-1 do if a[i]>a[i+1] then begin
r:=a[i]; a[i]:=a[i+1]; a[i+1]:=r; end;
read(m);
for j:=1 to m do readln(b[j]);
for j:=1 to m-1 do if a[j]<a[j+1] then begin
r:=a[j+1]; a[j+1]:=a[j]; a[j]:=r; end;
for i:=1 to n do for j:=1 to m do begin
if ((a[i]+b[j])=10000)and((a[i]+b[j])<>10000) then s:='yes' else
if (a[i]+b[j])=10000 then s1:='yes'
else if (a[i]+b[j])<>10000  then s2:='no'; end; if s='yes' then writeln(s)
else if s1='yes' then writeln(s1) else if s2='no' then writeln(s2); end.

Edited by author 08.11.2009 20:44