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>

void MergeSort(long *A,long r) {
    long *t,i=0,j=r,k=0,a;
    
    t=(long *)malloc((r+1)*sizeof(long));
    
    while(k<r+1) {
        if(A[i]<A[j]) a=i++;
        else a=j++;
        
        t[k]=A[a];
        k++;
        
        if(i==r) a=j;
        else if(j==r+1) a=i;
        else continue;
        
        while(k<r+1){
            t[k]=A[a];
            k++;
            a++;
        }
    }

    for(i=0;i<=r;i++) A[i]=t[i];
    free(t);
    return;
}

int main() {
    long A[200001];
    int B[200001];
    long N,i,k=1,s,m,e,c=0;

    scanf("%ld",&N);
    for(i=0;i<N;i++) {
        scanf("%ld",A+i);
        if(i!=0) MergeSort(A,i);
        B[i]=0;
    }

    while(k<A[N-1]) k*=2;
    k/=2;

    for(i=N-1;i>=1;i--) {
        if(B[i]==1) continue;
        while(k>A[i]) k/=2; 

        s=0;
        e=i-1;
        while(s<=e) {
            m=(s+e)/2;
            if(A[m]+A[i]==k*2) {
                if(B[m]==0) {
                    B[m]=1;
                    c++;
                }
                break;
            }
            if(A[m]+A[i]>k*2) e=m-1;
            else s=m+1;
        }
    }

    printf("%ld\n",c);
    return 0;
}