Codeforces 617E - XOR and Favorite Number(前缀异或 + 莫队)

E. XOR and Favorite Number

time limit per test4 seconds
memory limit per test256 megabytes
input standard input
outpu tstandard 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 pair l i and r i 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 a i, a i + 1, …, a j 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 a i (0 ≤ a i ≤ 1 000 000) — Bob’s array.

Then m lines follow. The i-th line contains integers l i and r 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.

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.

题目大意:

输入n m k ,然后第二行输入n个数,再输入m行数,每行输入两个数l ,r 表示有n个数,m个询问,对于每个询问,你应该输出区间[l, r] 中ai ^ … ^ aj = k 这样的i, j 有多少对。

解题思路:

一道很好的莫队算法题,只有查询没有修改,可以离线用莫队去做,我们需要输出有几对,开ans 数组表示答案,belong 数组表示每个数属于哪个块(莫队需要把n 个数分块,每块根号n),结构体q存每次的询问,cnt[i] 表示前缀异或值为 i 的数出现了几次,套用莫队算法模板,改一下add和del函数。
因为求的是ai ^… ^ aj == k,我们可以维护一个前缀异或,a[j] ^ a[i - 1] 即为 ai ^… ^ aj(这里的数组a表示前缀异或,和前面的意义不同) ,异或还有一个性质,我们要求的是a[i - 1] ^ a[j] = k 那么 a[j] ^ k = a[i - 1],对于每次答案的更新,实质上就是加上a[j] 和 k 去找对应匹配的a[i - 1]的值出现的次数。也就是说 每次 now += cnt[a[x] ^ k] ,del同理。要注意的是这里cnt[0] 要初始化为 1,拿样例来说,1 2 1 1 0 3 区间 1 6 ,前两个数的异或值为3,这时 cnt[a[2] ^ 3] 而a[2] ^ 3 的值为0,实际上他的值应该为 1。之后套用莫队的模板即可 AC。

Code:

#pragma GCC optimize(2)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
const int N = 5e6 + 50;
ll cnt[N], a[N], belong[N], ans[N], now = 0;
int n, m, k;
struct ask { int l, r, id; }q[N];
bool cmp(ask a, ask b)
{
	return belong[a.l] ^ belong[b.l] ? belong[a.l] < belong[b.l] : ((belong[a.l] & 1) ? a.r < b.r : a.r > b.r);
}
void add(int x)
{
	now += cnt[a[x] ^ k];//这里要先更新now 再++,防止与a[x] 和 a[x] 进行匹配,匹配结果会出错,因为a[l - 1] ^ a[r] = a[l] ^ a[r] , a[x] a[x] 不合法
	cnt[a[x]] ++;
}
void del(int x)
{
	cnt[a[x]] --;
	now -= cnt[a[x] ^ k];
}
int main()
{
	scanf("%d%d%d", &n, &m, &k);
	for (int i = 1; i <= n; i ++)//输入的同时初始化前缀异或数组
	{
		scanf("%d", &a[i]);
		a[i] = a[i - 1] ^ a[i];
	}
	double block = sqrt(n);//数据分块
	int num = ceil(n / block);
	for (int i = 1; i <= num; i ++)
		for (int j = (i - 1) * block + 1; j <= i * block; j ++)
			belong[j] = i;
	for (int i = 1; i <= m; i ++)
	{
		scanf("%d%d", &q[i].l, &q[i].r);
		q[i].id = i;
	}
	sort(q + 1, q + 1 + m, cmp);
	int l = 1, r = 0;
	cnt[0] = 1;
	for (int i = 1; i <= m; i ++)
	{
		int ql = q[i].l, qr = q[i].r;
		while(l < ql) del(l - 1), l ++;
		while(l > ql) l --, add(l - 1);
		while(r < qr) r ++, add(r);
		while(r > qr) del(r), r --;
		ans[q[i].id] = now;
	}
	for (int i = 1; i <= m; i ++)
		cout << ans[i] << endl;
	return 0;
}

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