E. XOR and Favorite Number
time limit per test
4 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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.

Input

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.

Output

Print m lines, answer the queries in the order they appear in the input.

Examples
Input
6 2 3
1 2 1 1 0 3
1 6
3 5
Output
7
0
Input
5 3 1
1 1 1 1 1
1 5
2 4
1 3
Output
9
4
4
Note

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\n",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 */