Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution

本文介绍了一种解决特定问题的方法：给定一个整数数组和一个整数x，求解数组中元素对(i, j)，使得ai XOR aj = x的数量。通过使用哈希表记录每个可能的异或结果出现的次数，可以有效地解决该问题。

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题意：给你n个数，问有对数异或后为x。(n, x, a[i] <= 1e5)

思路：a[i] ^ a[j] = x --> a[j]^x = a[i], 读入的时候找到前面有多少个对应的a[j]^x加上即

可。找的时候最大可能是2^17-1，所以数组要开的大一点。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6+5;
typedef long long ll;
int book[maxn], n, x;

int main(void)
{
    while(cin >> n >> x)
    {
        memset(book, 0, sizeof(book));
        ll ans = 0;
        for(int i = 1; i <= n; i++)
        {
            int temp;
            scanf("%d", &temp);
            ans += book[temp^x];
            book[temp]++;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

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 $\text{[math]}$ , where $\text{[math]}$ is bitwise xoroperation (see notes for explanation).

$\text{[math]}$

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. $\text{[math]}$ so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since $\text{[math]}$ ) and i = 1, j = 5 (since $\text{[math]}$ ).

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.