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 <cstdio>
#include <math.h>
#include <algorithm>
 
const long long int MAX_N = 200000;
 
long long int p[MAX_N];

long long int n = 0;
long long int lowid, midid, highid;
long long int c = 0;
long long int target = 0;

long long int nearPow = 1;

void quickSort(long long int numbers[], long long int array_size) {
  std::sort(numbers, numbers + array_size);
}

void solve(long long int x) {
  if (x > 0){
    if ((p[x - 1] & (p[x - 1] - 1)) == 0) {
      if (p[x - 2] == p[x - 1]) {
        c += 1;
        solve(x - 2);
      } else {
        solve(x - 1);
      }
    } else if (p[x - 1] % 2 == 1) {
      solve(x - 1);
    } else {
      nearPow = pow(2.0, (floor(log2(p[x - 1] - 1)) + 1.0));
      target = nearPow - p[x - 1];
  
      highid = x;
      lowid = 0;

      while (lowid <= highid) {
        midid = (lowid + highid) / 2;
        if (p[midid] == target) {
          while (p[midid - 1] == target) {
            midid -= 1;
          }
          p[midid] -= 1;
          c += 1;
          break;
        } else if (p[midid] < target) {
          lowid = midid + 1;
        } else {
          highid = midid - 1;
        }
      }
      solve(x - 1);
    }
  }
}
 
int main() {
  scanf("%lld", &n);
  for (long long int i = 0; i < n; i++) {
    scanf("%lld", &p[i]);
  }
  for (long long int i = 0; i < n; i++) {
    p[i] *= 2;
  }
  quickSort(p, n);
  solve(n);
  printf("%lld\n", c);
 
  return 0;
}