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>
using namespace std;


int main()
{
    cin.sync_with_stdio(0);

    int n; cin>>n;
    vector<int> a(n);
    map<long, int> m;
    for(int i=0; i<n; ++i) {
        cin>>a[i];
        m[a[i]]++;
    }
    vector<int> c(n);
    for(int i=0; i<n; ++i) {
        for(long j=2; j<1L<<32; j*=2) {
            if (m.count(j-a[i])==0) continue;
            c[i]+=m[j-a[i]];
            if (a[i]==j-a[i]) c[i]--;
        }
    }

    vector<pair<int, int>> b;
    for(int i=0; i<n; ++i) b.emplace_back(c[i], i);
    sort(begin(b), end(b));

    int res=0;
    for(auto& e: b) {
        int i=e.second;
        if (m[a[i]]==0) continue;
        m[a[i]]--;
        for(long j=2; j<1L<<32; j*=2) {
            if (m.count(j-a[i])==0) continue;
            if (m[j-a[i]]) {
                ++res;
                m[j-a[i]]--;
                break;
            }
        }
    }
    cout<<res<<endl;
}