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<bits/stdc++.h>
using namespace std;
#define rep(i,j,k) for(int i = j;i <= k;++i)
#define repp(i,j,k) for(int i = j;i >= k;--i)
#define rept(i,x) for(int i = linkk[x];i;i = e[i].n)
#define P pair<int,int>
#define Pil pair<int,ll>
#define Pli pair<ll,int>
#define Pll pair<ll,ll>
#define pb push_back 
#define pc putchar
#define mp make_pair
#define file(k) memset(k,0,sizeof(k))
#define ll long long
int rd()
{
	int num = 0;char c = getchar();bool flag = true;
	while(c < '0'||c > '9') {if(c == '-') flag = false;c = getchar();}
	while(c >= '0' && c <= '9') num = num*10+c-48,c = getchar();
	if(flag) return num;else return -num;
}
int n;
int a[201000];
bool flag[201000];
int mi[100],tot; 
multiset<int>h;
multiset<int>::iterator it;
int main()
{
	n = rd(); rep(i,1,n) a[i] = rd();
	sort(a+1,a+n+1);
	reverse(a+1,a+n+1);
	for(ll i = 2;i <= 2e9;i <<= 1) mi[++tot] = i;
	rep(i,1,n) h.insert(-a[i]);
	
	int ans = 0;
	
	while(h.begin() != h.end())
	{
		int x = -(*h.begin());
		bool ck = false;
		rep(i,1,tot)
		{
			int y = x - mi[i];
		    it = h.lower_bound(y);
		    if(y == -x)
		    {
		    	if(it != h.end() && (*it) == y)
		    	{
		    		it++;
					if(it != h.end() && (*it) == y)
					{
						h.erase(h.begin());
						h.erase(h.begin());
						ck = true;
						break;
					}
				}
		    }
		    else
		    if(it != h.end() && (*it) == y) 
		    {
		    	h.erase(it);
		    	h.erase(h.begin());
		    	ans++;
		    	ck = true;
				break;
		    }
		}
		if(!ck) h.erase(h.begin());
	}
	
	printf("%d\n",ans);
	return 0;
}