Arpa’s obvious problem and Mehrdad’s terrible solution 思维
There are some beautiful girls in Arpa’s land as mentioned before.
Once Arpa came up with an obvious problem:
Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that , where
is bitwise xor operation (see notes for explanation).
Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.
Input
First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.
Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.
Output
Print a single integer: the answer to the problem.
Example
2 3
1 2
1
6 1
5 1 2 3 4 1
2
Note
In the first sample there is only one pair of i = 1 and j = 2. so the answer is 1.
In the second sample the only two pairs are i = 3, j = 4 (since ) and i = 1, j = 5 (since
).
A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.
这道题一开始没看懂啥意思,后来才明白原来是异或^,a^b=c,那么a^c=b;抓住这一点就好做了,所有的ai去异或x得到cnt,然后记录cnt,看看数组里面有几个cnt就好了,可以开个map,注意结果可能超过long 要用long long
代码:
#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
int n,x,a[];
map<int,int> mark;
void input()
{
cin>>n>>x;
for(int i=;i<n;i++)
{
cin>>a[i];
mark[a[i]]++;
}
}
long long check()
{
long long cnt,ans=;
for(int i=;i<n;i++)
{
cnt=a[i]^x;
///x如果是0 那么 任意一个数q q^0=q;
if(a[i]==cnt)ans+=mark[cnt]-;
else ans+=mark[cnt];
}
return ans/;
}
int main()
{
input();
cout<<check();
}
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