C - Arpa’s obvious problem and Mehrdad’s terrible solution
Codeforces Round #383 (Div. 2)
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
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
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.
In the second sample the only two pairs are i = 3, j = 4 (since
A bitwise xor takes two bit integers of equal length and performs the logical xoroperation 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：给你一个序列和一个数x，求这个序列中有多少个数对的异或结果为x
解法：数列中的数有重复的，而且暴力会超时，已知xy=z，则xz=y，所以只要判断一个数与目标数的异或结果在不在序列中即可
代码：
#include<iostream>
using namespace std;
int a[100010];
int main()
{
int n,x,t;
long long cnt=0;
cin>>n>>x;
for(int i=0;i<n;i++){
cin>>t;
if(a[x^t])
cnt+=a[x^t];
a[t]++;
}
cout<<cnt<<endl;
return 0;
}
