A rabbit walk is defined by a sequence of jumps which starts from a
location and returns back to the starting location. We can repesent such
a Rabbit walk by an array A of n numbers, where A[i] denotes the next
array index to which the rabbit should jumps from the location i.
For example, the sequence: 2 3 4 5 0 1 indicates that the Rabbit follows
the jump sequence: 0->2, 2->4, 4->0 and stops.

Your also given another array B of N numbers, such that
B[i] denotes the number of carrots available at location i.
Your task is to compute the number of carrots that a rabbit can collect
when it makes rabbit walk(s) from various start position(s).
A sample input is provided for more clarification(s).

Input:
The first line will contain a single integer N which is the size of the
array (2<= N <= 10000).
The next line contains a series of N integers which define the array A.
The next line contains a series of N integers which define the array B.
You can be assured that all the calculations fit into a 32 bit signed
integer type(int in C).

The input is given such that, a rabbit walk will always exist with
start position as 0.

Output:
It should contain the number of carrots collected
when 0 is treated as the start location.

Sample Input:
6
2 2 2 2 -4 -4
2 3 6 3 0 1

Sample Output:
8

• #include
  #include
  #include
  int rabbitWalk(int * arr,int n,int *Barr);
  int main()
  {
  int A[] = {2,3,5,4,0,1},n=6,B[]= {2,2,3,2,1,3};
  int numCarrots = rabbitWalk(A,n,B);
  printf("nNumber of carrots : %d n",numCarrots);
  return 0;
  }
  int rabbitWalk(int * arr,int n,int *Barr)
  {
  int i=0,count = 0 ;
  while(1)
  {
  //printf("%c -->",charArray[i]);
  i = arr[i];
  count = count + Barr[i];
  if(i == 0)
  {
  count = count + Barr[i]; // Counting the carrots at 0 position again . This can bre removed if required.
  //printf("%c",charArray[i]);
  break;
  }
  }
  return count;
  }
• 9 years agoHelpfull: Yes(2) No(0)
• void main()
  {
  int a[6]={2,3,5,4,0,1},n=6,b[] = {2,2,3,2,1,3},count=0,i=0;
  while(1){
  count += b[a[i]];
  i=a[i];
  if(i==0)break;
  }
  printf("%d",count);
  }
  
• 9 years agoHelpfull: Yes(0) No(0)