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 ch_A(int n,int m,long long int*A,long long int B){
  if(m==n){
    if(A[m]==B){
      return m;
    }else{
      return -1;
    }
  }
  if (B<A[n]||A[m]<B) {
    return -1;
  }
  int k=(int)((n+m)/2);
  if(A[k]>B){
    return ch_A(n,k-1,A,B);
  }
  if(A[k]<B){
    return ch_A(k+1,m,A,B);
  }
  if(A[k]==B){
    return k;
  }
  return 0;
}

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

  int B[N];
  int A_B[N];
  int t=1;
  int ind=0;
  qsort(A, N, sizeof(long long int), compare_int);

  A_B[0]=0;
  for (i = 1; i < N; i++) {
    if (A[i-1]==A[i]) {
      t++;
      A_B[i]=ind;
    }
    if (A[i-1]!=A[i]) {
      B[ind]=t;
      A_B[i]=ind+1;
      ind++;
      t=1;
    }
  }
  B[ind]=t;
  A_B[i]=ind+1;
  ind++;
  // printf("%d\n", ind);
  // for ( i = 0; i < ind; i++) {
  //   printf("%d ", B[i]);
  // }
  // printf("\n");
  // for (i = 0; i < N; i++) {
  //   printf("%d ", A_B[i]);
  // }
  // printf("\n");

  long long int k;
  int temp;
  int ans=0;
  for (i = N-1; i >=1; i--) {
    k=2;
    while(k<=A[i]){k=k*2;}
    temp=ch_A(0,i-1,A,k-A[i]);
    if(temp!=0){
      if(B[A_B[temp]]>=1 && B[A_B[i]]>=1){
        B[A_B[temp]]--;
        B[A_B[i]]--;
        ans++;
      }
    }
    //printf("%lld %lld %d %d\n",A[i],k-A[i],temp,ans);
  }

  printf("%d", ans);
  return 0;
}