Problem:
Score : 200 points

Problem Statement

It is November 18 now in Japan. By the way, 11 and 18 are adjacent Lucas numbers.

You are given an integer N. Find the N-th Lucas number.

Here, the i-th Lucas number L_i is defined as follows:

L_0=2

L_1=1

L_i=L_{i-1}+L_{i-2} (i≥2)

Constraints

1≤N≤86

It is guaranteed that the answer is less than 10^{18}.

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the N-th Lucas number.

Sample Input 1

5

Sample Output 1

11

L_0=2

L_1=1

L_2=L_0+L_1=3

L_3=L_1+L_2=4

L_4=L_2+L_3=7

L_5=L_3+L_4=11

Thus, the 5-th Lucas number is 11.

Sample Input 2

86

Sample Output 2

939587134549734843

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
using namespace std;

typedef long long ll;


ll lu[100];

ll lucas(ll l)
{
  ll l1, l2;
  if(l==0) return 2LL;
  if(l==1) return 1LL;
  if(lu[l-1]<0){lu[l-1]=lucas(l-1);}
  if(lu[l-2]<0){lu[l-2]=lucas(l-2);}
  return lu[l-1]+lu[l-2];
}

int main()
{
  for(int i = 0;i < 100; i++) {
    lu[i] = -1;
  }
  ll n;
  cin >> n;
  
  cout << lucas(n) << endl;
}