Codeforces Round #340 (Div. 2) E. XOR and Favorite Number【莫队算法】

E. XOR and Favorite Number
time limit per test4 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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：给你一个序列，K次询问，每次给出一个L，R；询问，从L到R之间有多少个子区间满足异或和为K；

分析：题目给的1e6，但是实际异或操作会超过1e6。区间暴力问题，莫队！
Ans和flag注意使用long long int。

时间复杂度：O(n^1.5)

学习资料参考于：bilibili 莫队算法 卿学姐。

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<algorithm>using namespace std;#define ll long long intconst int maxn = 1 << 20;int a[maxn]; int n, m, k;ll ans[maxn], flag[maxn];int pos[maxn];struct node {    int l, r;    int id;}Q[maxn];bool cmp(node a, node b) {    if (pos[a.l] == pos[b.l])        return a.r < b.r;    return pos[a.l] < pos[b.l];}int L = 1, R = 0;ll Ans = 0;void add(int x) {    Ans += flag[a[x] ^ k];    flag[a[x]]++;}void del(int x) {    flag[a[x]]--;    Ans -= flag[a[x] ^ k];}int main(){    scanf("%d%d%d", &n, &m, &k);    int sz = sqrt(n);    for (int i = 1; i <= n; i++) {        scanf("%d", &a[i]);        a[i] = a[i] ^ a[i - 1];        pos[i] = i / sz;    }    for (int i = 1; i <= m; i++) { //离线操作        scanf("%d%d", &Q[i].l, &Q[i].r);        Q[i].id = i;    }    sort(Q + 1, Q + 1 + m, cmp);    flag[0] = 1;    for (int i = 1; i <= m; i++) {        while (L < Q[i].l) {            del(L - 1);            L++;        }        while (L > Q[i].l) {            L--;            add(L - 1);        }        while (R < Q[i].r) {            R++;            add(R);        }        while (R > Q[i].r) {            del(R);            R--;        }        ans[Q[i].id] = Ans;    }    for (int i = 1; i <= m; i++) {        printf("%lld\n", ans[i]);    }    return 0;}