Minimum Cost Perfect Matching（牛客）

A、Minimum Cost Perfect Matching | 时间限制：1秒 | 内存限制：256M
You have a complete bipartite graph where each part contains exactly n nodes, numbered from 0 to n - 1 inclusive.

The weight of the edge connecting two vertices with numbers x and y is  (bitwise AND). 

Your task is to find a minimum cost perfect matching of the graph, i.e. each vertex on the left side matches with exactly one vertex on the right side and vice versa. The cost of a matching is the sum of cost of the edges in the matching.

 denotes the bitwise AND operator. If you're not familiar with it, see {https://en.wikipedia.org/wiki/Bitwise_operation#AND}.

 

输入描述:

The input contains a single integer n (1 ≤ n ≤ 5 * 105).

输出描述:

Output n space-separated integers, where the i-th integer denotes pi (0 ≤ pi ≤ n - 1, the number of the vertex in the right part that is matched with the vertex numbered i in the left part. All pi should be distinct.

Your answer is correct if and only if it is a perfect matching of the graph with minimal cost. If there are multiple solutions, you may output any of them.

示例1

输入

3

输出

0 2 1

说明

For n = 3, p0 = 0, p1 = 2, p2 = 1 works. You can check that the total cost of this matching is 0, which is obviously minimal.

#include <bits/stdc++.h>

using namespace std;

const int maxn = 5e5+5;

int main()
{
    int n;
    cin>>n;
    int now = n-1;
    int ans[maxn];
    while(1)
    {
        if(now<0)
            break;
        for(int i=now;i>=0;i--)
        {
            int t = i&now;
            if(t==0)
            {
                int cnt = (now-i+1)/2;
                for(int j=i,k=cnt;k>0;j++,k--)
                {
                    ans[j] = j+k*2-1;
                }
                for(int p=now,k=cnt;k>0;p--,k--)
                {
                    ans[p] = p-(k*2-1);
                }
                now = i-1;
                break;
            }
        }
    }
    for(int i=0;i<n;i++)
    {
        cout<<ans[i]<<" ";
    }
}