XOR——思维

题目描述 
Once there was a king called XOR, he had a lot of land. Because of his name, he likes to play XOR games.
One day, he whimpered and wanted to establish N cities on that vast expanse of land, numbered 0, 1, 2..., N-1. He wanted to connect all the cities. If city A can reach City B through zero or one or several cities, then A and B are connected. The cost of repairing a road in City A and City B is the XOR value of number of City A and number of City B. This King XOR wanted to figure out the minimum cost for connecting all of the N cities.
Of course, like a fairy tale read as a child, there will be corresponding rewards after helping the king. If you help the king solve his problems, he will improve your ranking in the competition.
输入描述:
There are multi test cases
each test cases contains an integer N (2 ≤N≤ 20000), the number of cities the king wants to establish.
输出描述:
For each test case, print the minimum cost for connecting all of the N cities in one line.
示例1
输入
4
输出
4
说明

The weightof the minimum cost is 1+2+1=4 In the Sample Example.

思路：找一个最小生成树，节点0~n-1之间距离的权值是抑或和~

        由此我们不难猜出，想要异或和最小，那么最好是差的不一样的位数是1个位最好，通过枚举找规律可知，每个距离的权值是最低位1决定的。例：

如样例 n=4 那就是0-3之间连接;
0=0000
1=0001  ->最低位1在0 所以和0连接 总值+1（2^0)
2=0010  ->最低位1在1 所以和0连接 总值+2（2^1)
3=0011  ->最低位1在1 所以和2连接 总值+1（2^1）

最后答案为4;

那么简单方法就是直接用lowbit函数（最低位1的和）：

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        long long sum=0;
    for(int i=0;i<n;i++)
    {
        sum+=i&(-i);
    }
    cout<<sum<<endl;
    }
}