codeforces 665E E. Beautiful Subarrays(trie树)

题目链接：

E. Beautiful Subarrays

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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.

Examples

 

input

3 1
1 2 3

output

5

input

3 2
1 2 3

output

3

input

3 3
1 2 3

output

2


题意：

给n个数,问有多少对数满足a[i]^a[i+1]^...^a[j]>=k;

思路：

处理出前缀异或和,再用trie树搞;

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(b));
typedef long long LL;
const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=0x3f3f3f3f;
const int N=2e7+6;
int n,k,cnt=2;
int tree[N][2],val[N];
void add_num(int num)
{
    int x=1;
    for(int i=30;i>=0;i--)
    {
        int y=(num>>i)&1;
        if(!tree[x][y])tree[x][y]=cnt++;
        val[x]++;
        x=tree[x][y];
    }
    val[x]++;
}
LL get_ans(int num,int fy)
{
    LL sum=0;
    int x=1;
    for(int i=30;i>=0;i--)
    {
        int y=(fy>>i)&1;
        int fx=(num>>i)&1^1;
        if(y==0)
        {
            sum+=(LL)val[tree[x][fx]];
            x=tree[x][fx^1];
        }
        else x=tree[x][fx];
    }
    return sum+(LL)val[x];
}
int main()
{
    scanf("%d%d",&n,&k);
    int pre=0;
    LL ans=0;
    add_num(0);
    Riep(n)
    {
        int a;
        scanf("%d",&a);
        pre^=a;
        ans+=(LL)get_ans(pre,k);
        add_num(pre);
    }
    printf("%I64d
",ans);

    return 0;
}