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 pb push_back
#define mp make_pair
#define fi first
#define se second
#define sz(a) (a).size() 
#define all(c) (c).begin(),(c).end()
#define rep(i,n) for(int i=0;i<n;i++)
#define forn(i,n) for(int i=1;i<=n;i++)
#define ford(i,n) for(int i=n;i>=0;i--) 
#define REP(i,a,n) for(int i=a;i<n;i++)
#define foreach(itr,c) for(__typeof((c).begin(),(c).end())itr=c.begin();itr!=c.end();itr++) 
#define fill(a,b) memset(a,b,sizeof(a))
#define re return
#define INF 1e9
#define LINF 1e18
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pi;
typedef vector<int> vi;
typedef vector<pi > vpi; 
long long pw[32];
void get_pw(){
	pw[0]=1;
	REP(i,1,32){
		pw[i]=pw[i-1]*2;
	}
	re;
}
int main(){
	int n;
	cin>>n;
	ll a[n];
	bool used[n];
	set<ll> s;
	map<ll,int> m;
	rep(i,n){
		cin>>a[i];
		s.insert(a[i]);
		m[a[i]]++;
	}
	sort(a,a+n);
	fill(used,0);
	int i=0;
	foreach(itr,m){
		itr->se=i;
		i++;
	}
	get_pw();
	int ans=0;
	ford(i,n-1){
		REP(j,1,32){
			if(a[i]>=pw[j-1]&&a[i]<pw[j]){
				if(s.find(pw[j]-a[i])!=s.end()){
					if(!used[m[pw[j]-a[i]]]){
						ans++;
						used[i]=1;
						used[m[pw[j]-a[i]]]=1;
						break;
					}
				}
			}
		}
	}
	cout<<ans<<endl;
	re 0;
}