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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)



#include <stdio.h>

int main()
{
  int num = 0,n;
  int a[200000];

  scanf("%d",&n);

  int i;

  for(i=0;i<n;i++)
    scanf("%d",&a[i]);


  int j,k,x;

      for(j=1;j<n;j++)
	for(k=j+1;k<n;k++)
	  if(a[j]<a[k])
	    {
	      x=a[j];a[j]=a[k];a[k]=x;
	    }
	
      
      for(j=0;j<n-1;j++)
	if(a[j]!=-1)
	for(k=j+1;k<n;k++)
	  {
	    if(a[k]!=-1)
	    if(a[k]%2==a[j]%2)
	      {
		x = a[k] + a[j];

		while(x%2 != 1)
		  {
		    x /= 2;
		  };

		if(x == 1)
		  {
		    num++;
		    a[k] = -1;
		    break;
		    
		  }
	      }
	  }
	

  printf("%d\n",num);
  return 0;
}