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: Accepted)

#include <iostream>
#include <cassert>
#include <climits>
#include <bitset>
#include <stack>
#include <queue>
#include <iomanip>
#include <limits>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>

using namespace std;

typedef long long ll;
typedef pair<ll,ll> P;

long long int INF = 1e18;
long long int mod = 1000000007;
double Pi = 3.1415926535897932384626;

vector<ll> G[500005];
vector<P> tree[500010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop

int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
char dir[] = "RULD";
//↓,→,↑,←

#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<x;
#define re(i,n) for(i=0;i<n;i++);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define deba(x) cout<< #x << " = " << x <<endl

ll rui(ll number1,ll number2){
    if(number2 == 0){
        return 1;
    }else{
        ll number3 = rui(number1,number2 / 2);
        number3 *= number3;
        number3 %= mod;
        if(number2%2==1){
            number3 *= number1;
            number3 %= mod;
        }
        return number3;
    }
}
ll gcd(ll number1,ll number2){
    if(number1 > number2){
        swap(number1 , number2);
    }
    if(number1 == 0 || number1 == number2){
        return number2;
    }else{
        return gcd(number2 % number1 , number1);
    }
}


ll i,j,k,ii,jj;
ll n,m;
ll a,b,c,d,e,g,h,r;
ll l,ans,num;
ll x[500005],y[500005],z[500005];
bool dame,flag;
char s[500005],t[500005];
bool used[500005];

int main(){
    cin >> n;
	map<ll,ll> kazu;
	for(i=0;i<n;i++){
		cin >> x[i];
		kazu[x[i]]++;
	}
	sort(x,x+n);
	reverse(x,x+n);
	for(i=0;i<n;i++){
		//pe(x[i]);p(kazu[x[i]]);
		if(kazu[x[i]] <= 0)continue;
		for(j=30;j>=0;j--){
			num = (1 << j);
			//p(num);
			if(num < x[i])break;
			if(num == x[i] * 2){
				if(kazu[x[i]] >= 2){
					kazu[x[i]] -= 2;
					ans++;
					break;
				}
			}else{
				if(kazu[num - x[i]] >= 1){
					kazu[x[i]]--;
					kazu[num-x[i]]--;
					ans++;
					break;
				}
			}
		}
	}
	
	p(ans);
    return 0;
}