## CodeForces - 1005C Summarize to the Power of Two

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### 题目描述

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.

### 代码实现

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[200000];
int main()
{
map<ll,ll> ma;
ll n,i,k,ans=0;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i];
ma[a[i]]++;//对a[i]出现的次数进行计数
}
for(int i=0;i<n;i++)
{
ll er=1,g=0,j=0;
for(int j=0;j<31;j++)
{
if(ma[er-a[i]]!=0)//ma[er-a[i]]不为零说明差值在数列中存在
{
if(er-a[i]==a[i])//如果差值为它自己，就需要接着判断
{
if(ma[a[i]]>1)
g++;
}
else    g++;
}
er*=2;//er就是2的N次方
}
if(g==0)    ans++;//g为0则说明差值不存在或者差值与a[i]相等且a[i]只出现一次，即a[i]不是符合条件的数。
}
cout<<ans<<endl;
return 0;
}