codeforces 617E. XOR and Favorite Number （莫队）

最新推荐文章于 2025-08-19 21:01:11 发布

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最新推荐文章于 2025-08-19 21:01:11 发布

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分类专栏：莫队

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本文介绍了一种使用莫队算法优化大量XOR查询的问题解决方法。通过预处理数组的异或前缀和，并利用离线查询技巧，有效地计算指定区间内元素异或等于特定值k的子数组对数量。

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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.

题解：莫队

cnt[i]表示异或前缀和为i的数的个数，那么答案就是cnt[i]*cnt[i^k]

k=0的时候需要特判,cnt[i]*(cnt[i]-1)/2

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#define N 1000003
#define LL long long 
using namespace std;
struct data{
	int l,r,id;
}q[N];
int n,m,k,a[N],cnt[N*10],c[N],belong[N];
LL ans[N],sum;
int cmp(data a,data b)
{
	return belong[a.l]<belong[b.l]||belong[a.l]==belong[b.l]&&a.r<b.r;
}
void change(int x,int val)
{
	int t=c[x];
	if (k) sum-=(LL)cnt[t]*cnt[k^t];
	else sum-=(LL)cnt[t]*(cnt[k^t]-1)/2;
	cnt[t]+=val;
	if (k) sum+=(LL)cnt[t]*cnt[k^t];
	else sum+=(LL)cnt[t]*(cnt[k^t]-1)/2;
}
int main()
{
	freopen("a.in","r",stdin);
	scanf("%d%d%d",&n,&m,&k);
	for (int i=1;i<=n;i++) 
	 scanf("%d",&a[i]),c[i]=c[i-1]^a[i];
	for (int i=1;i<=m;i++) {
	 scanf("%d%d",&q[i].l,&q[i].r),q[i].id=i;
	 q[i].l-=1;
    }
	int block=ceil(sqrt(n));
	for (int i=1;i<=n;i++) belong[i]=(i-1)/block+1;
	sort(q+1,q+m+1,cmp);
	int l=0,r=0; change(0,1);
	for (int i=1;i<=m;i++) {
		while (q[i].l<l) change(--l,1);
		while (q[i].l>l) change(l++,-1);
		while (q[i].r>r) change(++r,1);
		while (q[i].r<r) change(r--,-1);
		ans[q[i].id]=sum;
	}
	for (int i=1;i<=m;i++)
	 printf("%I64d\n",ans[i]);
 }