AtCoder Beginner Contest 098 D - Xor Sum 2

D - Xor Sum 2

---

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.

---

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

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.

---

Sample Input 2

Copy

9
0 0 0 0 0 0 0 0 0

Sample Output 2

Copy

45

---

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

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 using namespace std;
 9 #define  ll long long 
10 #define  N  200009
11 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
12 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
13 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
14 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
15 #define  mem(a,b)  memset(a,b,sizeof(a))
16 int n,a[N];
17 ll sum[N],b[N];
18 /*
19 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
20 例如  1 2 4 8
21 如果 [l,r] 不可以，那么l++
22 如果[l,r] 可以,那么 r++
23 */
24 int  main()
25 {
26     scanf("%d",&n);
27     gep(i,1,n) {
28         scanf("%d",&a[i]);
29         sum[i]=sum[i-1]+a[i];
30         b[i]=b[i-1]^a[i];
31     }
32     int l=1;
33     ll ans=0;
34     gep(r,1,n){
35         for(;sum[r]-sum[l-1]!=(b[r]^b[l-1]);l++);//!=的优先级大于^,因此要加()
36         ans+=(r-l+1);
37     //[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
38     //可以一定是连续的，一旦不可以了就要改变r,l
39     }
40     printf("%lld\n",ans);
41     return  0;
42 }