題目描述

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.

輸入描述:

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

輸出描述:

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

輸入

複製

3

輸出

複製

0 2 1

說明

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<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<queue>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll unsigned long long
#define MAX 1000000000
#define ms memset
#define maxn 500005
using namespace std;

int vis[maxn],idex[maxn];
int n;
int compute(int x)
{
    int cnt=1;
    while(cnt<=x)  cnt*=2;
    return x^(cnt-1);
}
/*
題目大意:給定一個n,要求0到n-1中，
的完全匹配代價最小的序列。（就是把0到n-1重拍然後使得對應位置和數相and均為零）。

從大到小遍歷，（不能從小到大，因為零可能對應1，或者111等等）。
對每個數按位取反（用全為1的數異或即可，詳見程式碼），
然後標記每個位置，類似篩法的思想，標過的不用再標記，然後繼續遍歷下去。
*/

int main()
{
    scanf("%d",&n);    n--;
    memset(idex,0xff,sizeof(idex));
    for(int i=n;i>=0;i--)
    {
        int sta=compute(i);
        if(idex[i]!=-1) continue;
        idex[sta]=i , idex[i]=sta;
    }

    for(int i=0;i<=n;i++) printf("%d%c",idex[i],i==n?'\n':' ');
    return 0;
}