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: Time Limit Exceeded)

import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;
import java.util.Comparator;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BPowersOfTwo solver = new BPowersOfTwo();
        solver.solve(1, in, out);
        out.close();
    }

    static class BPowersOfTwo {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // input
            int n = in.nextInt();
            ArrayList<Integer> a = new ArrayList<>();
            int ans = 0;
            for (int i = 0; i < n; i++) {
                a.add(in.nextInt());
            }
            a.sort(Comparator.naturalOrder());
            while (a.size() > 1) {
                int size = a.size() - 1;
                // 二分探索で一つ上のを探す
                // 一回当たりlogNだから全体で N*logN が上界
                int target = 1;
                while (target <= a.get(size)) {
                    target *= 2;
                }
                target -= a.get(size);
                a.remove(size);

                // System.out.println("size: " + a.size() + ", target: " + target);
                // targetを二分探索で探す
                int low = 0;
                int high = a.size() - 1;
                int mid = (low + high) / 2;
                while (high >= low) {
                    int search = a.get(mid);
                    if (search < target) {
                        //System.out.println("low");
                        low = mid + 1;
                    } else if (search > target) {
                        //System.out.println("high");
                        high = mid - 1;
                    } else {
                        // System.out.println("got it");
                        ans++;
                        a.remove(mid);
                        break;
                    }
                    mid = (low + high) / 2;
                }
            }
            out.println(ans);
        }

    }
}