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.
题意：给你一串数组A，让你计算其中有几对数满足：ai^aj=x
思路：这个式子可以化简一下。可化为：ai^x=aj
我们可以遍历这个数组a然后找出来每个ai对应的aj。从而计算组数
这题可以map来简化步骤
并且要注意下数据类型。有的地方要用long long
#include<bits/stdc++.h>
using namespace std;
#define LL long long
int num[100005];
int main()
{int n,x;while(~scanf("%d %d",&n,&x)){map<LL,LL>m;memset(num,0,sizeof(num));for(int i=0;i<n;i++){scanf("%d",&num[i]);m[num[i]]++;}LL sum=0;for(int i=0;i<n;i++){LL ans=num[i]^x;if(ans==num[i]){sum+=m[ans]-1;}else{sum+=m[ans];}}printf("%lld\n",sum/2);}return 0;
}