题目链接：http://codeforces.com/problemset/problem/617/E
题目：
Bob has a favorite number k and ai of length n. Now he asks you to answer m queries. Each query is given by a pair li and ri and asks you to count the number of pairs of integers i and j, such that l ≤ i ≤ j ≤ r and the xor of the numbers ai, ai + 1, ..., aj is equal to k.
The first line of the input contains integers n, m and k (1 ≤ n, m ≤ 100 000, 0 ≤ k ≤ 1 000 000) — the length of the array, the number of queries and Bob‘s favorite number respectively.
The second line contains n integers ai (0 ≤ ai ≤ 1 000 000) — Bob‘s array.
Then m lines follow. The i-th line contains integers li and ri (1 ≤ li ≤ ri ≤ n) — the parameters of the i-th query.
Print m lines, answer the queries in the order they appear in the input.
6 2 3
1 2 1 1 0 3
1 6
3 5
7
0
5 3 1
1 1 1 1 1
1 5
2 4
1 3
9
4
4
In the first sample the suitable pairs of i and j for the first query are: (1, 2), (1, 4), (1, 5), (2, 3), (3, 6), (5, 6), (6, 6). Not a single of these pairs is suitable for the second query.
In the second sample xor equals 1 for all subarrays of an odd length.
题意：给定n个元素，q次询问，每次询问【l，r】区间内ai^...^aj=k有几对。
题解：先预处理出前缀的ai。ai^...^aj=sum[i-1]^sum[j]=k。sum[i-1]^k=sum[j]。然后就是基本的莫队操作了。add和del那里两个东西哪个再前面，这个其实想一想k=0的情况就很好理解了
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define PI acos(-1.0) 5 #define INF 0x3f3f3f3f 6 #define FAST_IO ios::sync_with_stdio(false) 7 8 typedef long long LL; 9 10 const int N=3e6+10; 11 struct node{ 12 int l,r,id; 13 }Q[N]; 14 15 LL a[N],ans[N],cnt[N]; 16 int pos[N],BLOCK; 17 bool cmp(node x,node y){ 18 if(pos[x.l]==pos[y.l]) return x.r<y.r; 19 return pos[x.l]<pos[y.l]; 20 } 21 22 int n,m,k; 23 LL Ans=0; 24 25 void add(int x){ 26 Ans+=cnt[a[x]^k]; 27 cnt[a[x]]++; 28 } 29 30 void del(int x){ 31 cnt[a[x]]--; 32 Ans-=cnt[a[x]^k]; 33 } 34 35 int main(){ 36 scanf("%d%d%d",&n,&m,&k); 37 BLOCK=sqrt(n); 38 for(int i=1;i<=n;i++){ 39 scanf("%lld",&a[i]); 40 pos[i]=i/BLOCK; 41 a[i]^=a[i-1]; 42 } 43 for(int i=1;i<=m;i++){ 44 scanf("%d%d",&Q[i].l,&Q[i].r); 45 Q[i].id=i; 46 } 47 sort(Q+1,Q+1+m,cmp); 48 int L=1,R=0; 49 cnt[0]=1; 50 for(int i=1;i<=m;i++){ 51 while(L<Q[i].l){ 52 del(L-1); 53 L++; 54 } 55 while(L>Q[i].l){ 56 L--; 57 add(L-1); 58 } 59 while(R<Q[i].r){ 60 R++; 61 add(R); 62 } 63 while(R>Q[i].r){ 64 del(R); 65 R--; 66 } 67 ans[Q[i].id]=Ans; 68 } 69 for(int i=1;i<=m;i++) 70 printf("%lld\n",ans[i]); 71 return 0; 72 }
