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 Python (3.4.3) (judge verdict: Runtime Error)


from math import factorial
import bisect

n = int(input())
lst = list(map(int,input().split()))

lst = sorted(lst)

ans = (-1,-1)
val = -1

def nCr(n,r):
    return factorial(n)/factorial(n-r) * factorial(r)

if n==2:
    print(*lst)
    exit()

for i in range(1, n):
    n = lst[i]
    r1 = bisect.bisect_left(lst, n//2)
    r2 = bisect.bisect_right(lst, n//2)
    
    v1 = nCr(n, lst[r1])
    v2 = nCr(n, lst[r2])
    
    if v1 > val and n!=lst[r1]:
        val = v1
        ans = (n, lst[r1])
    if v2 > val and n!=lst[r2]:
        val = v2
        ans = (n, lst[r2])
    
print(*ans)