//An efficient way to calculate nth fibonacci number faster and simpler than O(nlogn) method of matrix exponentiation
//This works by using both recursion and dynamic programming.
//as 93rd fibonacci exceeds 19 digits, which cannot be stored in a single long long variable, we can only use it till 92nd fibonacci
//we can use it for 10000th fibonacci etc, if we implement bigintegers.
//This algorithm works with the fact that nth fibonacci can easily found if we have already found n/2th or (n+1)/2th fibonacci
//It is a property of fibonacci similar to matrix exponentiation.

#include <iostream>
#include<cstdio>
using namespace std;

const long long MAX = 93;
 

long long f[MAX] = {0};
 

long long fib(long long n)
{

    if (n == 0)
        return 0;
    if (n == 1 || n == 2)
        return (f[n] = 1);
 

    if (f[n])
        return f[n];
 
    long long k = (n%2!=0)? (n+1)/2 : n/2;
 
    f[n] = (n%2!=0)? (fib(k)*fib(k) + fib(k-1)*fib(k-1))
           : (2*fib(k-1) + fib(k))*fib(k);
    return f[n];
}


int main()
{
	//Main Function
	for(long long i=1;i<93;i++)
	{
		cout << i << " th fibonacci number is " << fib(i) << "\n";		
	}
	return 0;
}