Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution —— 异或
题目链接:http://codeforces.com/contest/742/problem/B
1 second
256 megabytes
standard input
standard output
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.
题解:
1.a^b = c,则:a = b^c,b = a^c。说明:a如果与某个数的异或为c, 那么这个数是唯一的,且可直接求出:b = a^c。
2.c用于记录某个数出现的次数,一边读取数一边操作:假设当前读取的数据为a,则与之对应的数为x^a,则表明a能够与c[x^a]个x^a组成一对,所以 ans += c[x^a],然后再更新a出现的次数,即c[a]++。
注意之处:
刚开始看到x和a的最大范围为1e5,所以想都不想就把数组也开成1e5,错了一发。
后来发现x^a可能大于1e5,但却不会超过2*1e5,所以应该把数组开成2e5,才不会溢出。
代码如下:
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 1e5+10; LL n, x;
LL c[maxn<<1]; //!!!! int main()
{
cin>>n>>x;
LL ans = 0, a;
for(int i = 1; i<=n; i++)
{
scanf("%lld",&a);
ans += c[x^a];
c[a]++;
}
cout<<ans<<endl;
}
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