Codeforce 1054 D. Changing Array —— 区间异或和

At a break Vanya came to the class and saw an array of n k-bit integers a1,a2,…,an on the board. An integer x is called a k-bit integer if 0≤x≤2k−1.

Of course, Vanya was not able to resist and started changing the numbers written on the board. To ensure that no one will note anything, Vanya allowed himself to make only one type of changes: choose an index of the array i (1≤i≤n) and replace the number ai with the number ai¯¯¯¯. We define x¯¯¯ for a k-bit integer x as the k-bit integer such that all its k bits differ from the corresponding bits of x.

Vanya does not like the number 0. Therefore, he likes such segments [l,r] (1≤l≤r≤n) such that al⊕al+1⊕…⊕ar≠0, where ⊕ denotes the bitwise XOR operation. Determine the maximum number of segments he likes Vanya can get applying zero or more operations described above.

Input
The first line of the input contains two integers n and k (1≤n≤200000, 1≤k≤30).

The next line contains n integers a1,a2,…,an (0≤ai≤2k−1), separated by spaces — the array of k-bit integers.

Output
Print one integer — the maximum possible number of segments with XOR not equal to 0 that can be obtained by making several (possibly 0) operations described in the statement.

Examples
inputCopy
3 2
1 3 0
outputCopy
5
inputCopy
6 3
1 4 4 7 3 4
outputCopy
19
Note
In the first example if Vasya does not perform any operations, he gets an array that has 5 segments that Vanya likes. If he performs the operation with i=2, he gets an array [1,0,0], because 3¯¯¯=0 when k=2. This array has 3 segments that Vanya likes. Also, to get an array with 5 segments that Vanya likes, he can perform two operations with i=3 and with i=2. He then gets an array [1,0,3]. It can be proven that he can’t obtain 6 or more segments that he likes.

In the second example, to get 19 segments that Vanya likes, he can perform 4 operations with i=3, i=4, i=5, i=6 and get an array [1,4,3,0,4,3].

题意：

给你一个长度为n的串，你可以将任意多个位置的值变成它对 $2^k-1$的异或，问最多能有多少个区间，这个区间中的值异或和不为0；

题解：

一遍做过去，每次我们只需要让前缀异或和中的0最少就好了比如说1到x区间的异或和为a，那么如果1到y这个区间的异或和为a的话，y到x区间的异或和就为0，那么我们只需要记录一下有多少前缀异或和为a，用一个map记录

#include<bits/stdc++.h>
using namespace std;
#define ll long long
map<ll,ll>map1;
int main()
{
    int x,n,k,num=0;
    scanf("%d%d",&n,&k);
    int ret=(1<<k)-1;
    ll sum=0;
    
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        int ne=num^x;
        int z1=0,z2=0;
        if(!x)
            z1++;
        if(!ne)
            z1++;
        z1+=map1[ne];
        x=ret-x;
        int ne2=num^x;
        if(!x)
            z2++;
        if(!ne2)
            z2++;
        z2+=map1[ne2];  
        if(z1<=z2)
            sum+=i-z1,map1[ne]++,num=ne;
        else
            sum+=i-z2,map1[ne2]++,num=ne2;
    }
    printf("%lld\n",sum);
}