Problem Statement

There is an integer sequence of length 2^N: A₀,A₁,…,A_2^N?1. (Note that the sequence is 0-indexed.)

For every integer K satisfying 1≤K≤2^N?1, solve the following problem:

• Let i and j be integers. Find the maximum value of A_i+A_j where 0≤i<j≤2^N?1 and (i or j)≤K. Here, or denotes the bitwise OR.

Constraints

• 1≤N≤18
• 1≤A_i≤10⁹
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A0 A1 … A2N?1

Output

Print 2^N?1 lines. In the i-th line, print the answer of the problem above for K=i.

Sample Input 1

2
1 2 3 1

Sample Output 1

3
4
5

For K=1, the only possible pair of i and j is (i,j)=(0,1), so the answer is A₀+A₁=1+2=3.

For K=2, the possible pairs of i and j are (i,j)=(0,1),(0,2). When (i,j)=(0,2), A_i+A_j=1+3=4. This is the maximum value, so the answer is 4.

For K=3, the possible pairs of i and j are (i,j)=(0,1),(0,2),(0,3),(1,2),(1,3),(2,3)

Sample Input 2

3
10 71 84 33 6 47 23 25

Sample Output 2

81
94
155
155
155
155
155

Sample Input 3

4
75 26 45 72 81 47 97 97 2 2 25 82 84 17 56 32

Sample Output 3

101
120
147
156
156
178
194
194
194
194
194
194
194
194
194

題意 ： 給你 一堆數，並設定一個變量 k，要求 k 是從 1 開始遞增的，每次從不超過 k 的序列中位置上取兩個數 i, j， 且要求 i or j <= k

思路分析：對於一個k, 尋找小於等於 k 中

代碼示例：

const int maxn = 1e6+5;
int n;
int pre[maxn];
int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    cin >> n;
    for(int i =0; i < (1<<n); i++){
        scanf("%d", &pre[i]);
    }
    
    int ans = 0;
    for(int i = 1; i < (1<<n); i++){
        int s1 = pre[0], s2 = 0;
        for(int j = i; j; j = (j-1)&i){
            if (pre[j] > s1) {s2 = s1, s1 = pre[j];}
            else if (pre[j] > s2) s2 = pre[j]; 
        }
        ans = max(ans, s1+s2);
        printf("%d\n", ans);
    } 
    
    return 0;
}

位運算上的小技巧 - AtCoder