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Another way is using divide and conquer algorithm,which in O(log n) running time.
+Tricky part is: for base_array[2][2]={{1,1},{1,0}}; => {{A2,A1},{A1,A0}} and for base_array^n, we can get {{An+1,An},{An,An-1}}, and simply return An is enough.
+Algorithm: (bottom to top) base_array=>base_array^2=>base_array^4=>....base_array^n, thus running time is O(log n).

The good part is the running time, but the code is a little bit more since need to calculate the matrix matrix multiplication and when compare to dynamic program, I still recommend dynamic programming method for its simplicity.

//fibonacci series program
//author : scarlet
#include<stdio.h>
main()
{
int a,b,n,c;
scanf("%d",&n);
a=0;
b=1;
printf("%d %d",a,b);
n=n-2;
while(n>0)
{
c=a+b;
a=b;
b=c;
printf("%d ",c);
n--;
}
}

void main()
{
cin>>n;
cout<<fibo(n);


}

int fibo(int n){
if(n==0) return 0;
if(n==1) return 1;
else return(fibo(n-1)+fibo(n-2));
}
}

void main()
{
cin>>n;
cout<<fibo(n);


}

int fibo(int n){
if(n==0) return 0;
if(n==1) return 1;
else return(fibo(n-1)+fibo(n-2));
}
}

use dynamic programming. Store the results of a computation during the first all and reuse subsequently instead of computing at every call.

/*dynamic programming version,using cache O(n), here array's scope is [0,n-1]*/
int fib_one(int n,int *array){

int rs1,rs2;
if(n==0){
return 0;
}
else if(n==1){
return 1;
}
else{
rs1=array[n-1];
if(rs1==-1){
rs1=fib_one(n-1,array);
}
rs2=array[n-2];
if(rs2==-1){
rs2=fib_one(n-2,array);
}
return rs1+rs2;
}
}

<pre lang="" line="1" title="CodeMonkey52582" class="run-this">
#include <iostream>
#include <cstdio>

using namespace std;

int fib(int n){
unsigned long long fib1=1,fib2=1,fibr;
for(int i=2;i<=n;i++){
fibr= fib1+fib2;
fib1 = fib2;
fib2 = fibr;
}
return fibr;

}

int main(){
int n;
cin>>n;
cout<<fib(n)<<endl;
return 0;
}
</pre><pre title="CodeMonkey52582" input="yes">
</pre>

static int fib(int n)
	{
		if(n==0)
		{
			return 0;
		}
		else
			if(n==1 || n==2)
				return 1;
		
		return fib(n-1)+fib(n-2);
	}
	
	static int fib1(int n)
	{
		int c=0,a=0,b=1;
		for(int i=2;i<=n;i++)
		{
			c=a+b;
			a=b;
			b=c;
		}
		
		return c;
	}

public class fibb {

	public static void main(String[] args) {
	int a= fib(10);
	System.out.print(a);


	}
	static int fib(int n)
	{
		if(n==0)
		{
			return 0;
		}
		else
			if(n==1)
				return 1;
		
		return fib(n-1)+fib(n-2);
	}
	
	
}

Python working code.

"""
8:29
@Python 2.7

write a code of fibonacci series.

- sonali on August 03, 2010 Report Duplicate | Flag 
"""

class Fbi(object):
  def __init__(self, n):
    if n is None:
      print 'invalid n'
      raise SystemExit
      
    self._n = n
    
  def find(self, number = None):
    if number is None:
      number = self._n
      
    if number == 0:
      return 0
    elif number == 1:
      return 1
    else:
      return self.find(number-1) + self.find(number-2)
      
if __name__ == '__main__':
  f = Fbi(30)
  print f.find()

//prints the fibonacci series


# include<stdio.h>
# include<conio.h>
void main()
{
int a=0,b=1,i=0;
for(i=0;i<=10;i++)
{
printf("%d",a);
a=a+b;
b=a-b;
}
getch();
}

public class Fibonacci{
	
	public static void main(String args[])
	{
		int n = 10;
		int a = 0;
		int b = 1;

		System.out.println(a);
		System.out.println(b);
		for(int i = 2; i<= n; i++){
			int c = a + b;
			System.out.println(c);
			a= b;
			b = c;

		}

	}
}