i guess this would work...
n=number of nodes
h=height;
pour=quantity poured;
capacity = capacity of each cup
n=pow(2,h+1) -1;
call(capacity,pour,0,n)
node* fillCup(float capacity,float pour,int left,int right)
{
node *root;
int mid;
if(left > right)
return NULL;
root=(node *)malloc(sizeof(node));
if(left==right)
{
if(pour >=capacity)
root->data=capacity;
else
root->data=pour;
root->left=root->right=NULL;
}
else
{
mid=left+(right-left)/2;
if(pour >= capacity)
{
root->data=capacity;
pour=pour-capacity;
pour=pour/2;
}
else
{
root->data=pour;
root->left=root->right=NULL;
return root;
}
root->left=fillCup(capacity,pour,left,mid-1);
root->right=fillCup(capacity,pour,mid+1,right);
}
return root;
}
On Sat, Feb 25, 2012 at 5:05 PM, Ravi Ranjan <ravi.cool2...@gmail.com>wrote:
> |_|
> |_| |_|
> |_| |_| |_|
> |_| |_| |_| |_|
> |_| |_| |_| |_| |_|
>
> Each cup has capacity C and once a cup gets full, it drops half extra
> amount to left child and half extra amount to right child
>
> for Eg : let' first cups get 2C amount of liquid then extra amount C(2C-C)
> will be divided equally to left and right child cup of next level
>
> i.e. C/2 to left child and C/2 to right child
>
> Write a function which takes input parameter as amount of liquid poured at
> top (L) and height of particular cup (h) index of that cup (i) and it
> should return amount of liquid absorbed in that cup.
>
> source
>
> http://www.careercup.com/question?id=12770661
>
>
> whats exactly the qestion???
>
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