Codeforces 617E XOR and Favorite Number (区间异或和 莫队算法 分块暴力)

E. XOR and Favorite Number

time limit per test
4 seconds

memory limit per test
256 megabytes

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.

题目链接：http://codeforces.com/problemset/problem/617/E

题目大意：求给定区间的子区间异或和等于k的子区间个数

题目分析：对于l到r的异或和根据异或的性质可以通过sum[l - 1] ^ sum[r]得到，sum是前缀异或和，预处理完前缀先把左区间减1方便后面的运算，题目的运算满足在知道[l,r]的条件下可以O(1)得到[l-1, r]，[l+1, r]，[l, r-1]，[l, r+1]，因此可以用莫队算法，一种分块优化的思想，先记录要查询的区间然后排序，这样每次只需要控制两端端点的移动即可更新答案，因为已经排过序，所以端点移动的次数比无序时大大减少，时间复杂为O(n * sqrt(n))，不做证明，再回到这题，根据异或性质sum[l
- 1] ^ sum[r] == k ==> sum[r] ^ k = sum[l - 1]，我们可以用一个cnt动态记录sum[l - 1]出现的次数即可

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#define ll long long
using namespace std;
int const MAXN = 100005;
int const MAXM = 10000005;
int const CON = sqrt(MAXN);
ll sum[MAXN], a[MAXN], ans[MAXN], num;
ll cnt[MAXM];
int n, m, k;

struct DATA
{
int l, r, id;
}d[MAXN];

bool cmp(DATA a, DATA b)
{
if(a.l / CON == b.l / CON)
return a.r / CON < b.r / CON;
return a.l / CON < b.l / CON;
}

void Add(int x) //加的时候先看其之前出现了多少个sum[l - 1]，再更新当前点
{
num += cnt[x ^ k];
cnt[x] ++;
}

void Sub(int x) //减的时候先去掉该点再看其之前sum[l - 1]出现的次数，和加时对应
{
cnt[x] --;
num -= cnt[x ^ k];
}

int main()
{
scanf("%d %d %d", &n, &m, &k);
for(int i = 1; i <= n; i++)
{
scanf("%I64d", &a[i]);
sum[i] = sum[i - 1] ^ a[i];
}
for(int i = 0; i < m; i++)
{
scanf("%d %d", &d[i].l, &d[i].r);
d[i].id = i;
d[i].l -= 1;
}
sort(d, d + m, cmp);
int l = 0, r = 0;
num = 0;
cnt[0] = 1;
for(int i = 0; i < m; i++)
{
while(l < d[i].l)
Sub(sum[l ++]);
while(l > d[i].l)
Add(sum[-- l]); //考虑到边界，左端点始终更新前一个
while(r < d[i].r)
Add(sum[++ r]); //右端点始终更新后一个
while(r > d[i].r)
Sub(sum[r --]);
ans[d[i].id] = num;
}
for(int i = 0; i < m; i++)
printf("%I64d\n", ans[i]);
}