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 <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <map>

using namespace std;
long long Pow[45];
long long all[200005];
struct A
{
    long long how;
    long long con;
}num[200005];
bool F(long long a,long long b)
{
    return a>b;
}
int main()
{
    long long N,i,j,now=0,ans=0,l,r,t;
    Pow[0]=1;
    for(i=1;i<=30;i++) Pow[i]=Pow[i-1]*2;
    scanf("%lld",&N);
    for(i=0;i<N;i++) scanf("%lld",&all[i]);
    sort(all,all+N,F);
    num[now].how=all[0];
    num[now].con=1;
    for(i=1;i<N;i++)
    {
        if(all[i]==all[i-1]) num[now].con++;
        else
        {
            now++;
            num[now].how=all[i];
            num[now].con=1;
        }
    }
    for(i=now;i>=0;i--)
    {
        for(j=30;j>=0;j--)
        {
            if(Pow[j]-num[i].how<=0) continue;
            if(Pow[j]-num[i].how==num[i].how)
            {
                ans+=num[i].con/2;
                num[i].con%=2;
            }
            else
            {
                l=0;
                r=i;
                while((r-l)>1)
                {
                    if(num[(r+l)/2].how>=Pow[j]-num[i].how) l=(r+l)/2;
                    else r=(r+l)/2;
                }
                if(num[l].how+num[i].how==Pow[j])
                {
                    //printf("%lld %lld %lld\n",num[l].how,num[i].how,Pow[j]);
                    t=min(num[l].con,num[i].con);
                    num[l].con-=t;
                    num[i].con-=t;
                    ans+=t;
                }
                else if(num[r].how+num[i].how==Pow[j])
                {
                    //printf("%lld %lld %lld\n",num[r].how,num[i].how,Pow[j]);
                    t=min(num[r].con,num[i].con);
                    num[r].con-=t;
                    num[i].con-=t;
                    ans+=t;
                }
            }
        }
    }
    printf("%lld\n",ans);
    return 0;
}