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: Wrong Answer)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		int n = int.Parse(Console.ReadLine());
		long[] nums = Array.ConvertAll(Console.ReadLine().Split(' '),long.Parse);
		Array.Sort(nums);
		long[] answers = new long[n];
		long myNum = 0;
		long partnerNum = n-1;
		long aimNum = 2;//和として作る数
		long answer = 0;
		
		while(true)//全ての2のべき乗を試す
		{
			if(n == 1) break;
			if(aimNum > nums[n-1] + nums[n-2]) break;//作れない数になった
			myNum = 0;
			partnerNum = n-1;

			while(true)//全ての組み合わせを試す
			{
      if(nums[myNum] == nums[partnerNum] && nums[myNum] + nums[partnerNum] == aimNum)//同じ数で和となる
      {
        for(int i = (int)myNum; i <= (int)partnerNum; i++)
        {
          answers[i] = partnerNum - myNum;
        }
        break;
      }
      
				if(nums[myNum] + nums[partnerNum] == aimNum)
				{
					answers[myNum]++;
					answers[partnerNum]++;
				}
				if(myNum == n-1 || partnerNum == 0) break;

				if(nums[myNum] + nums[partnerNum-1] >= aimNum) partnerNum--;
				else
        {
          while(true)
          {
            myNum++;
            if(nums[myNum] != nums[myNum-1]) break;
            else
            {
              answers[myNum] = answers[myNum-1];
              answers[partnerNum] += answers[myNum-1];
            }
          }
        }

        if(myNum == partnerNum) break;
        
			}

			aimNum *= 2;
		}

    //解答を減らす
    for(int i = 0; i < n; i++)
    {
      answer += answers[i];
    }
    answer /= 2;
    //Console.WriteLine(answer);
		Array.Sort(answers);
		Array.Reverse(answers);
		long answerMinusMemo = 0;
		for(int i = 0; i < n-1; i++)
		{
			if(answers[i] > 1)
			{
				if(answerMinusMemo == 0)
				{
					answer -= answers[i]-1;
					answerMinusMemo = answers[i]-1;
				}else
				{
					if(answerMinusMemo >= answers[i]-1) answerMinusMemo -= (answers[i]-1);
					else
					{
						answers[i] -= answerMinusMemo;
						answer -= answers[i]-1;
						answerMinusMemo = answers[i]-1;
					}
				}
			}
		}
		
		Console.WriteLine(answer);
	}
}