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.

Code in C#!!!
first of ALL just print and reverse print .... 3 times

     static string [,] reverse(string[,] arr)
        {
            string[,] tmp = new string [4,4];

            for (int i = 0; i < 4; i++)
            {
                for (int j = 0; j < 4 ; j++)
                {
                    tmp[i,j] = arr[4-j-1,i];
                }
            }
            return tmp;
        }
in this case i wrote code for 4x4 array!!!
you can change your code to tmp[i,j] = arr[3-j,i]
if u want reverse nxn array change all 4 to n
GooD Luck!
sorry for my english!!

Edited by author 17.10.2012 02:39

http://pastebin.com/rSjasMnd
Вот код, вполне рабочий, только не принимает тимус...

Look how to use STL algorithm rotate(). It's very easy: rotate(arr, arr + k, arr + N);

ok...
if the player 2 wins, then stop showing the second line?
pls give help

teach=Int32.Parse(Console.ReadLine());
String[] arrTeach = new String[teach]
for (Int32 i = 0; i < teach; i++)
            arrTeach[i] = Console.ReadLine();

//It no worked correct on timus compiler! Why?

Can anybody help me with WA 4? My program passes all tests given in discuss...

#include<stdio.h>
int main()
{
    int N;

    scanf("%d", &N);

    int start;
    scanf("%d", &start);

    int sum = 1;

    for(int i = 1; i < N; i++)
    {
        int a;
        scanf("%d", &a);
        if(a == start) { sum++; }
        else
        {
            printf("%d %d ", sum, start);
            start = a;
            sum = 1;
        }
    }

    printf("%d %d ", sum, start);

    return 0;
}

#include<stdio.h>

main()
{
int a, sum=0;
scanf("%d",&a);
if(a<0&&a>=-10000)
{
  int i;
  for(i=a;i<=1;i++)
  {
    sum=sum +i;

}
printf("%d",sum);
}

else if( a>=0&&a<=10000)
{
     int i;
     for(i=1;i<=a;i++)
     {
      sum=sum +i;
      }
     printf("%d",sum);

}

}