The number of Fibonacci numbers lesser than a given n - c

Question

#include <stdio.h>
int fibonacci(int n) {
int count, n1 = 0, n2 = 1, fib = 0;
printf("Given number: ");
scanf("%d", &n);
count = 0;
while (count < n) {
fib = n1 + n2;
n1 = n2;
n2 = fib;
++count;
if (n > fib)
printf("%d ", fib);
}
return 0;
}
int main() {
int szam;
fibonacci(szam);
return 0;
}
I've gotten this far, I just don't know how to count the numbers.
for example:
input: 10
output: 1 2 3 5 8
but it should be:
in: 10
out: 5

Answer 1

The stop condition in your code is incorrect: you stop after n fibonacci numbers have been computed instead of stopping when you have computed a fibonacci number larger than n.
Here is a corrected version:
#include <stdio.h>
int count_fibonacci(unsigned long long int n) {
int count = 0;
unsigned long long n1 = 1, n2 = 1, fib = 1;
while (fib < n) {
count++;
fib = n1 + n2;
n1 = n2;
n2 = fib;
}
return count;
}
int main(void) {
unsigned long long n = 0;
printf("Given number: ");
scanf("%llu", &n);
printf("%d\n", count_fibonacci(n));
return 0;
}
It prints 5 for an input of 10, because your fibonacci sequence is: 1 2 3 5 8....
But the standard sequence is usually defined as 1 1 2 3 5 8..., and it should return 6. You can get this behavior bu changing the initial state to n1 = 0, n2 = 1, fib = 1.

Answer 2

Added the variable fib_count that counts the Fibonacci numbers (did not test this...)
#include <stdio.h>
int fibonacci(int n)
{
int n1=0, n2=1, fib=0, fib_count;
printf("Given number: ");
scanf("%d",&n);
fib_count = 0;
while (fib<n)
{`
fib=n1+n2;
n1=n2;
n2=fib;
fib_count += 1;
printf("%d ",fib);
}
printf("Fibonacci numbers smaller than %d : %d ",n, fib_count);
return 0;
}
int main(){
int szam;
fibonacci(szam);
return 0;
}