1. Using recursion method

In recursion a lot work is done as it repeatedly calculates the number and solves same case more than one time.

For Example:

As it calculates Fibo(1) & Fibo(0) multiple time

Consider the program:

#include<iostream>
using namespace std;

int Fib(int n)
{
	if(n<=1)
	return n;
	return Fib(n-1)+Fib(n-2);	/*recusively add the numbers*/
}

int main()
{
	int n;  /*position of the element starting from 0...n*/
	cout<<"Enter the number :";
	cin>>n;
	cout<<"Number at "<<n<<"th place is "<<Fib(n)<<endl;
	return 0;
}

Output

Enter the number :10
Number at 10th place is 55

2. Using Dynamic programming method

A dynamic programming algorithm remembers the past result and uses them to find new result means it solve complex problems by breaking it down into a collection of simpler subproblems, then solving each of those subproblems only once ,and storing their solution for future use instead of recomputing their solutions again.

Algorithm:

Fib(n)
    if n=0
	    return 0
	else
	    prev_Fib=0,curr_Fib=1
	repeat n-1 times  /*if n=0 it will skip*/
	    next_Fib=prev_Fib+curr_Fib
	    prev_Fib=curr_Fib
	    curr_Fib=next_Fib
	return curr_Fib

Consider the program:

#include<iostream>
using namespace std;

int main()
{
	int n;
	cout<<"Enter the number :";
	cin>>n; /*position of the element starting from 0...n*/

	int Fibo[n+1],i;
	Fibo[0]=0,Fibo[1]=1;

	for(i=2;i<=n;i++)
	{
		Fibo[i]=Fibo[i-1]+Fibo[i-2];  /*It will store the sum of previous two elements */
	}
	cout<<"Number at "<<n<<"th place is "<<Fibo[n]<<endl;

	return 0;
}

Output

Enter the number :10
Number at 10th place is 55

3. Without using array

Well it is the easiest way to calculate Fibonacci number as we just have to add two previous numbers to calculate next number.

Consider the program:

#include<iostream>
using namespace std;

int Fib(int n)
{
    if(n==0)
	    return 0;
	else
	{
	    int prev_Fib=0,curr_Fib=1,next_Fib;
	    if(n==0)
	    return n;
	    while(n>=2)
	    {
			next_Fib=prev_Fib+curr_Fib;
	        prev_Fib=curr_Fib;
	        curr_Fib=next_Fib;
	        n--;
		}
	return curr_Fib;
	}
}

int main()
{
	int n;   /*position of the element starting from 0...n*/
	cout<<"Enter the number :";
	cin>>n;
	cout<<"Number at "<<n<<"th place is "<<Fib(n)<<endl;
	return 0;
}

Output

Enter the number :10
Number at 10th place is 55