Codeforces Round #496 (Div. 3), problem: (C) Summarize to the Power of Two

#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define pii pair<int,int>
#define pli pair<long long int,int>
#define pbi pair<bool,int>
#define pob pop_back
#define all(a) a.begin(),a.end()
#define fio ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0);
#define i64 long long int
#define mem(x,y) memset(x,y,sizeof(x))
#define fill(arr,b) fill(arr, arr+sizeof(arr)/sizeof(arr[0]), b)
#define maxarr(n,arr,m) m =-1e9;for(int i=0;i<n;i++)if(arr[i]>m) m=arr[i];

const double pi = acos(-1.0);
const int maxn = (i64)2e9+5;
int arr[120005];
i64 power[40];
map<i64,int>mp;

void f()
{
    for(int i = 0 ; i<36;  i++)
    {
        power[i] = pow(2,i);
    }
}
int main()
{
    //freopen("input.txt","r",stdin);
    f();
    int n;
    scanf("%d",&n);
    for(int i = 0 ; i<n ; i++)  {scanf("%d",&arr[i]);mp[arr[i]]++;}
    int cnt = 0;
    for(int i = 0 ; i<n ; i++)
    {
        bool fg = false;
        for(int j = 0 ; j<36 ; j++)
        {
            i64 x = power[j]-arr[i];
            if(x==arr[i])
            {
                if(mp[x]>1) {fg=true;break;}
            }
            else if(mp[x]!=0) {fg=true;break;}
        }
        if(!fg) cnt++;
    }
    printf("%d",cnt);
    return 0;
}