CodeForces 2018-6-1 div3 D. Points and Powers of Two-白红宇

CodeForces 2018-6-1 div3 D. Points and Powers of Two

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发布时间：2019-05-26

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There are $n$ distinct points on a coordinate line, the coordinate of $i$ -th point equals to $x_{i}$ . Choose a subset of the given set of points such that the distance between each pair of points in a subset is an integral power of two. It is necessary to consider each pair of points, not only adjacent. Note that any subset containing one element satisfies the condition above. Among all these subsets, choose a subset with maximum possible size.

In other words, you have to choose the maximum possible number of points $x_{i_{1}}, x_{i_{2}}, \dots, x_{i_{m}}$ such that for each pair $x_{i_{j}}$ , $x_{i_{k}}$ it is true that $| x_{i_{j}} - x_{i_{k}} | = 2^{d}$ where $d$ is some non-negative integer number (not necessarily the same for each pair of points).

Input

The first line contains one integer $n$ ( $1 \leq n \leq 2 \cdot 10^{5}$ ) — the number of points.

The second line contains $n$ pairwise distinct integers $x_{1}, x_{2}, \dots, x_{n}$ ( $- 10^{9} \leq x_{i} \leq 10^{9}$ ) — the coordinates of points.

Output

In the first line print $m$ — the maximum possible number of points in a subset that satisfies the conditions described above.

In the second line print $m$ integers — the coordinates of points in the subset you have chosen.

If there are multiple answers, print any of them.

Examples

Input

Copy

63 5 4 7 10 12

Output

Copy

37 3 5

Input

Copy

5-1 2 5 8 11

Output

Copy

Note

In the first example the answer is $[7, 3, 5]$ . Note, that $| 7 - 3 | = 4 = 2^{2}$ , $| 7 - 5 | = 2 = 2^{1}$ and $| 3 - 5 | = 2 = 2^{1}$ . You can't find a subset having more points satisfying the required property.

这题主要难点在于数学推理，得出最多只有3个点，

不妨设最小的数 x , x + 2^d1, x+2^d1+2^d2 (d1 <= d2)

那么 x + 2^d3 = x+2^d1+2^d2 -> 2^d3 = 2^d1 +2^d2 提取 2^d1 可得 2^d1*(2^(d2-d1)+1) = 2^d3

可得d2-d1必须为0 即d1 == d2 所以 d2 = d1 = d3 - 1

这是证明了可以满足3个点的，可以类似的证明当有四个点的时候 2^d1( 2^(d2-d1) +2^(d3-d1) +1 ) = 2^d4

也就是需要内部的 2^(d2-d1) +2^(d3-d1) +1 为2的幂次又 (假设d1 <= d2 <= d3) 则 2^(d2-d1) +2^(d3-d1) +1式始终为奇数，不是2的幂次，也就是上式永远不会满足，故最多只可能是三个点满足题意。

实现：

#include
   
    #include
    
     #include
     
      #include
      #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #include
            
             using namespace std;typedef long long LL;const int N = 200000+5;int a[N];int f[40];int pow2(int a,int b){ int ans = 1; while(b&1 != 1){ b >>= 1; a *= a; } while(b){ if(b&1 == 1){ ans *= a; } b >>= 1; a *= a; } return ans;}bool bin_ser(int a[],int key, int n){ int l = 1, r = n; while(l <= r){ int m = l + (r-l)/2; if(a[m] < key) l = m+1; else if(a[m] > key) r = m-1; else return true; } return false;}int main() { int n; f[0] = 1; for(int i = 1; i < 31; ++i){ f[i] = f[i-1]*2; } scanf("%d",&n); for(int i = 1; i <= n; ++i){ scanf("%d",&a[i]); } bool flag = false; sort(a+1,a+n+1); if(!flag) for(int i = 1; i <= n;++i){ for(int j = 0; j < 30; ++j){ if(bin_ser(a,a[i]+f[j],n) && bin_ser(a,a[i]+f[j+1],n)){ flag = true; printf("3\n%d %d %d\n",a[i],a[i]+f[j],a[i]+f[j+1]); break; } } if(flag) break; } if(!flag) for(int i = 1; i <= n;++i){ for(int j = 0; j < 31; ++j){ if(bin_ser(a,a[i]+f[j],n)){ flag = true; printf("2\n%d %d\n",a[i],a[i]+f[j]); break; } } if(flag) break; } if(!flag) printf("1\n%d\n",a[1]); return 0; }

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