AtCoder Beginner Contest 098 D - Xor Sum 2

D - Xor Sum 2

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Time limit : 2sec / Memory limit : 1024MB

Score : 500 points

Problem Statement

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

 

Constraints

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

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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.

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Sample Input 1

Copy

4
2 5 4 6

Sample Output 1

Copy

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.

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Sample Input 2

Copy

9
0 0 0 0 0 0 0 0 0

Sample Output 2

Copy

45

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Sample Input 3

Copy

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

Sample Output 3

Copy

37

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <deque>
using namespace std;
#define  ll long long 
#define  N  200009
#define  gep(i,a,b)  for(int  i=a;i<=b;i++)
#define  gepp(i,a,b) for(int  i=a;i>=b;i--)
#define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
#define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
#define  mem(a,b)  memset(a,b,sizeof(a))
int n,a[N];
ll sum[N],b[N];
/*
要想相加和等于异或和，就要区间的数二进制的每一位1只有一个
例如  1 2 4 8
如果 [l,r] 不可以，那么l++
如果[l,r] 可以,那么 r++
*/
int  main()
{
    scanf("%d",&n);
    gep(i,1,n) {
        scanf("%d",&a[i]);
        sum[i]=sum[i-1]+a[i];
        b[i]=b[i-1]^a[i];
    }
    int l=1;
    ll ans=0;
    gep(r,1,n){
        for(;sum[r]-sum[l-1]!=(b[r]^b[l-1]);l++);//!=的优先级大于^,因此要加()
        ans+=(r-l+1);
    //[l,r] :每一次的可以就是多了[l,r],[l+1,r],[l+2,r]……[r,r]共(r-l+1)个区间
    //可以一定是连续的，一旦不可以了就要改变r,l
    }
    printf("%lld\n",ans);
    return  0;
}