~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
题目描述
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.
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;
}
原文链接: https://www.cnblogs.com/zhgyki/p/9452637.html
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