CodeForces742b

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.

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. 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.

异或的逆运算还是异或
坑是特判0的情况

#include<iostream>
#include<cmath>
using namespace std;
int tu[1000001];
int yingshe[1000001];
int main()
{
#define int long long int n, x;cin >> n >> x;for (int a = 1;a <= n;a++){scanf("%I64d", &tu[a]);yingshe[tu[a]] ++;}int jieguo = 0;for (int a = 1;a <= n;a++){if ((tu[a] ^ x) > 100000)continue;if (x == 0){jieguo += (yingshe[x^tu[a]]-1);continue;}if (yingshe[x^tu[a]]){jieguo +=yingshe[x^tu[a]];}}cout << jieguo / 2 << endl;return 0;
}

CodeForces742b

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