( window | A    | aisle |B     C| aisle |    D | window )-- Premium   (1-2)
( window | A   B| aisle |C     D| aisle |E   F | window )-- Business  (3-20)
( window | A B C| aisle |D E F G| aisle |H G K | window )-- Economy   (21-64)




Edited by author 30.05.2022 16:38

VAR
s:string;
k,n,l,c:Longint;
begin
read(s);      l:=length(s);
val(s,n,c);
if (n<3) and (n>0) then
begin
k:=0;
    case s[l] of
    'a','A': write('window');
    'D','d': write('window');
    'b','B': write('aisle');
    'c','C': write('aisle');
    end;
end;
if (n>=3) and (n<=20) then
begin
k:=0;
    case s[l] of
    'a','A': write('window');
    'f','F': write('window');
    'b','B': write('aisle');
    'd','D': write('aisle');
    'E','e': write('aisle');
    'c','C': write('aisle');
    end;
end;
if (n>=21) and (n<=65) then
begin
k:=0;
    case s[l] of
    'a','A': write('window');
    'k','K': write('window');
    'g','G': write('aisle');
    'd','D': write('aisle');
    'h','H': write('aisle');
    'c','C': write('aisle');
    'b','B': write('neither');
    'e','E': write('neither');
    'f','F': write('neither');
    'J','j': write('neither');
    end;
end;
if n>65 then  write('neither');
end.

Edited by author 20.11.2012 23:14

As i read the problem, the configuration of two first rows are :

Window | A | Aisle | B | C | Aisle | D | Window

I check with my program and for 1A 'window' is AC while 'aisle' gives WA6


Edited by author 30.10.2011 20:54

Edited by author 30.10.2011 20:54

del

Edited by author 11.03.2018 11:13

Edited by author 11.03.2018 11:13

output limit exceeded!
what's the case 4 like?
3Q

whats wrong?

program fly;
const okn='window';
      ser='neither';
      proh='aisle';
var i,r,n,bb:integer;
    s,s1:string;
    m:char;
begin
readln(s);
n:=length(s);
m:=s[n];
s1:=s;
delete(s1,n,1);
val(s1,r,bb);
if r<3 then
 if (m='A') or (m='D') then writeln(okn) else writeln(proh) else
if r<21 then
 if (m='A') or (m='F') then writeln(okn) else writeln(proh) else
if r<65 then
if (m='A') or (m='K') then writeln(okn) else
if (m='C') or (m='D') or (m='G') or (m='H') then writeln(proh) else writeln(ser);
end.

window | ******| window
  window?|****|window?
The distance between window and last chairs is 1 chair but it's there!
Windows can be only in Business and Econom cause these parts are in different decks.

can someone tell me what is test #5???

import java.util.Scanner;


public class A380 {
    public static void main(String agrs[]){
        Scanner sys=new Scanner(System.in);
        String input=sys.next();
        String[] part=input.split("(?<=\\d)(?=\\D)");
        int num=Integer.parseInt(part[0]);
        if(num==1 || num==2){
            if(part[1].equals("A")||part[1].equals("D"))
                System.out.println("window");
            else if(part[1].equals("B")||part[1].equals("C"))
                System.out.println("aisle");

        }
        else if(num>=3 && num<=20){
            if(part[1].equals("A")||part[1].equals("F"))
                System.out.println("window");
            else if(part[1].equals("B")||part[1].equals("C") || part[1].equals("D")||part[1].equals("E"))
                System.out.println("aisle");

        }
        else
            if(num>=21 && num<=65){
            if(part[1].equals("A")||part[1].equals("K"))
                System.out.println("window");
            else if(part[1].equals("C")||part[1].equals("D") || part[1].equals("H")||part[1].equals("G"))
                System.out.println("aisle");
            else if(part[1].equals("B")||part[1].equals("E") || part[1].equals("F")||part[1].equals("J"))
                System.out.println("niether");
        }

    }

}

What is 62th testcase?

del

Edited by author 06.01.2015 21:37

Try to "draw" the organization of seats, it helps a lot in writing algorithm solution.

Brute force this question! There's really no reason to do anything else.

Edited by author 03.03.2013 02:41

Edited by author 03.03.2013 02:41

#include <iostream>
using namespace std;
    int r;
    char n;
int main()
{
    cin >> r;
    cin >> n;

    if(r >= 1 && r <= 2    )
    {
        if(n == 'A' || n == 'D' )
            cout << "window";
        else if(n == 'B' || n == 'C')
            cout << "aisle";
    }
    else if( r >= 3 && r <= 20 )
    {
        if(n=='A' || n=='F' )
            cout << "window";
        else if(n == 'B' || n == 'C' || n=='D' || n=='E' )
            cout <<"aisle";
    }

    else if(r >= 21 && r <= 65)
    {
        if (n=='A' || n=='K')
            cout<<"window";
        else if(n=='B' || n=='E' || n=='F' || n=='J')
            cout <<"neither";
        else if(n=='C' || n=='D' || n=='G' || n=='H')
        cout << "aisle";
    }


    return 0;
}

#include<iostream>
using namespace std;
int n;
char c;
int main()
{
cin>>n>>c;
if (n>=1 && n<=2)
{ if(c=='A')
cout<<"window";
if(c=='D')
cout<<"window";
if(c=='B')
cout<<"aisle";
if(c=='C')
cout<<"aisle";
}
if(n>=3 && n<=20)
{ if(c=='A')
cout<<"window";
if(c=='F')
cout<<"window";
if(c=='B')
cout<<"aisle";
if(c=='E')
cout<<"aisle";
if(c=='C')
cout<<"aisle";
if(c=='D')
cout<<"aisle";
}
if(n>=21 && n<=65)
{ if(c=='A')
cout<<"window";
if(c=='K')
cout<<"window";
if(c=='B')
cout<<"neither";
if(c=='C')
cout<<"aisle";
if(c=='D')
cout<<"aisle";
if(c=='E')
cout<<"neither";
if(c=='F')
cout<<"neither";
if(c=='G')
cout<<"aisle";
if(c=='H')
cout<<"aisle";
if(c=='J')
cout<<"neither";
}
return 0;
}
si asta a lui rad

Edited by author 25.10.2012 13:44

What is different with case #15.
My program runs good for 14 tests.