Summarize to the Power of Two（CodeForces - 1005C ）

最新推荐文章于 2018-12-31 23:08:06 发布

Baiyi_destroyer

最新推荐文章于 2018-12-31 23:08:06 发布

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分类专栏：水题集容器

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本文介绍了一种算法，旨在通过删除最少数量的元素使序列达到特定条件：对于序列中的每个元素，都存在另一个不同位置的元素使得两者的和为2的幂。

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A sequence a1,a2,…,ana1,a2,…,an is called good if, for each element aiai, there exists an element ajaj (i≠ji≠j) such that ai+ajai+aj is a power of two (that is, 2d2d for some non-negative integer dd).

For example, the following sequences are good:

[5,3,11][5,3,11] (for example, for a1=5a1=5 we can choose a2=3a2=3. Note that their sum is a power of two. Similarly, such an element can be found for a2a2 and a3a3),
[1,1,1,1023][1,1,1,1023],
[7,39,89,25,89][7,39,89,25,89],
[][].

Note that, by definition, an empty sequence (with a length of 00) is good.

For example, the following sequences are not good:

[16][16] (for a1=16a1=16, it is impossible to find another element ajaj such that their sum is a power of two),
[4,16][4,16] (for a1=4a1=4, it is impossible to find another element ajaj such that their sum is a power of two),
[1,3,2,8,8,8][1,3,2,8,8,8] (for a3=2a3=2, it is impossible to find another element ajaj such that their sum is a power of two).

You are given a sequence a1,a2,…,ana1,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 nn (1≤n≤1200001≤n≤120000) — the length of the given sequence.

The second line contains the sequence of integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤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 nn elements, make it empty, and thus get a good sequence.

Examples

Input

6
4 7 1 5 4 9

Output

Input

5
1 2 3 4 5

Output

Input

1
16

Output

Input

4
1 1 1 1023

Output

Note

In the first example, it is enough to delete one element a4=5a4=5. The remaining elements form the sequence [4,7,1,4,9][4,7,1,4,9], which is good.

题解：任意两个数相加为2的n次方，刚开始用两个数组做wa了，因为没考虑两个重复数构成的情况。。。

代码如下：

#include <iostream>
#include <cstdio>
#include <stdlib.h>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <string.h>
#include <vector>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <ctime>
#define maxn 55557
#define N 100005
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lowbit(x) (x&(-x))
#define eps 0.000000001
#define read(x) scanf("%d",&x)
#define put(x) printf("%d\n",x)
#define Debug(x) cout<<x<<" "<<endl
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
typedef long long ll;

int a[120007];
map<ll,ll>m;
int main()
{
    ll x[120007];
    x[0]=2;
    for(int i=1; i<31; i++)
        x[i]=x[i-1]*2;
    int n;
    cin>>n;
    for(int i=0; i<n; i++)
    {
        cin>>a[i];
        m[a[i]]++;
    }
    sort(a,a+n);
    int sum=0;
    for(int i=0; i<n; i++)
    {
        int flag=0;
        for(int k=0; k<30; k++)
        {
            if(a[i]<x[k])
            {
                int asd=x[k]-a[i];
                if(m[asd]>0)
                {
                    if(a[i]==asd)
                    {
                        if(m[asd]>1)
                        {
                            flag=1;
                            break;
                        }
                    }
                    else
                    {
                        flag=1;
                        break;
                    }
                }

            }
        }
        if(!flag)
            sum++;
    }
    cout<<sum<<endl;
    return 0;
}