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>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back
#define DEBUG(x) cout << #x << " = " << x << endl;

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 2e5 + 100;

int N, A[Maxn];
int pw2[40];
map <int, int> Map;

inline void Solve ()
{
	pw2[0] = 1;
	for (int i = 1; i <= 31; ++i) pw2[i] = pw2[i - 1] * 2;
	sort (A + 1, A + N + 1, greater <int>());
	int ans = 0;
	for (int i = 1; i <= N; ++i)
	{
		if (!Map[A[i]]) continue;
		for (int j = 1; j <= 31; ++j)
			if (Map[pw2[j] - A[i]])
			{
				++ans;
				--Map[pw2[j] - A[i]];
				break;
			}
	}
	cout << ans << endl;
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) ++Map[A[i] = read<int>()];
}

int main()
{

#ifdef hk_cnyali
	freopen("B.in", "r", stdin);
	freopen("B.out", "w", stdout);
#endif

	Input ();
	Solve ();

	return 0;
}