C. Summarize to the Power of Two

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 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

  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 a4=5

  . The remaining elements form the sequence [4,7,1,4,9], which is good.

AC代码：

#include
#include

#define mod 1000000007
const int INF = 0x3f3f3f3f;
typedef long long ll;
const int maxn = 120050;
using namespace std;

int n;
int biao[31] = {1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824};
map mp;
bool judge(int num)
{
    for(int i = 0; i < 31; i++)
    {
        if(num == biao[i])
            return true;
    }
    return false;
}
int main()
{
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
    {
        int temp;
        scanf("%d", &temp);
        mp[temp]++;
    }
    int ans = 0;
    for(auto i : mp)
    {
        int t = i.first;
        bool flag = false;
        for(int j = 0; j <31; j++)
        {
            if(biao[j]-t == t)
            {
                if(mp.count(t) && mp[t] > 1)
                {
                    flag = true;
                    break;
                }
            }
            else if(mp.count(biao[j]-t))
            {
                flag = true;
                break;
            }
        }
        if(!flag)
            ans += i.second;
    }
    printf("%d\n", ans);
	return 0;
}