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 xoroperation (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.
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.
Print a single integer: the answer to the problem.
2 3 1 2
1
6 1 5 1 2 3 4 1
2
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.
思路:若a^b=c,有a^c=b和b^c=a,于是便容易解决了。
# include <stdio.h> int a[140000]={0};//数组不要开小了 int main() { int n, x, t; long long ans = 0; scanf("%d%d",&n,&x); while(n--) { scanf("%d",&t); ans += a[x^t]; ++a[t]; } printf("%I64d\n",ans); return 0; }