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: Time Limit Exceeded)

#include <iostream>
#include <sstream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <climits>
#include <iomanip>
#include <bitset>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#include <deque>
#include <set>
#include <list>
#include <map>

using namespace std;

#define int long long

int32_t main() {
  int n;
  cin >> n;
  int arr[n + 5];
  map <int, int> cnt;
  vector < pair <int, int> > vec;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    cnt[arr[i]]++;
  }
  for (int i = 0; i < n; i++) {
    int el = arr[i];
    int here = 0;
    for (int j = 0; j <= 32; j++) {
      here += cnt[(1ll << j) - el];
    }
    if (((el * 2) & (el * 2 - 1)) == 0) here--;
    vec.push_back({here, el});
  }
  sort(vec.begin(), vec.end());
  int ans = 0;
  multiset <int> s;
  for (int i = 0; i < n; i++) {
    bool done = 0;
    for (int j = 31; j >= 0 && !done; j--) {
      if (s.find((1ll << j) - vec[i].second) != s.end()) {
        s.erase(s.find((1ll << j) - vec[i].second));
        ans++;
        done = 1;
      }
    }
    if (!done) s.insert(vec[i].second);
  }
  cout << ans << endl;
  return 0;
}