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//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;
}