B. Arpa’s obvious problem and Mehrdad’s terrible solution

B. Arpa’s obvious problem and Mehrdad’s terrible solution
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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
inputCopy
2 3
1 2
outputCopy
1
inputCopy
6 1
5 1 2 3 4 1
outputCopy
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.

#include 
#include 
#include
int a[10000000],f[10000000];
int main()
{
    long long int x,n,sum=0;
    long long int i;
    memset(f,0,sizeof(f));
    scanf("%lld%lld",&n,&x);
    for(i=0;iscanf("%d",&a[i]);
    }
    for(i=0;i//这个就是把每个a[i]的配对都写出来，并且对应的f[a[i]]++,如果这一对是一个异或组合的话，那么这个a[i]^x即a[j]在后面肯定会出现的，这样的话就会把a[i]还原出来，就是说f数组里的数就是a[i]了，同时，f[a[i]]之前已经加过了，直接用就可以，而且这一对不会重复，因为前面的a[j]配对的a[i]并没有出现，当后面出现a[j]在a[i]行列中出现的时候，a[i]已经出现过了，就会加一了
        f[a[i]]++;
    }
    printf("%lld\n",sum);
    return 0;
}

思路：考察的是数学知识，求异或，异或有一个性质，A^B=X,则A^X=B；
见上面的代码