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 <bits/stdc++.h>
#define MAXN 1000000005
using namespace std;
typedef long long ll;


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    
    long n,i,j;
    cin>>n;
    ll a[n+1];
    map<ll,ll> c;
    ll ans = 0;

    for(i=1;i<=n;i++)
    {
        cin>>a[i];
        c[a[i]]++;
    }
    vector<ll> po2,vec;
    for(i=1;i<MAXN;i<<=1)
    {
        po2.push_back(i);
        // cout<<i<<" ";
    }
    for(auto it: c)
    {
        vec.push_back(it.first);
    }
    sort(vec.begin(),vec.end());
    for(i=vec.size()-1;i>=0;i--)
    {
        // cout<<c[vec[i]]<<": ";
        if(c[vec[i]]>0)
        {
            for(j=po2.size()-1;j>=0;j--)
            {
                if(po2[j]-vec[i]>=0 && c[po2[j]-vec[i]]>0)
                {
                    // cout<<po2[j]-vec[i]<<endl;
                    if(vec[i]!=po2[j]-vec[i])
                    {
                        ll z = min(c[vec[i]],c[po2[j]-vec[i]]);
                        ans += z;
                        c[vec[i]] -= z;
                        c[po2[j]-vec[i]] -= z;
                        if(c[vec[i]]<=0)
                        break;
                    }
                    else
                    {
                        if(c[vec[i]]>1)
                        {
                            ll z = c[vec[i]]/2;
                            ans += z;
                            c[vec[i]] -= z*2;
                            if(c[vec[i]]<=0)
                            break;
                        }
                    }
                }
            }
        }
        // cout<<endl;
    }
    cout<<ans;
    
    return 0;
}