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;
using Int = long long;
signed main(){
    Int N;
    cin >> N;
    vector<Int>A(N);
    for(auto& e : A)
        cin >> e;
    map<Int,Int>mp,cnt;
    for(auto e : A){
        for(Int i = 1; i < 32; ++i){
            Int bit = (1LL << i);
            if( bit > e and cnt.find(bit - e) != cnt.end()){
                mp[bit - e]++;
            }
        }
        mp[e]++;
        cnt[e]++;
    }
    using tp = tuple<Int,Int>;
    vector<tp>data;
    for(auto e : A)
        data.emplace_back(mp[e],e);
    sort(data.begin(),data.end());
    Int ans = 0;
    for(auto p :  data){
        Int rank, key;
        tie(rank,key) = p;
        if(cnt[key] == 0) continue;
        Int trank = N + 1,tkey = -1;
        for(Int i = 1; i < 32; ++i){
            Int bit = (1LL << i);
            if(bit > key){
                Int dif = bit - key;
                if(mp.find(dif) != mp.end() and ( dif == key ? cnt[key] > 1 : cnt[dif] > 0))
                    trank =  mp[dif], tkey = dif;
            }
        }
        if(tkey == -1) continue;
        cnt[tkey]--; cnt[key]--;
        mp[key]--; mp[tkey]--;
        ++ans;
    }
    cout << ans << endl;
}