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.

没有搞清楚异或的用法 a^b=x, a=b^x.
而且另一个需要知道的是 异或的优先级比==低 所以好多次都没出结果。穷举连答案都出不了，虽然这道题不应该用穷举。
错误代码

#include <iostream>#include <cstring>#include <cstdio>using namespace std;int a[250010];int main(){    int n,x;    while(cin>>n>>x)    {        for(int i=0;i<n;i++)            cin>>a[i];        for(int i=0;i<n;i++)        {            for(int j=0;j<i;j++)            {                if(a[i]^a[j]==x)   //这里错了，应该改为if((a[i]^a[j])==x)                      cout<<a[i]<<" "<<a[j]<<endl;            }        }    }}

正确代码 注意10的5次方会到10万用LL

#include <iostream>#include <cstring>#include <cstdio>using namespace std;typedef long long LL;LL a[250010];LL cnt[250010];int main(){    int n,x;    while(cin>>n>>x)    {        LL ans=0;        for(int i=0;i<n;i++)        {            cin>>a[i];            cnt[a[i]]++;        }        for(int i=0;i<n;i++)        {            cnt[a[i]]--;            LL t=a[i]^x;            ans+=cnt[t];        }        cout<<ans<<endl;    }}