Uva10795 - A Different Task

The (Three peg) Tower of Hanoi problem is a popular one in computer science. Brie
y the problem is
to transfer all the disks from peg-A to peg-C using peg-B as intermediate one in such a way that at
no stage a larger disk is above a smaller disk.
Normally, we want the minimum number of moves required for this task. The problem is used as an
ideal example for learning recursion. It is so well studied that one can nd the sequence of moves for
smaller number of disks such as 3 or 4. A trivial computer program can nd the case of large number
of disks also.
Here we have made your task little bit difficult by making the problem more
exible. Here the disks
can be in any peg initially.
If more than one disk is in a certain peg, then they will be in a valid arrangement (larger disk will
not be on smaller ones). We will give you two such arrangements of disks. You will have to nd out the
minimum number of moves, which will transform the rst arrangement into the second one. Of course
you always have to maintain the constraint that smaller disks must be upon the larger ones.
Input
The input le contains at most 100 test cases. Each test case starts with a positive integer N (1
N 60), which means the number of disks. You will be given the arrangements in next two lines. Each
arrangement will be represented by N integers, which are 1, 2 or 3. If the i-th (1 i N) integer is
1, you should consider that i-th disk is on Peg-A. Input is terminated by N = 0. This case should not
be processed.
Output
Output of each test case should consist of a line starting with `Case #: ' where # is the test case
number. It should be followed by the minimum number of moves as speci ed in the problem statement.
Sample Input
3
1 1 1
2 2 2
3
1 2 3
3 2 1
4
1 1 1 1
1 1 1 1
0
Sample Output
Case 1: 7
Case 2: 3
Case 3: 0

#include <iostream>
#include <stdio.h>
using namespace std;
const int MAXN = 65;

long long f(int p[],int i,int final)
{
    if(i==0)return 0;
    if(p[i]==final) return f(p,i-1,final);
    return f(p,i-1,6-final-p[i]) + ((long long)1<<(i-1));
}
int main()
{
    int n, cs=1, i;
    int start[MAXN],final[MAXN];
     
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)break;
        for(i=1; i<=n; i++) scanf("%d",&start[i]);
        for(i=1; i<=n; i++) scanf("%d",&final[i]);
        int k = n;
        while(k>0&&final[k]==start[k])k--;
        long long ans = 0;
        if(k>0)
        {
            ans = f(start,k-1,6-final[k]-start[k]) + f(final,k-1,6-final[k]-start[k]) + 1;
        }
        printf("Case %d: %lld\n",cs++,ans);
    }
}