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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;

public class Main{
    
    static int[] v,d,c,result,used;
    
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.next());
        int[] a = new int[n];
        
        ArrayList<Integer> nibeki = new ArrayList<>();
        for(long i=2; i<2_000_000_000; i*=2){
            nibeki.add((int)i);
        }
        
        List<Integer> s = new ArrayList<>();
        Map<Integer,Integer> map = new TreeMap<>();
        Map<Integer,Integer> snt = new TreeMap<>();
        for(int i=0; i<n; i++){
            a[i] = Integer.parseInt(sc.next());
            if(a[i]>1 && Integer.bitCount(a[i])==1){
                map.merge(a[i], 1, Integer::sum);
            }else{
                s.add(a[i]);
                snt.merge(a[i], 1, Integer::sum);
            }
        }
        Collections.sort(s, Comparator.reverseOrder());
        int ans = 0;
        for(int i=0; i<s.size(); i++){
            int tmp = s.get(i);
            for(int j=0; j<nibeki.size(); j++){
                if(snt.get(nibeki.get(j)-tmp)!=null
                    && snt.get(nibeki.get(j)-tmp)>=1){
                    
                    snt.merge(tmp, -1, Integer::sum);
                    snt.merge(nibeki.get(j)-tmp, -1, Integer::sum);
                    ans++;
                }
            }
            snt.merge(tmp, 1, Integer::sum);
        }
        for(int key : map.keySet()){
            int tmp = map.get(key);
            ans += tmp/2;
        }
        System.out.println(ans);
    }
}

 
class BipartiteMatching{
    
    public static final int INF = (int)1e9;
    
    public static int n;
    private static List<Integer>[] edge;
    private static int[] match;
    private static boolean[] used;
    
    public BipartiteMatching(int n){
        this.n = n;
        edge = new List[n];
        for(int i=0; i<n; i++){
            edge[i] = new ArrayList<>();
        }
        match = new int[n];
        used = new boolean[n];
    }
    
    public int bipartiteMatching(){
        int res = 0;
        Arrays.fill(match, -1);
        for(int v=0; v<n; v++){
            if(match[v] < 0){
                Arrays.fill(used, false);
                if(dfs(v)){
                    res++;
                }
            }
        }
        return res;
    }
    
    public boolean dfs(int v){
        used[v] = true;
        for(int i=0; i<edge[v].size(); i++){
            int u = edge[v].get(i);
            int w = match[u];
            if( w < 0 || (!used[w] && dfs(w)) ){
                match[v] = u;
                match[u] = v;
                return true;
            }
        }
        return false;
    }
    
    public void addEdge(int u, int v){
        edge[u].add(v);
        edge[v].add(u);
    }
    
}