Problem:
Score : 600 points

Problem Statement

Takahashi has N balls with positive integers written on them. The integer written on the i-th ball is A_i.
He would like to form some number of pairs such that the sum of the integers written on each pair of balls is a power of 2.
Note that a ball cannot belong to multiple pairs.
Find the maximum possible number of pairs that can be formed.

Here, a positive integer is said to be a power of 2 when it can be written as 2^t using some non-negative integer t.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq A_i \leq 10^9

A_i is an integer.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible number of pairs such that the sum of the integers written on each pair of balls is a power of 2.

Sample Input 1

3
1 2 3

Sample Output 1

1

We can form one pair whose sum of the written numbers is 4 by pairing the first and third balls.
Note that we cannot pair the second ball with itself.

Sample Input 2

5
3 11 14 5 13

Sample Output 2

2

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    static int n;
    static int[] A;
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        A = new int[n];
        for(int i=0;i<n;i++) A[i] = sc.nextInt();
        Arrays.sort(A);
        HashMap<Integer,Integer> map = new HashMap<>();
        int[] dic = new int[30];
        dic[0] = 2;
        for(int i=1;i<30;i++) dic[i] = dic[i-1]*2;
        for(int w:A){
            map.put(w,map.getOrDefault(w,0)+1);
        }
        HashSet<Integer> used = new HashSet<>();
        int ans = 0;
        for(int i=n-1;i>=0;i--){
            if(!used.add(A[i])) continue;
            int idx = Arrays.binarySearch(dic,A[i]);
            if(idx>=0){
                int cur = map.get(A[i]);
                ans += cur/2;
            } else{
                idx= -(idx+1);
                int target = dic[idx];
                if(!map.containsKey(target-A[i])) continue;
                int cur = Math.min(map.get(target-A[i]),map.get(A[i]));
                ans += cur;
                map.put(target-A[i],map.get(target-A[i])-cur);
            }
        }
        PrintWriter out = new PrintWriter(System.out);
        out.println(ans);
        out.flush();
    }
}