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: Accepted)

#include <bits/stdc++.h>
#define INF (1<<30)
#define MOD 1000000007
#define VM 200000
#define l_ength size
using ll = long long;
using namespace std;

// ペアの候補となる数を返す
// 具体的には ( x の 2 進数表記の最上位の桁の 1 つ左に 1 が立っていて残り 0 であるような数 ) - x
int f( int x ){
	int k = 0;
	while( true ){
		if( x == (1<<k) ){
			return x;
		} else if( x < (1<<k) ){
			return (1<<k)-x;
		} else {
			++k;
		}
	}
}

int main(){

	int n; cin >> n;
	if( n == 1 ){
		cout << 0 << endl;
		return 0;
	}
	vector<int> a(n);
	for( auto &k : a ) cin >> k;
	sort( a.begin(), a.end() );
	// 番兵
	a.push_back(INF);
	
	// uniq した上で個数を記録
	int cnt = 1;
	vector<int> b;
	vector<int> g;
	for( int i = 1; i <= n; ++i ){
		if( a[i-1] == a[i] ){
			++cnt;
		} else {
			b.push_back( a[i-1] );
			g.push_back( cnt );
			cnt = 1;
		}
	}
	// 番兵を消す
	a.pop_back();
	
	/*
	for( auto &k : b ) cout << k << " "; cout << endl;
	for( auto &k : g ) cout << k << " "; cout << endl;
	*/
	
	// 大きい方から貪欲に
	int ans = 0;
	for( int i = n-1; i >= 0; --i ){
		int aindex = lower_bound( b.begin(), b.end(), a[i] ) - b.begin();
		if( g[aindex] == 0 ) continue;
		int phi = f(a[i]);
		int phindex = lower_bound( b.begin(), b.end(), phi ) - b.begin();
		if( phi > 0 && phindex != b.l_ength() && b[phindex] == phi ){
			--g[aindex];
			if( g[phindex] >= 1 ){
				--g[phindex];
				++ans;
			} else {
				++g[aindex];
			}
		}
		/*
		cout << "==" << endl;
		for( auto &k : b ) cout << k << " "; cout << endl;
		for( auto &k : g ) cout << k << " "; cout << endl;
		*/
	}

	cout << ans << endl;

}