位运算上的小技巧 - AtCoder

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.

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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.

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Sample Input 1

Copy

2
1 2 3 1

Sample Output 1

Copy

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) . When (i,j)=(1,2), A_i+A_j=2+3=5. This is the maximum value, so the answer is 5.

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Sample Input 2

Copy

3
10 71 84 33 6 47 23 25

Sample Output 2

Copy

81
94
155
155
155
155
155

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Sample Input 3

Copy

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

Sample Output 3

Copy

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

题意 ： 给你 一堆数，并设定一个变量 k，要求 k 是从 1 开始递增的，每次从不超过 k 的序列中位置上取两个数 i, j， 且要求 i or j <= k

思路分析：对于一个k, 寻找小于等于 k 中

代码示例：

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;
}