AtCoder Regular Contest E - Or Plus Max

Time limit : 2sec / Memory limit : 1024MB

Score : 700 points

Problem Statement

There is an integer sequence of length 2N: A0,A1,…,A2N−1. (Note that the sequence is 0-indexed.)

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

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

Constraints

• 1≤N≤18
• 1≤Ai≤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 2N−1 lines. In the i-th line, print the answer of the problem above for K=i.

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), Ai+Aj=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), Ai+Aj=2+3=5. This is the maximum value, so the answer is 5.

Sample Input 2

Copy

3
10 71 84 33 6 47 23 25

Sample Output 2

Copy

81
94
155
155
155
155
155

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要求max(Ai+Aj) (0≤i<j≤2N−1 && i|j≤k)，显然i <= k && j <= k,只要求max(Ai+Aj) (0≤i<j≤2N−1 && i|k=k && j|k=k),然后，就可以得出想要的结果，i|k=k说明i是在k的基础上二进制位中的0保持不变，改变1，把1变为0，则i < k,这样求前缀，然后排着往后更新最大值即可。代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#define Max 1 << 19
using namespace std;
int n,e,s[Max];
int r[Max][2];///记录最大值下标
bool cmp(int a,int b){
    return s[a] < s[b];
}
int main(){
    scanf("%d",&n);
    e = 1 << n;///总数个数
    for (int i = 0;i < e;i ++)
        scanf("%d",&s[i]);
    s[e] = -1;
    r[0][0] = 0;///初始化
    r[0][1] = e;
    for (int i = 1;i < e;i ++){
        r[i][0] = i;///初始化
        r[i][1] = e;
        int x[4];
        for (int j = 0;j < n;j ++)
            if (i & (1 << j)){///此位为1
                int d = i ^ (1 << j);///把这位变成0 对于d的情况前面已经确定过 这里直接用
                x[0] = r[i][0];
                x[1] = r[i][1];
                x[2] = r[d][0];
                x[3] = r[d][1];
                sort(x,x + 4,cmp);///按照对应位置值从小到大排列下标
                r[i][0] = x[3];///存最大值下标
                r[i][1] = x[2] == x[3] ? x[1] : x[2];///存非重复下标
            }
    }
    int ans = 0;
    for (int i = 1;i < e;i ++){
        ans = max(ans,s[r[i][0]] + s[r[i][1]]);///前缀更新i的最大值
        printf("%d\n",ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/8023spz/p/9456664.html