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

最新推荐文章于 2021-09-18 16:30:44 发布

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本文介绍了一种解决特定数组中异或对数量问题的方法。给定一个整数数组及一个整数x，任务是计算所有满足条件的索引对(i, j)，使得元素ai与aj之间的异或操作结果等于x。文章提供了完整的代码实现，并解释了关键步骤。

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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 这里写图片描述 , 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.

思路：
ai^aj = x >> ai^x = aj;

代码：

#include <cstdio>
#include <iostream>

using namespace std;

int board[1000005];//记录数据 
int book[1000005];//记录哪些数出现过和出现次数。  

int main(){
    int N,X;
    cin>>N>>X;
    for(int i=0 ; i<N ; i++){
        scanf("%d",&board[i]);
        book[board[i]]++;
    }

    long long sum=0;
    for(int i=0 ; i<N ; i++){
        int mid = board[i]^X;//mid可能超过1E5，所以上边数组开大一点。 
        if(mid!=board[i])sum += (long long)book[board[i]^X];
        else sum += (long long)(book[board[i]^X]-1);
    }

    cout<<sum/2<<endl;
    return 0;
}