B. Arpa’s obvious problem and Mehrdad’s terrible solution

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output
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 $B. Arpa’s obvious problem and Mehrdad’s terrible solution$ , where $B. Arpa’s obvious problem and Mehrdad’s terrible solution$ is bitwise xoroperation (see notes for explanation).
$B. Arpa’s obvious problem and Mehrdad’s terrible solution$
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.
Examples
input
2 3
1 2
output
1
input
6 1
5 1 2 3 4 1
output
2
Note

In the first sample there is only one pair of i = 1 and j = 2. $B. Arpa’s obvious problem and Mehrdad’s terrible solution$ so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since $B. Arpa’s obvious problem and Mehrdad’s terrible solution$ ) and i = 1, j = 5 (since $B. Arpa’s obvious problem and Mehrdad’s terrible solution$ ).

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.

题意：

找出给定数列里两两异或值为x的组合个数

a^b=x; x^a=b; x^b=a;

x已经给定我们只需循环a找是否有b与之对应。

附AC代码：

 1 #include<bits/stdc++.h> 
 2 using namespace std;
 3 int a[100010];
 4 int main(){
 5     map<int ,int > m;
 6     int n,x;
 7     cin>>n>>x;
 8     for(int i=1;i<=n;i++){
 9         scanf("%d",a+i);
10     }
11     long long ans=0;
12     for(int i=1;i<=n;i++){
13         if(m.count(x^a[i])) //找到返回1 否则返回0 
14         ans+=m[x^a[i]];
15         m[a[i]]++;
16     }
17     cout<<ans<<endl;
18     return 0;
19 }