2018牛客网暑期ACM多校训练营（第七场）A Minimum Cost Perfect Matching

链接：https://www.nowcoder.com/acm/contest/145/A
来源：牛客网
 

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld

题目描述

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.

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.

输入

复制

3

输出

复制

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.

与运算，二进制找下规律

AC代码：

#include<bits/stdc++.h>
using namespace std;
int a[500010];

int get(int temp){
	int j=1;
	while(temp){
		temp/=2;
		j*=2;				
	}
	return j-1;
}

int main(){
	int n;
	scanf("%d",&n);
	if(n%2==1){
		
		int num=get(n-1);
		int up=n-1;
		while(num){
			int j=num-up;
			int temp=num-up;
			for(int i=up;i>=num-up;i--){
				a[i]=j++;
			}
			num=get(num-up-1);
			up=temp-1;
		}
		
				
	}
	else{
		a[n-1]=0;
		a[0]=n-1;
		int num=get(n-2);
		int up=n-2;
		while(num){
			int j=num-up;
			int temp=num-up;
			for(int i=up;i>=num-up;i--){
				a[i]=j++;
			}
			num=get(num-up-1);
			up=temp-1;
		}
	}
	for(int i=0;i<n-1;i++){
		printf("%d ",a[i]);
	}
	printf("%d\n",a[n-1]);
}