Printing 1 to n using Fibonacci recursion [closed] - c++

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I want to print Fibonacci series from 1 to n in my function.
I know that I can do it by writing a regular Fibonacci and using it in a for block to print 1 to N. Like this:
#include <iostream>
using namespace std;
int fibo(int);
int main(){
for (int i = 0; i < 5; i++)
cout << fibo(5);
system("pause");
return 0;
}
int fibo(int n){
if (n == 1 || n == 2)
return 1;
else
return fibo(n - 1) + fibo(n - 2);
}
but my problme is that I can't do it without for,IN my function
I mean I want to Print it with a recursive algorithm
Here is my code up to now
#include <iostream>
using namespace std;
int fibo(int, bool);
int main(){
fibo(5, false);
system("pause");
return 0;
}
int fibo(int n, bool IsPrinted){
if (n == 1 || n == 2){
if (!IsPrinted)
cout << 1 << endl;
return 1;
}
else{
int temp = fibo(n - 1, IsPrinted) + fibo(n - 2, IsPrinted);
if (!IsPrinted){
cout << temp << endl;
IsPrinted = true;
}
return temp;
}
}

Answer 1

long fibo(int N, bool print) {
long value = 0;
if(1 == N)
value = 1;
if(1 < N)
value = fibo(N-1, print) + fibo(N-2, false);
if(print)
std::cout << N << " => " << value << std::endl;
return value;
}
int main(){
fibo(5, true);
return 0;
}
What you should realize is that the calls to the fibo function makes a tree. The root of the tree is the call to fibo(5, true) in the main(). As you only want to print each value once, the solution is to decide to print the value of the function only on the leftmost branch of that tree. The rule is then simply:
never print when on a right branch (hence the call to fibo(N-2, false)
never print if the parent didn't print (to avoid printing when on a child left branch of a right branch)

Answer 2

A common solution is to use memoization:
int fibo(int n)
{
static std::map<int,int> memo;
auto it=memo.find(n);
if(it!=std::end(memo))
{
return it->second;
}
int ret=1;
if (n > 2)
{
ret = fibo(n - 1) + fibo(n - 2);
}
memo[n]=ret;
return ret;
}
Then you can safely loop over the input parameters without recomputing the values over and over again:
for(int i=0;i<20;++i)
{
std::cout<<i<<" "<<fibo(i)<<std::endl;
}
Note that this is not only advantageous in printing but also for the calculation itself (at least as long as you call the function more than once).
Beside the above, you should also consider using long or double for the return type, as int will overflow more quickly.
EDIT: Ok, after your edit I don't know whether my answer exactly fits to your question, but I think it's a good advice anyway.
But here is another quick alternative which comes close, I guess:
int fibo(int n, bool first=true)
{
int ret=0;
if(n>2)
{
ret=fibo(n-1,false)+fibo(n-2,false);
}
else
{
ret=1;
}
if(first)
{
std::cout<<ret<<std::endl;
}
return ret;
}
DEMO