Sorry line number 7 is a[n]=fibo(n-1)+fibo(n-2);
On Mon, Jun 6, 2011 at 6:25 PM, saurabh singh <saurab...@gmail.com> wrote:
> long a[10000000]={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 <aakashj....@gmail.com>wrote:
>
>> 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 <kumarg...@gmail.com> 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 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 efficent algorithm.
>>>
>>>
>>>
>>> --
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>>
>>
>>
>> --
>> -Aakash Johari
>> (IIIT Allahabad)
>>
>>
>>
>>
>>
>> --
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>
>
>
> --
> Saurabh Singh
> B.Tech (Computer Science)
> MNNIT ALLAHABAD
>
>
>
--
Saurabh Singh
B.Tech (Computer Science)
MNNIT ALLAHABAD
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