Codeforces Round #383 (Div. 2) B. Arpa’s obvious problem and Mehrdad’s terrible solution 数论、易错

B. Arpa’s obvious problem and Mehrdad’s terrible solution

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
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.

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. 

 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.

Source

Codeforces Round #383 (Div. 2)

My Solution

题意：找出多少组ai和aj 使 ai ^ aj == x.

数论、易错

 ai ^ aj == x.   =>   ai ^ x  == aj

这样把sz[ai] 为数ai出现的次数，对于每个ai，ans += sz[ai] * sz[ai ^ x] ;  sz[ai ^ x] = 0; //即每一对 ai 和 aj 只处理一次。

很显然还有一个特殊情况，即x == 0 的时候， ai ^ ai == 0, 故此时特殊处理  ans += sz[ai] * (sz[ai] - 1) / 2;

然后 中间过程有int相乘(sz[ai] * sz[ai ^ x] )(sz[ai] * (sz[ai] - 1) / 2) ，故 int 很可能会溢出，故用long long

复杂度 O(n)

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
typedef long long LL;
const int maxn = 1e5 + 8;

map<int, LL> mp;
int a[maxn];
int main()
{
#ifdef LOCAL
freopen("b.txt", "r", stdin);
//freopen("b.out", "w", stdout);
int T = 4;
while(T--){
#endif // LOCAL
ios::sync_with_stdio(false); cin.tie(0);

int n, x;
LL ans = 0;
cin >> n >> x;
for(int i = 0; i < n; i++){
cin >> a[i];
mp[a[i]]++;
}
if(x == 0){
//ans 可能会溢出所以用LL
for(auto i = mp.begin(); i != mp.end(); i++){
ans += (i->second) * (i->second - 1) / 2;
}
}
else{
//ans 可能会溢出所以用LL
for(auto i = mp.begin(); i != mp.end(); i++){
ans += (i->second) * mp[(i->first) ^ x];
mp[(i->first) ^ x] = 0;
i->second = 0;
}
}

cout << ans << endl;

#ifdef LOCAL
mp.clear();
cout << endl;
}
#endif // LOCAL
return 0;
}

  Thank you!

                                                                                                           
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