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 static System.Math;
using System.Collections.Generic;
using System;

public class Hello
{
    public static void Main()
    {
        var b = new long[32];
        b[0] = 1;
        for (int i = 1; i <= 31; i++) b[i] = b[i - 1] * 2;
        var n = int.Parse(Console.ReadLine().Trim());
        string[] line = Console.ReadLine().Trim().Split(' ');
        var a = Array.ConvertAll(line, int.Parse);
        Array.Sort(a);
        Array.Reverse(a);
        var d = new Dictionary<long, int>();
        for (int i = 0; i < n; i++)
            if (d.ContainsKey(a[i])) d[a[i]]++;
            else d[a[i]] = 1;
        var ans = getAns(d, a, b);
        Console.WriteLine(ans);
    }
    static int getAns(Dictionary<long, int> d, int[] a, long[] b)
    {
        var count = 0;
        foreach (var x in a)
            if (d[x] > 0)
            {
                var t = b[getB(x)];
                var w = t - x;
                if (x == w)
                {
                    if (d.ContainsKey(w) && d[w] > 1) { count++; d[w] -= 2; }
                }
                else
                {
                    if (d.ContainsKey(w) && d[w] > 0) { count++; d[w]--; }
                }
            }
        return count;
    }
    static int getB(long a) => (int)(Log10(2 * a) / Log10(2));
}