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: Wrong Answer)

#include "stdio.h"
#include <stdlib.h>

int compare_int(const void *a, const void *b)
{
    return *(long long int *)a - *(long long int*)b;
}

int sarch(int n,int m,long long int*A,long long int B){
  if(m-n<=5){
    for (size_t i = n; i <= m; i++) {
      if(A[i]==B){return i;}
    }
    return -1;
  }
  int k=(int)((n+m)/2);
  if(A[k]>B){sarch(n,k,A,B);}
  if(A[k]<B){sarch(k,m,A,B);}
  if(A[k]==B){return k;}
  return 0;
}

int main(int argc, char const *argv[]) {
  int N;
  scanf("%d", &N);
  long long int A[N];
  for (size_t i = 0; i < N; i++) {scanf("%lld", &A[i]);}
  qsort(A, N, sizeof(long long int), compare_int);
  long long int B[N];
  long long int k;

  for (size_t i = 0; i < N; i++) {
    k=2;
    while(k<=A[i]){k=k*2;}
    B[i]=k-A[i];
  }
  int ans=0;
  int check[N];
  int temp;
  for (size_t i = 0; i < N; i++) {check[i]=0;}
  for (size_t i = N-1; i >=1 ; i--) {
    temp=sarch(0,i-1,A,B[i]);
    if(temp>=0 && check[temp]==0){check[temp]=1;ans++;}
  }
  printf("%d", ans);
  return 0;
}