~按位取反

~是按位取反运算符

这里先说一下二进制在内存的存储：二进制数在内存中以补码的形式存储

另外，正数的原码、补码和反码都相同

 

负数的反码与原码符号位相同，数值为取反；补码是在反码的基础上加1

比如：

~9的计算步骤：

转二进制：0 1001

计算补码：0 1001

按位取反：1 0110

转为原码：1 0110

按位取反：1 1001   反码

末位加一：1 1010   补码

符号位为1是负数，即-10

规律：~x=-（x+1）；

因此，t=~9（1001）并不能输出6（0110），而是-10；

 

牛客网暑假训练第七场A题：

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

Minimum Cost Perfect Matching

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 * 10

⁵

).

输出描述:

Output n space-separated integers, where the i-th integer denotes p

_i

 (0 ≤ p

_i

 ≤ n - 1, the number of the vertex in the right part that is matched with the vertex numbered i in the left part. All p

_i

 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

题意：给你一个n,让你输出一行与1~n-1 &起来 的值的和最小，如：0 2 1 （0&0+1&2+2&1=0

 

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define mod 1000000007
const int N=5e+5;
int a[N];
int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    int n;
    cin>>n;
    memset(a,-1,sizeof(a));
    for(int i=n-1;i>=0;i--)
    {
        int t=~i,k=0;//此时t是以补码输出的，是负数
     /***********************
        9      1001
       取反    0110  
          &    1111   
         得:   0110    
     */cout<<t;
        while(1<<k<i)k++;
        cout<<(1<<k)-1;
        t=t&((1<<k)-1);cout<<t<<endl;
        if(a[i]==-1)
            a[i]=t,a[t]=i;
    }
    for(int i=0;i<n;i++)
        cout<<a[i]<<" ";
    cout<<endl;
    return 0;
}