Educational Codeforces Round 12 E. Beautiful Subarrays 字典树

E. Beautiful Subarrays

题目连接：

http://www.codeforces.com/contest/665/problem/E

Description

One day, ZS the Coder wrote down an array of integers a with elements a1,  a2,  ...,  an.

A subarray of the array a is a sequence al,  al  +  1,  ...,  ar for some integers (l,  r) such that 1  ≤  l  ≤  r  ≤  n. ZS the Coder thinks that a subarray of a is beautiful if the bitwise xor of all the elements in the subarray is at least k.

Help ZS the Coder find the number of beautiful subarrays of a!

Input

The first line contains two integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 109) — the number of elements in the array a and the value of the parameter k.

The second line contains n integers ai (0 ≤ ai ≤ 109) — the elements of the array a.

Output

Print the only integer c — the number of beautiful subarrays of the array a.

Sample Input

3 1
1 2 3

Sample Output

5

Hint

题意

问你有多少个区间，异或起来大于等于k

题解：

显然求个前缀和之后，就等于有多少对数异或起来大于等于k了

这个玩意儿我们每次暴力爬字典树就好了

当k这一位等于0的时候，我们可以直接加上另外一边1的子树大小，因为爬那边之后，我怎么爬都是大于等于k的

然后就这样直接暴力莽一波就好了~

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e7+6;
struct Tri
{
    int ch[maxn][2];
    int sz[maxn];
    int tot;
    void init()
    {
        memset(ch,0,sizeof(ch));
        memset(sz,0,sizeof(sz));
        tot=2;
    }
    void insert(int x)
    {
        int u=1;
        for(int i=30;i>=0;i--)
        {
            int p = (x>>i)&1;
            if(!ch[u][p])ch[u][p]=tot++;
            sz[u]++;
            u=ch[u][p];
        }
        sz[u]++;
    }
    int get(int x,int y)
    {
        int u=1;
        long long ans = 0;
        for(int i=30;i>=0;i--)
        {
            int p = (x>>i)&1^1;
            int q = (y>>i)&1;
            if(q==0)ans+=sz[ch[u][p]],u=ch[u][p^1];
            else u=ch[u][p];
        }
        return ans+sz[u];
    }
}T;
int main()
{
    T.init();
    int n,k;
    scanf("%d%d",&n,&k);
    int pre = 0;
    long long ans = 0;
    T.insert(0);
    for(int i=1;i<=n;i++)
    {
        int x;scanf("%d",&x);
        pre^=x;
        ans+=T.get(pre,k);
        T.insert(pre);
    }
    cout<<ans<<endl;
}