题目
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 ≤ 10^5, 0 ≤ x ≤ 10^5) — the >number of elements in the array and the integer x.Second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 10^5) — 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. 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.
题意
输入n和x,和n个数字,问n个数字两两异或结果为x的组数
解题思路
数据范围是1 ≤ n ≤ 10^5,直接用双重循环,时间复杂度大概是n^2会超时。
a^b=c
c^a=b
所以,每输入一个数,就把它和x异或,将异或出现结果的次数存起来
代码
#include <iostream>
#include <vector>
#include <map>
using namespace std;
int main()
{
int n, x;
cin >> n >> x;
map<int, long long>num;
long long ans = 0;
for (int i = 0; i < n; i++)
{
int a;
cin >> a;
ans += num[a];
num[a^x]++;//将异或结果出现的次数记下来
}
cout << ans << endl;
return 0;
}