Codeforces 742B

计数异或对算法

最新推荐文章于 2021-09-18 16:30:44 发布

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最新推荐文章于 2021-09-18 16:30:44 发布

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文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/Littlewhite520/article/details/54100118

本文介绍了一种解决特定算法问题的方法：给定一个整数数组和一个整数x，计算数组中所有满足条件a[i] ^ a[j] = x的元素对(i, j)的数量。文中提供了一个错误示例和正确的解决方案，并解释了为何使用正确的实现。

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.

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. You can read more about bitwise xor operation here: https://en.wikipedia.org/wiki/Bitwise_operation#XOR.

没有搞清楚异或的用法 a^b=x, a=b^x.
而且另一个需要知道的是异或的优先级比==低所以好多次都没出结果。穷举连答案都出不了，虽然这道题不应该用穷举。
错误代码

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
int a[250010];
int main()
{
    int n,x;
    while(cin>>n>>x)
    {
        for(int i=0;i<n;i++)
            cin>>a[i];
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(a[i]^a[j]==x)   //这里错了，应该改为if((a[i]^a[j])==x)  
                    cout<<a[i]<<" "<<a[j]<<endl;
            }
        }
    }
}

正确代码注意10的5次方会到10万用LL

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
typedef long long LL;
LL a[250010];
LL cnt[250010];
int main()
{
    int n,x;
    while(cin>>n>>x)
    {
        LL ans=0;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            cnt[a[i]]++;
        }
        for(int i=0;i<n;i++)
        {
            cnt[a[i]]--;
            LL t=a[i]^x;
            ans+=cnt[t];
        }
        cout<<ans<<endl;
    }
}