Sorry line number 7 is a[n]=fibo(n-1)+fibo(n-2);
On Mon, Jun 6, 2011 at 6:25 PM, saurabh singh wrote:
> long a[1000]={0,1};
> long fibo(long n)
> {
> if(a[n]) return a[n];
> else
> {
> a[n]=fibo(n-1)+sum(n);
> return a[n];
>
> }
> }
>
> On Mon,
long a[1000]={0,1};
long fibo(long n)
{
if(a[n]) return a[n];
else
{
a[n]=fibo(n-1)+sum(n);
return a[n];
}
}
On Mon, Jun 6, 2011 at 6:19 PM, Aakash Johari wrote:
> Memoize your solution for nth fibonacci and use this memoized value in
> further
Memoize your solution for nth fibonacci and use this memoized value in
further computations.
On Mon, Jun 6, 2011 at 5:42 AM, kumar vr wrote:
> The Fibonacci series Recursion using
> F(n) = F(n-1) + F(n-2)
> Will of exponential complexity.
> This occurs because each of the Term is calculated twic
The Fibonacci series Recursion using
F(n) = F(n-1) + F(n-2)
Will of exponential complexity.
This occurs because each of the Term is calculated twice
eg
F5= F4+F3
F4= F3+F2.
So F3 calculation is done twice.
Can someone come up with an algorithm to minimize these computation and come
up with effice
