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：给你一个长度为n的序列 q个查询[l,r] 问[l,r] 有多少个子区间的异或和为k
题解：莫队+前缀异或和
1 #include<iostream> 2 #include<cstdio> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<map> 7 #include<queue> 8 #include<stack> 9 #include<vector> 10 #include<set> 11 #define ll __int64 12 using namespace std; 13 int n,m,k; 14 struct node 15 { 16 int l,r,id; 17 }N[100005]; 18 int p[100005]; 19 int block; 20 int a[100005]; 21 int x[100005]; 22 int mp[5000006]; 23 ll ans=0; 24 ll re[100005]; 25 int cmp(struct node aa,struct node bb) 26 { 27 if(p[aa.l]==p[bb.l]) 28 return aa.r<bb.r; 29 else 30 return p[aa.l]<p[bb.l]; 31 } 32 void update(int w,int h) 33 { 34 if(h==1){ 35 ans=ans+mp[x[w]^k]; 36 mp[x[w]]++; 37 } 38 else 39 { 40 mp[x[w]]--; 41 ans=ans-mp[x[w]^k]; 42 } 43 } 44 int main() 45 { 46 scanf("%d %d %d",&n,&m,&k); 47 for(int i=1;i<=n;i++) 48 scanf("%d",&a[i]); 49 x[0]=0; 50 mp[0]=1; 51 for(int i=1;i<=n;i++) 52 x[i]=a[i]^x[i-1]; 53 for(int i=1;i<=m;i++){ 54 scanf("%d %d",&N[i].l,&N[i].r); 55 N[i].id=i; 56 } 57 block=(int)sqrt((double)n); 58 for(int i=1;i<=n;i++) 59 p[i]=(i-1)/block+1; 60 sort(N+1,N+1+m,cmp); 61 ans=0; 62 for(int i=1,l=1,r=0;i<=m;i++) 63 { 64 for(;r<N[i].r;r++) update(r+1,1); 65 for(;l>N[i].l;l--) update(l-2,1);// 取异或的原因 66 for(;r>N[i].r;r--) update(r,-1); 67 for(;l<N[i].l;l++) update(l-1,-1);// 68 re[N[i].id]=ans; 69 } 70 for(int i=1;i<=m;i++) 71 printf("%I64d ",re[i]); 72 return 0; 73 } 74 /* 75 10 2 0 76 0 0 0 0 0 0 0 0 0 0 77 2 8 78 2 8 79 */