Having trouble printing a series - c++

Question

Here's the problem statement for what I'm supposed to do:
Write a program to print the following series up to the term input by user.
0, 1, 1, 2, 3, 5, 8, 13, ….
Where 0 is 1st term and 13 is 8th term.
Hint: 0, 1
0+1 = 1
0, 1, 1
1+1 = 2
0, 1, 1, 2
And here's my code:
int prev_i = 0;
cout << "Enter a number: " << endl;
cin >> number;
for (i = 0; i <= number; i++)
{
cout << prev_i + i << " ,";
prev_i = i;
}
I do get what is wrong with my code though. It adds i to prev_i then prev_i is set to i. So in the next iteration when i is 1 thats i + prev_i = 1 so now prev_i = 1 and here's the problem i is 2 now so i + prev_i = 3. And I really can't seem to figure out how to get 1 instead of 3 as the output here and so on.
Oh and don't worry about i not declared properly. I just didn't copy that part.
pls help!

Answer 1

You're trying to generate a fibonacci sequence (starts with two terms (0,1), and each subsequent term is the addition of the prior two). Therefore, i should not be part of the calculation; it is only there to control looping.
A simple generation of the first ten numbers in the sequence is simply this:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a=0, b=1;
for (int i=0; i<10; ++i)
{
printf("%d ", a);
int c = a+b;
a = b;
b = c;
}
fputc('\n', stdout);
return EXIT_SUCCESS;
}
That's it. The code above will generate the following:
0 1 1 2 3 5 8 13 21 34
I leave applying the above logic to generate whatever your final requirements are up to you, but that's how the sequence is iteratively generated.
PS: Apologies in advance for writing C code. I totally spaced the language tag, but nonetheless the algorithm is the same.

Answer 2

The shown series is the fibonacci sequence.
Look at its definition and find out: Which numbers do you need to compute the current one?
In your current code, you only have one previous number available.
If that's not enough, what else might you need?

Answer 3

Here are my three cents.:)
#include <iostream>
#include <utility>
int main()
{
while (true)
{
std::cout << "Enter a non-negative number (0 - exit): ";
unsigned int n;
if (!( std::cin >> n ) || ( n == 0 )) break;
unsigned long long int first = 0;
unsigned long long int second = 1;
std::cout << '\n';
for (unsigned int i = 0; i < n; i++)
{
if (i != 0) std::cout << ' ';
std::cout << first;
second += std::exchange( first, second );
}
std::cout << "\n\n";
}
}
The program output might look like
Enter a non-negative number (0 - exit): 1
0
Enter a non-negative number (0 - exit): 2
0 1
Enter a non-negative number (0 - exit): 3
0 1 1
Enter a non-negative number (0 - exit): 4
0 1 1 2
Enter a non-negative number (0 - exit): 5
0 1 1 2 3
Enter a non-negative number (0 - exit): 6
0 1 1 2 3 5
Enter a non-negative number (0 - exit): 7
0 1 1 2 3 5 8
Enter a non-negative number (0 - exit): 8
0 1 1 2 3 5 8 13
Enter a non-negative number (0 - exit): 0
Fibonacci numbers grow very quickly. So in general you need to check whether an overflow can occur in the for loop or not.

Answer 4

Your code is printing the sum of current and just previous element. But the above question asks for the Fibonacci number which is define as:
Fib[i] = Fib[i - 1] + Fib[i - 2]; Fib[0] = 0, Fib[1] = 1
Now it can be solved through recursion or 1D DP.
But a simple solution can be constructed by knowing the above relation. We can define the current Fibonacci number is the sum of just previous and previous of just previous.
The code is:
int prev1 = 0; // Fib[0]
int prev2 = 1; // Fib[1]
int curr;
cout << prev1 << ' ' << prev2 << ' ';
for (int i = 2; i <= n; i++)
{
// Fib[i] = Fib[i - 1] + Fib[i - 2];
curr = prev2 + prev1;
cout << curr << ' ';
prev1 = prev2;
prev2 = curr;
}