Codeforces Round #383 (Div. 2) Arpa’s obvious problem and Mehrdad’s terrible solution 数学

题目：

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.

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 be0 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.

就是给你一个序列，问有多少个二元组(a,b)满足a^b=c(c给定).

1,异或有一个性质 a^b^a=b.那么a^b=c -->  a^b^a=c^a -->a^c=b

2,a^b最大值可以近似取到max(a,b)*2 例如 011^100=111开数组时不注意会RE的

3，int*int可能会返回一个负数,也就是溢出。如a=100000,res=a*a, 那么res=-727379968。

4，这种题主要是是考虑重复，已经使用过的二元组用visit数组标记，那么问题来了，如果x=0,序列是1 1 1 1时该怎么办，我们不难发现这个就是cnt[i]*(cnt[i]-1)/2(这个地方可能会发生第三点的情况)

code：

#include<cstdio> const int MAXN=2e5+5; int a[MAXN]; long long  cnt[MAXN]; long long  visit[MAXN]; int main(){     int n,x;scanf("%d%d",&n,&x);     for(int i=0;i<n;++i){         scanf("%d",a+i);         cnt[a[i]]++;     }     long long res=0;     for(int i=1;i<=100000;++i){         if(!visit[i]&&cnt[i]&&cnt[x^i]){             if(i==(x^i))res+=cnt[i]*(cnt[i]-1)/2;             else res+=cnt[i]*cnt[x^i];             visit[i]=visit[x^i]=1;         }     }     printf("%I64d\n",res); }

还要继续加油呀！