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, ..., ajis 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
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
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.
题解:首先要知道 a^b=c 等价于 a=c^b; 我们用a[i]记录前I个数的异或和,然后离线处理所有区间(对于所有区间我们按L所在块为第一排序,该询问的r为第二排序,对所有询问区间排序);
对于新增加的一个数我们加上前区间异或等于k^s的数的个数,然后更新一下异或为s的数量,对于一个需要去掉的数,我需要先新更新这个数的数量,然后减去k^s的数量即可;
离线跑一遍,
参考代码为:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include<cstdio> 2 #include<cstring> 3 #include<cmath> 4 #include<iostream> 5 #include<algorithm> 6 #define LL long long 7 using namespace std; 8 const int Max = 1100000; 9 const int MAXM = 1<<22; 10 struct Point 11 { 12 int L ,R,Id; 13 } a[Max]; 14 15 LL sum[Max],ans[Max],k; 16 int n,m,L,R; 17 LL cnt[MAXM],ant; 18 bool cmp(Point b,Point c) 19 { 20 return b.L/400==c.L/400? b.R<c.R:b.L<c.L; 21 } 22 void Dec(LL s) 23 { 24 --cnt[s]; 25 ant-=cnt[s^k]; 26 } 27 void Inc(LL s) 28 { 29 ant += cnt[s^k]; 30 cnt[s]++; 31 } 32 int main() 33 { 34 scanf("%d %d %lld",&n,&m,&k); 35 LL data; 36 for(int i=1;i<=n;i++) scanf("%lld",&sum[i]),sum[i]^=sum[i-1]; 37 for(int i=1;i<=m;i++) scanf("%d %d",&a[i].L,&a[i].R),a[i].Id = i,a[i].L--; 38 sort(a+1,a+m+1,cmp); 39 L=0,R=0,cnt[0]=1,ant=0; 40 for(int i=1;i<=m;i++) 41 { 42 while(R<a[i].R) Inc(sum[++R]); 43 while(R>a[i].R) Dec(sum[R--]); 44 while(L<a[i].L) Dec(sum[L++]); 45 while(L>a[i].L) Inc(sum[--L]); 46 ans[a[i].Id]=ant; 47 } 48 for(int i=1;i<=m;i++) printf("%lld ",ans[i]); 49 return 0; 50 }