eric bre rom delete there set while == []

A sequence a1,a2,…,an is called good if, for each element ai, there exists an element aj (i≠j) such that ai+aj is a power of two (that is, 2d for some non-negative integer d).

For example, the following sequences are good:

[5,3,11] (for example, for a1=5 we can choose a2=3. Note that their sum is a power of two. Similarly, such an element can be found for a2 and a3),

[1,1,1,1023],
[7,39,89,25,89],
[].
Note that, by definition, an empty sequence (with a length of 0) is good.

For example, the following sequences are not good:

[16] (for a1=16, it is impossible to find another element aj such that their sum is a power of two),
[4,16] (for a1=4, it is impossible to find another element aj such that their sum is a power of two),

[1,3,2,8,8,8] (for a3=2, it is impossible to find another element aj such that their sum is a power of two).
You are given a sequence a1,a2,…,an. What is the minimum number of elements you need to remove to make it good? You can delete an arbitrary set of elements.

Input
The first line contains the integer n (1≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers a1,a2,…,an (1≤ai≤109).

Output
Print the minimum number of elements needed to be removed from the given sequence in order to make it good. It is possible that you need to delete all n elements, make it empty, and thus get a good sequence.

Examples
Input
6
4 7 1 5 4 9
Output
1
Input
5
1 2 3 4 5
Output
2
Input
1
16
Output
1
Input
4
1 1 1 1023
Output
0
Note
In the first example, it is enough to delete one element a4=5. The remaining elements form the sequence [4,7,1,4,9], which is good.

#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,x,n) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long ll;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const ll LNF = 1e18;
const int N = 1e3 + 20;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
int dir[4][2] = {{0,1},{0,-1},{-1,0},{1,0}};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n;
int a[maxm];
map<int,int> m;
/*
*/
int main()
{
    while(cin>>n)
    {
        int f,c=0;
        m.clear();
        ms(a,0);

        rep(i,0,n)
        {
            scanf("%d",&a[i]);
            m[a[i]]++;
        }
        rep(i,0,n)
        {
            f=0;
            rep(j,0,31)
            {
                int t = 1<<j;
                int tt = t - a[i];
                if(m[tt])
                {
                    if(t - a[i] == a[i]) //找到相同的比如4 4/8 8
                    {
                        if(m[tt] > 1) //相同的不止有一個
                        {
                            f = 1;
                            break;
                        }
                    }
                    else
                    {
                        f = 1;
                        break;
                    }
                }
            }
            if(!f) c++;
        }
        cout<<c<<endl;
    }
}
/*
1 2 4 8 16 32 64
1 3 5
2 4

4 7 1 5 4 9
1 4 4 5 7 9

*/

CF 1005C Summarize to the Power of Two 【hash/STL-map】