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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <set>
#include <cmath>
#include <vector>
using namespace std;
int main()
{
	long pow_2_10_9 = 2*pow(10,9),finded;
	set< long>pow_2_num;
	set<long>num_sequ;
	for(long i = 1;1;++i)//setting power of 2 number series table
	{
		if(pow(2,i)<pow_2_10_9)
		{
			pow_2_num.insert(( long)pow(2,i));
		}
		else break;
	}
	long col_num,single_num,count;
	while(cin>>col_num)//for each group
	{
		num_sequ.clear();
		count = 0;
		for(int i = 0;i<col_num;++i)//for each number
		{
			cin>>single_num;
			if(i == 0)
			{
				num_sequ.insert(single_num);
				continue;
			}
			for(set<long >::iterator iter = num_sequ.begin();iter!=num_sequ.end();)
			{
				if(pow_2_num.find(( long)(single_num+*iter))!=pow_2_num.end())
				{
					count++;finded = *iter;
					num_sequ.erase(finded);
					break;
				}
				++iter;
				if(iter == num_sequ.end())//not find
				{
					num_sequ.insert(single_num);
					break;
				}
			}
		}
		cout<<count<<endl;
	}
	return 0;
}