output_description: Print the N-th Lucas number.
  
  * * *
input_description: Input is given from Standard Input in the following format:
  
      
      
      N
attempt: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using ld = long double;
  using P = pair<ll, ll>;
  using Pld = pair<ld, ld>;
  using Vec = vector<ll>;
  using VecP = vector<P>;
  using VecB = vector<bool>;
  using VecC = vector<char>;
  using VecD = vector<ld>;
  using VecS = vector<string>;
  using Graph = vector<VecP>;
  #define REP(i, m, n) for(ll (i) = (m); (i) < (n); ++(i))
  #define REPR(i, m, n) for(ll (i) = (m); (i) > (n); --(i))
  #define rep(i, n) REP(i, 0, n)
  #define R cin>>
  #define repr(i, n) REPR(i, n, 0)
  #define all(s) (s).begin(), (s).end()
  #define pb push_back
  #define mp make_pair
  #define fs first
  #define sc second
  #define in(a) insert(a)
  #define P(p) cout<<(p)<<endl;
  #define ALL(x) (x).begin(),(x).end()
  #define ALLR(x) (x).rbegin(),(x).rend()
  #define SORT(a) sort((a).begin(), (a).end())
  typedef vector<int> vi;
  typedef vector<vi> vvi;
  typedef vector<long long int> vll;
  typedef vector<string> vs;
  typedef pair<int, int> pii;
  typedef long long ll;
  typedef pair<ll, ll> pll;
  void sonic(){ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);}
  void setp(const ll n){cout << fixed << setprecision(n);}
  const ll INF = 1e9+1;
  const ll LINF = 1e18+1;
  const ll MOD = 1e9+7;
  //const ll MOD = 998244353;
  const ld PI = acos(-1);
  const ld EPS = 1e-11;
  template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return true;}return false;}
  template<class T>bool chmin(T &a,const T &b){if(a>b){a=b;return true;}return false;}
  template<typename T> void co(T e){cout << e << "\n";}
  template<typename T> void co(const vector<T>& v){for(const auto& e : v)
      { cout << e << " "; } cout << "\n";}
  ll gcd(ll a, ll b) {
      if (a < b)swap(a, b);
      if (b == 0) return a;
      unsigned r;
      while ((r = a % b)) {
          a = b;
          b = r;
      }
      return b;
  }
  ll lcm(ll a, ll b) {
      ll g = gcd(a, b);
      return a * b / g;
  }
  bool prime(ll n) {
      for (ll i = 2; i <= sqrt(n); i++) {
          if (n%i == 0)return false;
      }
      return n != 1;
  }
  
  Vec fac, finv;
  
  ll PowMod(ll a, ll n){
      if(n == 1) return a;
      if(n%2 == 0) return PowMod(a*a%MOD, n/2);
      return a*PowMod(a*a%MOD, n/2)%MOD;
  }
  
  ll combi(ll n, ll k){
      if(k>n) return 0;
      return fac[n]*finv[k]%MOD*finv[n-k]%MOD;
  }
  
  int main(){
      int n;
      cin >> n;
      ll l[2] = {2, 1};
      rep(i, n - 1){
          l[i % 2] = l[0] + l[1];
      }
      cout << l[n % 2] << '\n';
  }
problem_description: 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)
samples: [{"input": "5", "output": "11\n    \n\n  * L_0=2\n  * L_1=1\n  * L_2=L_0+L_1=3\n  * L_3=L_1+L_2=4\n  * L_4=L_2+L_3=7\n  * L_5=L_3+L_4=11\n\nThus, the 5-th Lucas number is 11.\n\n* * *"}, {"input": "86", "output": "939587134549734843"}]