Fibonacci series - c++

Question

I am trying to do an exercise with the Fibonacci series.
I have to implement with a recursive function, a succession of the prime n number of Fibonacci and print them
in the same function. The problem is that my function print also the intermediate number.
The results, for example, for n = 6, should be : 1 1 2 3 5 8.
Any solutions?
Thanks
#include<iostream>
using namespace std;
int rec(int n)
{
int a, b;
if (n == 0 || n == 1)
{
return n;
}
else
{
a = rec(n - 1);
b = rec(n - 2);
cout << a + b << endl;
return a + b;
}
}
int main()
{
int n = 6;
rec(n);
return 0;
}

Answer 1

I have taken help of static int. That worked the way you wanted.
void rec(int n)
{
static int a=0,b=1,sum;
if(n>0)
{
sum = a+b;
a=b;
b= sum;
cout<<sum<<" ";
rec(n-1);
}
}
Though you have to print the first Fibonacci number yourself in main().
cout<<"0 ";
rec(n);

Answer 2

You can use this:
#include<iostream>
using namespace std;
#define MAXN 100
int visited[MAXN];
int rec(int n)
{
if(visited[n])
{
return visited[n];
}
int a, b;
if (n == 0|| n==1)
{
return n;
}
else
{
a = rec(n - 1);
b = rec(n - 2);
cout << " " <<a + b;
return visited[n] = a + b;
}
}
int main()
{
int n = 6;
cout<< "1";
rec(n);
cout<<endl;
return 0;
}
This implementation uses dynamic programming. So it reduces the computation time :)

Answer 3

Because you are printing in rec, its printing multiple times because of recursion. No need to print in the recursive function. Instead print the result in main
#include<iostream>
using namespace std;
int rec(int n)
{
int a, b;
if (n == 0 || n == 1)
{
return n;
}
else
{
a = rec(n - 1);
b = rec(n - 2);
//cout << a + b << endl;
return a + b;
}
}
int main()
{
int n = 6;
for (int i = 1; i <= n; i++)
{
cout << rec(i) << endl;
}
system("pause");
return 0;
}

Answer 4

I'm pretty sure you have gotten working solutions but I have a slightly different approach that doesn't require you to use data structures:
/* Author: Eric Gitangu
Date: 07/29/2015
This program spits out the fibionacci sequence for the range of 32-bit numbers
Assumption: all values are +ve ; thus unsigned int works here
*/
#include <iostream>
#include <math.h>
#define N pow(2.0,31.0)
using namespace std;
void fibionacci(unsigned int &fib, unsigned int &prevfib){
unsigned int temp = prevfib;
prevfib = fib;
fib += temp;
}
void main(){
int count = 0;
unsigned int fib = 0u, prev = 1u;
while(fib < N){
if( fib ==0 ){
fib = 0;
cout<<" "<< fib++ <<" \n ";
continue;
}
if( fib == 1 && count++ < 2 ){
fib = 1;
cout<< fib <<" \n ";
continue;
}
fibionacci(fib, prev);
cout<< fib <<" \n ";
}
}

Answer 5

Try this recursive function.
int fib(int n)
return n<=2 ? n : fib(n-1) + fib(n-2);
It is the most elegant solution I know.