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;
#define ll long long
int l[200000] = {0};

int main() {
    int N;
    map<int, int> d;

    scanf("%d", &N);
    for(int i=0;i<N;i++) {
        scanf("%d", &l[i]);
    }

    for(int i=0;i<N;i++) {
        auto itr = d.find(l[i]);
        if(itr != d.end()) {
            d[l[i]]++;
        }
        else {
            d[l[i]] = 1;
        }
    }

    int ans = 0;
    for(auto i=d.begin();i!=d.end();i++) {
        while(i->second > 0) {
            int isFinished = 0;
            for(int j=31;j>=1;j--) {
                int v = pow(2, j) - i->first;
                if (d.find(v)!=d.end()) {
                    if ((v!=i->first && d[v]>=1) || (v==i->first && d[v]>=2)) {
                        i->second--;
                        d[v]--;
                        ans++;
                        isFinished = 1;
                        break;
                    }
                }
            }
            if(isFinished==0) {
                break;
            }
        }
    }
    cout << ans << endl;

    return 0;
}