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;
typedef long long ll;
map<int,int>mp;
int a[200000+100];
int power[40];
int len=0;
int main()
{
    int n;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
        scanf("%d",&a[i]),mp[a[i]]++;
    sort(a,a+n);
    for(ll i=2;i<=2000000000;i*=2)
        power[len++]=i;
    ll sum=0;
    for(int i=n-1;i>=0;i--)
    {
        for(int k=0;k<len;k++)
        {
            
            if(power[k]>a[i])
            {
                int q=a[i];
                int b=power[k]-q;
                int c,d;
                
                c=mp[q];
                d=mp[b];
                if(d!=0 && c!=0)
                {
                    //    printf("%d %d %d %d\n",q,b,c,d);
                    if(b==q)
                        sum+=((ll)c*(ll)(d-1));
                    else
                        sum+=(ll)c*(ll)d;
                    mp[q]=mp[b]=0;
                }
            }
        }
    }
    printf("%lld\n",sum);
    return 0;
}