Xor Sum 2 （atcoder） (前缀异或和技巧)

There is an integer sequence A of length N.

Find the number of the pairs of integers l and r (1≤l≤r≤N) that satisfy the following condition:

• Al xor Al+1 xor … xor Ar=Al + Al+1 + … + Ar

Here, xor denotes the bitwise exclusive OR.

Definition of XOR

The XOR of integers c1,c2,…,cm is defined as follows:

• Let the XOR be X. In the binary representation of X, the digit in the 2k's place (0≤k; k is an integer) is 1 if there are an odd number of integers among c1,c2,…cm whose binary representation has 1 in the 2k's place, and 0 if that number is even.

For example, let us compute the XOR of 3 and 5. The binary representation of 3 is 011, and the binary representation of 5 is 101, thus the XOR has the binary representation 110, that is, the XOR is 6.

Constraints

• 1≤N≤2×105
• 0≤Ai<220
• All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A1 A2 … AN

Output

Print the number of the pairs of integers l and r (1≤l≤r≤N) that satisfy the condition.

Sample Input 1

4
2 5 4 6

Sample Output 1

5

(l,r)=(1,1),(2,2),(3,3),(4,4) clearly satisfy the condition. (l,r)=(1,2) also satisfies the condition, since A1 xor A2=A1 + A2=7. There are no other pairs that satisfy the condition, so the answer is 5.

Sample Input 2

9
0 0 0 0 0 0 0 0 0

Sample Output 2

45

Sample Input 3

19
885 8 1 128 83 32 256 206 639 16 4 128 689 32 8 64 885 969 1

Sample Output 3

37

首先我们要知道s[R]^s[L-1]=s[L]^s[L+1]………^s[R]，这里用到了前缀和技巧，以及a^a=0.

#include
using namespace std;
const int maxn=2e5+5;
long long f[maxn],s[maxn];
long long ans;
int main()
{
    int n,x;
    scanf("%d",&n);
    f[0]=0,s[0]=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&x);
        f[i]=f[i-1]+x;
        s[i]=s[i-1]^x;
    }
    int L=1;
    ans=0;
    for(int R=1;R<=n;R++)
    {
        while((f[R]-f[L-1])!=(s[R]^s[L-1])) L++;
        ans+=R-L+1;
    }
    printf("%lld\n",ans);
}