Codeforces 617E：XOR and Favorite Number 莫队算法

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.

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 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.

题意是给出n个数，然后给出m个区间，问在每一个区间内 有多少个子区间，其中子区间所有的数异或等于给定的K。

首先计算前缀异或(是这么说的么？。。。)，即pre[i]=pre[i-1]^val[i]。这样在给定的[le,ri]区间内询问有多少个子区间所有异或的数，就等于在[le-1,ri]区间内，找到两个pre[x] pre[y]，使得pre[x]^pre[y]=K。查询有多少个数对。

通过这个题学习了莫队算法，之前没有接触过。莫队算法的思想和之前51nod找朋友那道题有一点点思想上的类似。对于固定的离线区间查询能够减少时间复杂度。

先占个坑，毕竟做莫队算法的题目比较少，认识感觉还比较浅。当下对于莫队算法思想的感觉就是将区间排序，找寻单个元素对于结果的贡献，然后利用相邻区间更新结果。当前的认识就是这样。

代码：

#pragma warning(disable:4996)  #include <iostream>  #include <algorithm>#include <cstring>#include <cstdio>#include <vector>  #include <string>  #include <cmath>#include <queue>#include <map>using namespace std;typedef long long ll;#define INF 0x3fffffffconst int mod = 1e9 + 7;const int maxn = 2000100;ll n, m, K, bk;ll val[maxn], pos[maxn], num[maxn], res[maxn], pre[maxn], ans;struct no{ll le;ll ri;ll id;}qu[maxn];bool cmp(no a, no b){if (pos[a.le] == pos[b.le]){return a.ri < b.ri;}else{return pos[a.le] < pos[b.le];}}void add(ll x){ans += num[pre[x] ^ K];num[pre[x]]++;}void del(ll x){num[pre[x]]--;ans -= num[pre[x] ^ K];}void input(){ll i;scanf("%I64d%I64d%I64d", &n, &m, &K);memset(num, 0, sizeof(num));bk = ceil(sqrt(1.0*n));for (i = 1; i <= n; i++){scanf("%I64d", &val[i]);pre[i] = pre[i - 1] ^ val[i];pos[i] = (i - 1) / bk;}for (i = 0; i < m; i++){scanf("%I64d%I64d", &qu[i].le, &qu[i].ri);qu[i].le--;qu[i].id = i;}}void solve(){ll i, j, id;sort(qu, qu + m, cmp);ll pr = -1, pl = 0;ans = 0;for (i = 0; i < m; i++){id = qu[i].id;if (qu[i].le == qu[i].ri){res[id] = 0;continue;}if (pr < qu[i].ri){for (j = pr + 1; j <= qu[i].ri; j++){add(j);}}else{for (j = pr; j > qu[i].ri; j--){del(j);}}pr = qu[i].ri;if (pl < qu[i].le){for (j = pl; j < qu[i].le; j++){del(j);}}else{for (j = pl - 1; j >= qu[i].le; j--){add(j);}}pl = qu[i].le;res[id] = ans;}for (i = 0; i < m; i++){printf("%I64d\n", res[i]);}}int main(){//freopen("i.txt", "r", stdin);//freopen("o.txt", "w", stdout);input();solve();return 0;}