Problem:
Score : 400 points

Problem Statement

Let {\rm comb}(n,r) be the number of ways to choose r objects from among n objects, disregarding order.
From n non-negative integers a_1, a_2, ..., a_n, select two numbers a_i > a_j so that {\rm comb}(a_i,a_j) is maximized.
If there are multiple pairs that maximize the value, any of them is accepted.

Constraints

2 \leq n \leq 10^5

0 \leq a_i \leq 10^9

a_1,a_2,...,a_n are pairwise distinct.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n
a_1 a_2 ... a_n

Output

Print a_i and a_j that you selected, with a space in between.

Sample Input 1

5
6 9 4 2 11

Sample Output 1

11 6

\rm{comb}(a_i,a_j) for each possible selection is as follows:

\rm{comb}(4,2)=6

\rm{comb}(6,2)=15

\rm{comb}(6,4)=15

\rm{comb}(9,2)=36

\rm{comb}(9,4)=126

\rm{comb}(9,6)=84

\rm{comb}(11,2)=55

\rm{comb}(11,4)=330

\rm{comb}(11,6)=462

\rm{comb}(11,9)=55

Thus, we should print 11 and 6.

Sample Input 2

2
100 0

Sample Output 2

100 0

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

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <utility>
#include <cmath>

using namespace std;
using llong = long long;

using P = pair<int,int>;
int a[100000];

double logC(P p){
  return double(p.first-p.second) * std::log(double(p.first)/double(p.first-p.second))
    + (std::log(double(p.first)/p.second) - std::log(2.0*M_PI*(p.first-p.second)))/2.0;
}

bool cmp(P l,P r){
  return logC(l) <= logC(r);
}

int main(){
  int N;
  cin >> N;
  for(int i=0;i<N;++i){
    cin >> a[i];
  }

  sort(a,a+N);
  
  P max_p(0,0);
  for(int i=1;i<N;++i){
    int left = 0 , right = i;
    int center ;
    while(left < right){
      center = ( left + right )/ 2;
      // cout << left << " " << right << " " << center << endl;
      if(a[center] < (a[i]+1)/2){
        left = center+1;
      }else if(a[center] > (a[i]+1)/2){
        right = center;
      }else{
        break;
      }
    }
    int r = a[center];
    if(center>0 && std::abs(a[center-1]-(a[i]+1)/2) < std::abs(r-(a[i]+1)/2)){
      r = a[center-1];
    }
    if(center<i && std::abs(a[center+1]-(a[i]+1)/2) < std::abs(r-(a[i]+1)/2)){
      r = a[center+1];
    }

    P p(a[i],r);
    if(max_p.first==0||cmp(max_p,p)){
      max_p = p; 
      // cout << max_p.first << " " << max_p.second << endl;
    }
  }

  cout << max_p.first << " " << max_p.second << endl;
  return 0;
}