Modified Fibonacci in C - c

Question

I need to generate a modified Fibonacci series and it must be completely dynamic. Here f0 and f1 will be given, i.e f0=1 and f1=3 after generating the series. I should print the resulting value at a particular index.
Ex: f0 = 1, f1 = 3, testcase(n) = 3 (This can change not a particular value)
t1 = 4 t2 = 8 t3 = 11 and so on. Series should be generated for 11 elements by adding current element and previous element using: f[i] = f[i-1] + f[i-2]
It can be represented as:
0=>1
1=>3
2=>4
3=>7
4=>11
5=>18
6=>29
7=>47
8=>76
9=>123
10=>199
11=>322
I should print the values at indices 4,8 and 11 (which must be the output of my program), i.e. 11 76 322.
Input Format:
f0,f1 and n (where n is the no of indices)
where ti=[t1,t2,....tn-1] (which specifies the index for R-Fibonacci series).
Output Format:
Print the values from the R-fibonacci series based on the given indices.
Sample Input:
1 3 3 4 8 11
Sample Output:
11 76 322
I have the code that generates the Fibonacci series for the above program but I want to display the value at 4,8,11 indices. Here is the code:
int fib(int n)
{
int f[n+1];
int i;
f[0]=1;
f[1]=3;
for(i=2;i<=n;i++)
{
f[i]=f[i-1]+f[i-2];
}
return f[n];
}
int main()
{
int n=11
printf("%d ",fib(n));
getchar();
return 0;
}

Answer 1

Like this? The array is defined in main and passed to the function as an argument. The function fills in the array, returns nothing, and then in main you can print the elements you want.
You will need a loop to do that, with another dynamic array holding the indices you are asked to print.
#include <stdio.h>
void fib(int n, int *f)
{
int i;
f[0] = 1;
f[1] = 3;
for(i = 2; i <= n; i++)
{
f[i] = f[i-1] + f[i-2];
}
}
int main()
{
int n = 11;
int f[n+1];
fib(n, f);
printf("%d ", f[8]);
printf("%d ", f[11]);
printf("\n");
getchar();
return 0;
}
Program output:
76 322
I will leave you some code to write, but suppose you make a dynamic array of the index values required, such as
int index[m];
index[0] = 4;
index[1] = 8;
index[2] = 11;
you can print the series term with such as
printf("%d ", f[ index[i] ]);

Answer 2

If I understand correctly, this question really has little to do with fibonacci and is about scope in C. You are declaring and defining an array in a function, fib, and filling it within that function (and returning a single element value). What you want is to have access to the entire array from the caller.
A straightforward way of doing this is to declare the array in the calling method, and pass a pointer to it to the fib function:
#include <stdio.h>
int fib(int f[], int n)
{
int i;
f[0]=1;
f[1]=3;
for(i=2;i<=n;i++)
{
f[i]=f[i-1]+f[i-2];
}
return f[n];
}
int main()
{
int n=11;
int f[12];
fib(f, n);
printf("%d ", f[4]);
printf("%d ", f[8]);
printf("%d ", f[11]);
getchar();
return 0;
}

Answer 3

Here is a code to generate a set of fibonacci numbers less than a given number N using recursive algorithm :
#include<stdio.h>
int fibo(int n)
{
if(n<2)
return n;
else
return (fibo(n-1)+fibo(n-2));
}
void main()
{
int n,i;
printf("\n Enter number : ");
scanf("%d",&n);
printf("\n Fibonacci series is : ");
for(i=0;i<n;i++)
printf("\n %d",fibo(i));
}