加法要遍历两遍所有元素，那就用减法，只遍历一遍即可（思想）

C. Summarize to the Power of Two

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A sequence $a_1,a_2,\dots ,a_n$ is called good if, for each element $a_i$, there exists an element $a_j$ ($i\ne j$) such that $a_i+a_j$ is a power of two (that is, $2^d$ for some non-negative integer $d$).

For example, the following sequences are good:

• $[5,3,11]$ (for example, for $a_1=5$ we can choose $a_2=3$. Note that their sum is a power of two. Similarly, such an element can be found for $a_2$ and $a_3$),
• $[1,1,1,1023]$,
• $[7,39,89,25,89]$,
• $\left[\right]$.

Note that, by definition, an empty sequence (with a length of $0$) is good.

For example, the following sequences are not good:

• $\left[16\right]$ (for $a_1=16$, it is impossible to find another element $a_j$ such that their sum is a power of two),
• $[4,16]$ (for $a_1=4$, it is impossible to find another element $a_j$ such that their sum is a power of two),
• $[1,3,2,8,8,8]$ (for $a_3=2$, it is impossible to find another element $a_j$ such that their sum is a power of two).

You are given a sequence $a_1,a_2,\dots ,a_n$. 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 $n$ ($1\le n\le 120000$) — the length of the given sequence.

The second line contains the sequence of integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^9$).

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

Copy

6
4 7 1 5 4 9

output

Copy

1

input

Copy

5
1 2 3 4 5

output

Copy

2

input

Copy

1
16

output

Copy

1

input

Copy

4
1 1 1 1023

output

Copy

0

Note

In the first example, it is enough to delete one element $a_4=5$. The remaining elements form the sequence $[4,7,1,4,9]$, which is good.

用加法，用暴力一个个找能匹配的元素，并标记找到的元素，三重循环，超时

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#include<bits/stdc++.h>
using namespace std;
int a[120005];
map<int,int>vis;
int main()
{
    int n;
    int i,j,k;
    for(i=0; i<n; i++)
    {
        cin>>a[i];
    }
    int flag=0;
    for(i=0; i<n; i++)
    {
        flag=0;
        if(vis[a[i]]!=1)
        {

            for(j=0; j<n; j++)
            {
                if(flag==1)
                    break;
                for(k=1; k<34; k++)
                {
                    if(a[i]+a[j]==pow(2,k))
                    {
                        flag=1;
                        vis[a[i]]=1;
                        vis[a[j]]=1;
                        break;
                    }
                }

            }
        }
    }
int sum=0;
for(i=0;i<n;i++)
    if(vis[a[i]]!=1)
    sum++;
cout<<sum;
}

只是判断对应的匹配元素存不存在，不用管是多少，只需要两重循环，还是用map，用减法

#include<bits/stdc++.h>
using namespace std;
int a[120005];
map<int,int >vis;
int main()
{
    int n;
    scanf("%d",&n);
    int i,j,k;
    for(i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
        vis[a[i]]++;
    }
    int sum=0;
   for(i=0;i<n;i++)
   {
       
    for(j=1;j<32;j++)
   {
       if(vis[pow(2,j)-a[i]]>1||(vis[pow(2,j)-a[i]]==1&&pow(2,j)!=2*a[i]))
        {sum++;
          break;
        }
   }
   }
   printf("%d",n-sum);
}