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,,n) {
         scanf("%d",&a[i]);
         sum[i]=sum[i-]+a[i];
         b[i]=b[i-]^a[i];
     }
     int l=;
     ll ans=;
     gep(r,,n){
         for(;sum[r]-sum[l-]!=(b[r]^b[l-]);l++);//!=的优先级大于^,因此要加()
         ans+=(r-l+);
     //[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  ;
 }