is this because I am using java ?? Java is slower compared to c or c++. This program is normal stack operation isn't it? or some other algo is to be followed??

can any one please describe the input-output given?

Может ли между двумя соседними по стороне комнатами не быть пустоты (пробел)? то есть верно ли, что каждая комната всегда окружена какими-то стенами и углами?

What's the condition they kill themselves?

the total power > their resistance upper bound
or
the total power >= their resistance upper bound
?

Thanks
Ade

Test 4: crash (integer division by zero)
int main()
{
    int n;
    scanf("%d",&n);
    if (n>1000000 || n<1)
        return 1/(n-n);
    return 0;
}

Any suggestions about test 8 data?

It would be great, if somebody can provide corner case test.
Thanks.

Edited by author 14.10.2012 16:33

Edited by author 14.10.2012 16:33

It's not in the FAQ.
Is it completely arbitrary, and assigned by the problem's authors?
Is it updated as new solutions are submitted?

I have got problem with this task from my university algorithm course (translated from german):

> John have got a party. He prepared $x$ glasses of vodka and $k*x$
> glasses of rum.
>
> Number of glasses is n. Each guests drink one
> round of alcohol (one round is $k$ glasses of rum and $one$ of vodka) but
> guest must chose glass that every full glass must have got one full
> neighbor.
>
> Write the program which simulate John's party. The party ends when all
> glasses are empty.
>
> In input we have got $n$ and $k$ which means number of glasses and number
> of glasses of rum drunk in single turn).

Simple Input :  (assume that r is rum and v - vodka ).

> 16 3 vvrrrvrvrrrrrrrr

Simple Output : (we numbered glass 1...n)

> 1 14 15 16
> 2 11 12 13
> 3 4 5 6
> 7 8 9 10

Explain :

Guest 1 drink 1, 14, 15 and 16 glass.
Guest 2 drink 2, 11, 12, 13 glass.
etc.

We write it in ascending order. Notice that we have got $\frac{n}{(k+1)}$ guests, $n$ glasses, $k$ glasses of rum in every round.  And in all inputs $k+1$ divides $n$. And $n$ can be very big - about 2 000 000. And we have about 1-2s to answer the question.

Find the fastest as possible algorithm to solve this problem.

Thanks for every help. If you have got any questions, i will answer it.
Edger.

What does the program output for n=0?

Hi All,

1. 1st player Take 2 stones   left over is 8-2=6
2. 2nd player  takes 1 stone  left over  is 6-1=5
3. 1st player takes either 1 (left over is 4)0r 4 stones (left over is 1)
4. Finally 2nd player takes remiang stones.  Thus second player wins.

Did I understand the problem statements properly? According to the output given in sample 1st player has won but how? Please could you explain.

Regards
Anupam

4 3 1 1 3 1 4 or 1 3 4 1 3 1 4
sum of all is 17
and first player wins not second,
can somebody explan?


Edited by author 18.10.2012 17:08

Edited by author 18.10.2012 17:11

Program nitochka;
var n,r,i:integer;
x: array [1..100] of real;
y: array [1..100] of real;
s,sg:real;
begin
s:=0;
readln(n,r);
For i:=1 to n do
  readln(x[i],y[i]);
sg:=2*pi*r;
For i:=1 to n-1 do  s:=s+sqrt(sqr(x[i]-x[i+1])+sqr(y[i]-y[i+1]));
s:=s+sqrt(sqr(x[n]-x[1])+sqr(y[n]-y[1]));
s:=s+sg;
writeln(s:1:2);
end.

The example calculates correct

can you write hear some examples?

96 96 2 2

ans: 1

There are only one trick in this problem, it's the position of the first and last digit, the first position can be smaller 0 and the last position can be larger than the length, so you can consider the following segment of code:

if(first<0 || last > 0)
{
//stop the code and print output
}

#include <iostream.h>
int main()
{
    int N,M,i,t;
    cin>>N>>M;
    int *mas=new int [N];
    int *mas2=new int [N];
    for(i=1;i<=N;i++)
        cin>>mas[i];
    for(i=1;i<=N;i++)
    {
        t=i+M;
        if(t>N)
        {
            t-=N;
        }
        mas2[t]=mas[i];
    }
    if(N>10)
    {
        for(i=3;i<=N;i++)
        cout<<mas2[i];
    }
    else
    {
        for(i=1;i<=N;i++)
        cout<<mas2[i];
    }
    return 0;
}

1
ab

1
abcd

2
a
b

2
ab
bc

4
aa
ab
ba
bb

6
a
ba
bb
ca
cb
cc


Edited by author 15.10.2012 02:26

Edited by author 15.10.2012 02:40

New tests were added to the problem. All possible inputs are checked now :)
Problem statement was updated: limitation "0 < K <= 50" was replaced by "2 <= K <= 50". There wasn't a case K=1 in the tests.
AC submissions were rejudged, 44 of them lost AC.