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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections;
using System.Collections.Generic;
using System.ComponentModel;
using System.IO;
using System.Linq;
using System.Numerics;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			int n = int.Parse(Console.ReadLine());
			var a = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();

			Array.Sort(a);
			long max = a.Max();
			var done = new bool[n];

			long target = 1;
			while (target < max) {
				target *= 2;
			}

			int ans = 0;
			while (target > 0) {
				var dic = new Dictionary<long, Queue<int>>();
				for (int i = 0; i < n; i++) {
					if (a[i] > target * 2) {
						break;
					}

					if (done[i]) {
						continue;
					}

					if (a[i] == target) {
						if (dic.ContainsKey(0) && dic[0].Count > 0) {
							ans++;
							int c = dic[0].Dequeue();
							done[c] = true;
							done[i] = true;
						} else {
							if (dic.ContainsKey(0) == false) {
								dic[0] = new Queue<int>();
							}

							dic[0].Enqueue(i);
						}
					} else if (a[i] < target) {
						long num = target - a[i];
						if (dic.ContainsKey(num) == false) {
							dic[num] = new Queue<int>();
						}

						dic[num].Enqueue(i);
					} else {
						long num = a[i] - target;
						if (dic.ContainsKey(num) && dic[num].Count > 0) {
							ans++;
							int c = dic[num].Dequeue();
							done[c] = true;
							done[i] = true;
						}
					}
				}

				target /= 2;
			}

			Console.WriteLine(ans);
		}
	}
}