Find nth fibonacci number naive and DP Based

Fibonacci numbers form a sequence such that each number is the sum of the two preceding ones.
Fibonacci numbers start from 0, 1.

 and

${\displaystyle F_{n}=F_{n-1}+F_{n-2},}$

 for n > 1

So the series is

To find nth fibonacci number

Recursive Solution -

#include<stdio.h>
int count;
int fib(int no)
{
    count++;
    if(no<=1)
        return no;
    else
        return fib(no-1)+fib(no-2);
}
void main()
{
    int num;
    printf("Enter which fibbonacci number you want to find out\n");
    scanf("%d",&num);
    printf("The %dth fibbonacci number is %d",num,fib(num-1));
    printf("Recursion=%d",count);
}

This program has high complexity we can use DP to reduce complexity.
DP Based Solution -

#include<stdio.h>
int lookup[1000];
void initialize()
{
    int i;
    for(i=0;i<1000;i++)
        lookup[i]=-1;
}
int fibo(int num)
{
    if(lookup[num]!=-1)             //if number found in look up table than return
        return lookup[num];
    else        //number not found in look up table
    {
        if(num<=1)
            lookup[num]=num;
        else            //calculate the number and store in look up table than return
            lookup[num]=fibo(num-1)+fibo(num-2);
    }

    return lookup[num];
}
void main()
{
    int num,i;
    printf("Enter which fibbonacci number you want to find out\n");
    scanf("%d",&num);
    initialize();
    printf("The %dth fibbonacci number is %d",num,fibo(num-1));
}