Codeforces 742B-Arpa’s obvious problem and Mehrdad’s terrible solution

最新推荐文章于 2018-01-20 02:17:26 发布

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最新推荐文章于 2018-01-20 02:17:26 发布

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本文介绍了一种解决特定数组中寻找异或等于给定值x的元素对数量的方法。利用哈希表（map）记录x异或a[i]的结果出现次数，实现快速查找配对。示例代码展示了如何通过这一技巧高效解决问题。

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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 $\text{[math]}$ , where $\text{[math]}$ is bitwise xoroperation (see notes for explanation).

$\text{[math]}$

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
 
     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. $\text{[math]}$ so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since $\text{[math]}$ ) and i = 1, j = 5 (since $\text{[math]}$ ).

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：给出含有n个数的数组a[]，求在数组中有多少对 a[i]^a[j]==x。

解题思路：若a xor b=c，则c xor b=a 且c xor a=b，用map来记录x^a[i]的答案在数组a[]中出现过几次

#include <iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <algorithm>
#include <cstring>
#include <stack>
#include <map>

using namespace std;

#define ll long long

ll a[100009];

int main()
{
    ll n,x;
    while(~scanf("%lld %lld",&n,&x))
    {
        map<ll,ll>p;
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a[i]);
            p[a[i]]++;
        }
        ll sum=0;
        for(int i=0;i<n;i++)
        {
            ll xx=a[i]^x;
            if(xx!=a[i]) sum+=p[xx];
            else sum+=p[xx]-1;
        }
        printf("%lld\n",sum/2);
    }
    return 0;
}