What could it be?

What's the test number 2 ?

3
3 3 1
out:3
3
1 1 1
out:0
3
1 2 3
out:-1
3
2 2 2
out:7

import java.util.Scanner;
public class Hanoi_Tower  {
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int d[] = new int [n+1];
        for (int i = 1; i <= n; i++) {
            d[i] = sc.nextInt();
        }
        int x =n%2;
        int a[][] ={{3, 1, 2, 3}, {2, 1, 3, 2}};
        double ans = 0;
        for (int i = n; i > 0; i--) {
            int y = (i+x)%2;
            double rep = (ans/Math.pow(2, i) + 1)%3;
                if (d[i] != a[y][(int)rep]){
                    if (d[i] != a[y][(int)(rep+1)]){
                        System.out.println("-1");
                        return;
                    }
                    ans += Math.pow(2, (i-1));
                }
        }
        System.out.printf("%.0f",ans);
    }
}

3
1 1 1

Ans=0

Tried all test cases i know but it gives wrong answer for test#1.
Can anyone please tell me what case is failing in my code?

#include<stdio.h>
int count=0,ni;
int seq[31][31];
int a=-1,b=-1,c=-1,move=-1,movea=-1,moveb=-1,movec=-1,flag=0,go=0;
void tower(int n,char beg,char aux,char end)
{
     if(n==1)
     {
            int begin,ends,k,m;
            begin=(int)(beg-48)-1;
            ends=(int)(end-48)-1;
            count++;
            for(k=ni-1;k>0;k--)
                seq[ends][k]=seq[ends][k-1];
            seq[ends][0]=seq[begin][0];
            for(k=0;k<ni-1;k++)
                        seq[begin][k]=seq[begin][k+1];
            seq[begin][ni-1]=0;
            /*for(k=0;k<3;k++)
            {
                            for(m=0;m<ni;m++)
                                             printf("%d",seq[k][m]);
                            printf("\n");
                            }
                            printf("\n");
            */for(k=0;k<ni;k++)
            {
                    if(seq[a][k]==1)
                    {
                          movea=1;
                          break;
                    }
            }
            for(k=0;k<ni;k++)
            {
                    if(seq[b][k]==2)
                    {
                          moveb=1;
                          break;
                    }
            }
            for(k=0;k<ni;k++)
            {
                    if(seq[c][k]==3)
                    {
                          movec=1;
                          break;
                    }
            }
            if(movea==1&&moveb==1&&movec==1&&flag==0)
            {
                          flag=1;
                          move=count;
            }
            movea=-1;
            moveb=-1;
            movec=-1;
            return;
     }
     else
     {
            tower(n-1,beg,end,aux);
            tower(1,beg,aux,end);
            tower(n-1,aux,beg,end);
     }
}
void assign()
{
     int k,j;
     for(k=1;k<3;k++)
     {
             for(j=0;j<ni;j++)
                     seq[k][j]=0;
     }
             for(j=0;j<ni;j++)
                     seq[0][j]=(j+1);
}
int main()
{
    scanf("%d",&ni);
    assign();
    scanf("%d %d %d",&a,&b,&c);
    if(a==1)
            a--;
    else
    if(a==3)
            a=1;
    if(b==1)
            b--;
    else
    if(b==3)
            b=1;
    if(c==1)
            c--;
    else
    if(c==3)
            c=1;
    tower(ni,'1','2','3');
    printf("%d",move);
    return 0;
}

Hints:
-  Use 3variable which means first,too and last and
-  do for n downto 1 ...
-  use "1 shr i" to get 2^i..
-  eachtime just swap 2 of (from,temp or too)...

and...
  for i:=n downto 1 do begin
    if p[i]=temp then
      begin writeln(-1); exit; end
    else
      if p[i]=from then
        swap(temp,too)
      else begin
        inc(ans,(1 shl (i-1)));
        swap(from,temp);
      end;
  end;

For more information:
  aidin_n7@hotmail.com

c++ code as below
===================

#include <cstdio>
#include <iostream>

#define MAX_NUM 31

using namespace std;

int g_Dseq[MAX_NUM] = {1};
int g_Dtargseq[MAX_NUM];
int g_count = 0;
int N = 1;

int g_done = 0;

void movedisk(int n, int dsrc, int ddesc);
void hanoi_proc(int n,int dsrc, int daid, int ddesc);
int isdone();

void main()
{

int i=0;

cin >> N;

if(N <1 || N > 31)
{
    cout<<-1;

}else
{


    for(i=0;i<N;++i)
    {
    cin>> g_Dtargseq[i];
    g_Dseq[i] = 1;

    }



    hanoi_proc(N, 1, 3, 2);


    if(g_done ==0)
    cout<<-1;

}


}


int isdone()
{
    for(int i=0;i<N;++i)
    {
        if(g_Dseq[i] != g_Dtargseq[i]) return 0;
    }

    return 1;
}

void movedisk(int n, int dsrc, int ddesc)
{

    g_Dseq[n-1] = ddesc;
    ++g_count;



    if(g_done ==0 && isdone() == 1)
    {
        cout<<g_count;
        g_done = 1;
    }
}



void hanoi_proc(int n,int dsrc, int daid, int ddesc)
{

    if(n == 1)
    {
        movedisk(n, dsrc, ddesc);

    }else
    {
        hanoi_proc(n-1, dsrc, ddesc, daid);

        movedisk(n, dsrc, ddesc);

        hanoi_proc(n-1, daid, dsrc, ddesc);
    }
}


=============================

This Program Got AC within 0.01 Sec!
----------------------------------------------------------------------
program answer1054;
const
     st:array[0..1,0..2]of 1..3=((1,2,3),(1,3,2));
var
   d:array[1..31]of 1..3;
   e:array[1..31]of 0..1;
   n,i,s,f,delta,j,s1:longint;
function mi2(x:integer):longint;{mi2=2^x}
         var
            s:longint;
            i:integer;
         begin
              s:=1;
              for i:=1 to x do s:=s*2;
              mi2:=s;
         end;
function c(x:integer):longint;{N-dished Hanoi tower step uses}
         begin
              if x=31 then c:=maxlongint else c:=mi2(x)-1;
         end;
begin
     readln(n);
     for i:=1 to n do read(d[i]);
     s:=0;f:=c(n);
     fillchar(e,sizeof(e),0);
     for i:=n downto 1 do begin
         s1:=1;delta:=0;
         for j:=i+1 to n do begin
             delta:=delta+e[j]*s1;
             s1:=s1*2;
         end;
         if d[i]=st[(n+i) mod 2,delta mod 3] then begin
         f:=f-mi2(i-1);
         end else
            if d[i]=st[(n+i) mod 2,(delta+1) mod 3] then
               begin
                    s:=s+mi2(i-1);
                    e[i]:=1;
               end else begin
                   writeln(-1);
                   halt;
               end;
     end;
     writeln(s);
end.

Can anyone tell me what are the input data for this test?
The program seems to work OK here are some test results:

3
3 3 1          3

3
2 2 2          7

3
1 2 2          6

2
2 1            -1

//I have delete the code.
if you want it ,i will send it.

Edited by author 13.02.2005 07:00

What can I make with this formula to receive is Accepted?
This formula calculates quantity of rearrangements of disks:
  T (n) =2^n-1
Help me, to deduce the formula which calculates quantity of steps.

Help!!! Many thanks, beforehand!

program hanoi;
var
step:longint;
i,n:integer;
goal:array[1..31] of byte;
procedure p(k:integer;a,b,c:integer);
begin
if k=0 then exit;
if goal[k]=b then
 begin
  step:=step+trunc(exp((k-1)*ln(2)));
  p(k-1,c,b,a);
 end
else if goal[k]=c then begin writeln(-1); halt; end
else p(k-1,a,c,b);
end;
begin
readln(n);
for i:=1 to n do read(goal[i]);
step:=0;
p(n,1,2,3);
writeln(step);
end.

/*
    Prog: acm_1054
    results:
*/

#include <stdio.h>
#include <math.h>
int a = 1,b = 2,c = 4;
int disks[5];
int numbers[5][101];
int dest[5][101];
int destdisks[5];
int count;
int found;
int num;

int finish()
{
    int i,j;
    found = 1;

    for(i = 1;i<5;i++)
    {
        for(j = 0;j<=num;j++)
        {
            if(numbers[i][j]!=dest[i][j])
            {
                found = 0;
                break;
            }
        }
        if(!found)
            break;
    }

    if(found)return 1;
    if(disks[b]==num)
        return 1;
    return 0;
}
void move(int s,int n, int d)
{
    int k ;
    if(found||finish())return;
    if(n==1)
    {
        disks[d]++;
        numbers[d][disks[d]] = numbers[s][disks[s]];
        numbers[s][disks[s]] = 0;
        disks[s]--;

        count++;

    }else {
          k = 7^(s|d);

        move(s,n-1,k);

        move(s,1,d);
        move(k,n-1,d);
        finish();
    }
}

int main(int argc, char *argv[]) {
    int pos[100],i,j;
    int tmp[101];
    scanf("%d",&num);
    disks[a] = num;
    for(i = 0;i<num;i++)
        numbers[a][i+1] = num-i;

    for(i = 1;i<=num;i++)
    {
        scanf("%d",&pos[i]);
        pos[i]  =(int)pow(2,pos[i]-1);
        destdisks[pos[i]]++;
        dest[pos[i]][destdisks[pos[i]]] = i;
    }
    for(i = 1;i<5;i++)
    {
        for(j = 1;j<=destdisks[i];j++)
            tmp[destdisks[i]-j+1] = dest[i][j];
        for(j = 1;j<=destdisks[i];j++)
            dest[i][j] = tmp[j];
    }



    move(a,num,b);
    if(found)
        printf("%d",count);
    else printf("%d",-1);


    return 0;
}
// Cheers!

I've written program that finds the number of steps but Can't
determine whether it's possible or not.

var
   N:integer;
   s,d:array[1..31] of integer;
   i,j,t:integer;
   K:longint;
   n2:array[0..30] of longint;

begin

   readln(N);
   for I:=1 to n do read(s[i]);
   n2[0]:=1;
   for I:=1 to 30 do n2[i]:=n2[i-1]*2;
   for I:=1 to n do d[i]:=1;

   k:=0;
   for I:=n downto 1 do
   begin
      IF D[I]<>S[I] then
      begin
          K:=K+N2[i-1];
         for j:=1 to 3 do if (s[i]<>j) and (d[i]<>j) then
         begin
            t:=j;
            break;
         end;
         for J:=1 to i-1 do if d[j]=d[i] then
         begin
            d[j]:=t;
         end;
         d[i]:=s[i];
         end;
   end;
    writeln(k);

end.

Can anyone of those who have solved the problem tell me
how ? If you use recursion 31 is a extremely enormous limit
and it works years. So a formula? How?