【CF 617E】 XOR and Favorite Number （Mo's algorithm）

【CF 617E】  XOR and Favorite Number （Mo's algorithm）

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 a_i of length n. Now he asks you to answer m queries. Each query is given by a pairl_i andr_i and asks you to count the number of pairs of integersi and j, such thatl ≤ i ≤ j ≤ r and the xor of the numbersa_i, a_i + 1, ..., a_j is equal tok.

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 a_i (0 ≤ a_i ≤ 1 000 000) — Bob's array.

Then m lines follow. The i-th line contains integers l_i andr_i (1 ≤ l_i ≤ r_i ≤ n) — the parameters of the i-th query.

Output

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

Sample test(s)

Input

6 2 31 2 1 1 0 31 63 5

Output

70

Input

5 3 11 1 1 1 11 52 41 3

Output

944

Note

In the first sample the suitable pairs of i andj 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.

题目大意是找某个区间内连续异或能得到K的区间数

现在先考虑单组区间查询。对于区间[L,R] 可以求个前缀和pre[i] 表示第1～i的数的异或和。这样[L,R]的异或和即为pre[L-1]^pre[R] 换种思路 预存一下pre[L-1]^k 就可以延长它的生命周期。

这样求[L,R]区间内能异或出k的区间数 从L遍历到R 遍历过程中统计遍历过的前缀异或和的个数 当遍历到i时 pre[i]^k出现的次数即为[L,i]中到i的连续的能异或得到k的区间数

这样如果区间向左或向右就可以O(1)的修改 对于多组查询 就可以上莫队算法了

代码如下：

#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <cstring>#include <queue>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread() freopen("in.in","r",stdin)#define fwrite() freopen("out.out","w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int msz = 100100;const double eps = 1e-8;//L的分块int pos[msz];struct Range{int l,r,id;bool operator < (const struct Range a)const{return pos[l] == pos[a.l]? r < a.r: pos[l] < pos[a.l];}};Range rg[msz];int a[msz];//统计出现过的异或数LL cnt[1048576];LL ans[msz];LL num;int n,m,k;//移动区间边界时更新void update(int x,int d){if(d < 0)cnt[k^a[x]] +=d;num += d*cnt[a[x]];if(d > 0)cnt[k^a[x]] +=d;}int main(){scanf("%d%d%d",&n,&m,&k);int dm = ceil(sqrt(n*1.0));a[0] = 0;a[n+1] = 0;for(int i = 1; i <= n; ++i){scanf("%d",&a[i]);a[i] ^= a[i-1];}for(int i = 0; i < m; ++i){scanf("%d%d",&rg[i].l,&rg[i].r);rg[i].l--;pos[rg[i].l] = rg[i].l/dm;rg[i].id = i;}sort(rg,rg+m);int l = 1,r = 0;num = 0;memset(cnt,0,sizeof(cnt));for(int i = 0; i < m; ++i){int id = rg[i].id;if(rg[i].l == rg[i].r-1){ans[id] = (a[rg[i].r]^a[rg[i].l]) == k;continue;}while(r < rg[i].r) update(++r,1);while(r > rg[i].r) update(r--,-1);while(l < rg[i].l) update(l++,-1);while(l > rg[i].l) update(--l,1);ans[id] = num;}for(int i = 0; i < m; ++i)printf("%lld\n",ans[i]);return 0;}