CodeForces - 742B

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.
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.
题目的意思：
给出n和m,n是接下来输入数字的数量，m是下面一行数字取出两个进行某种位运算后的数值，求出这样的对数的个数。某种位运算指相同为0，不同为1
如1-> 0001,2->0010. 1?2=0011,即3，有一对，输出1.

补了一下基础知识，位运算，桶排序。
& 按位与只有对应的两个二进位均为1时，结果位才为1 ，否则为0
| 按位或参与运算的两数各对应的二进位相或。只要对应的二个二进位有一个为1 时，结果位就为1
^ 按位异或参与运算的两数各对应的二进位相异或，当两对应的二进位相异时，结果为1
~按位取反 1变0，0变1.

很明显上面的某种运算是指按位异或，然而直接暴力会超时。
所以要用到按位运算的一些式子
异或满足交换律：a^b^c=a^c^b。
还有 a^b=c,b=a^c

下面是代码

#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<stdlib.h>
#include<map>
using namespace std;
#define maxn 100005
long long int s[maxn];
long long int arr[maxn * 10];
int main()
{long long n, x;long long sum, cnt;while (cin>>n>>x){sum = 0;cnt = 0;memset(arr, 0, sizeof(arr));for (int i = 0; i<n; i++){scanf("%lld", &s[i]);arr[s[i]]++;}for (int i = 0; i<n; i++){cnt = x^s[i];if (cnt == s[i])sum += arr[cnt] - 1;elsesum += arr[cnt];}cout << sum / 2 << endl;}return 0;
}

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