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.
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：
给了一个长度为N的序列和M次询问，每次询问有多少个满足
使得a[l]^...^a[r]为k。
思路：
考虑到异或的一个性质，a[1]^....^a[l-1]^a[1]^...^a[r]=a[l]^...a[r]。
所以可以先处理出一个异或的前缀和。
接下来问题变成了：
a[l-1]^a[r]=k，满足这个的l和r有多少个。
两边同异或一个值，
a[l-1]=k^a[r] 或 a[r]=k^a[l-1]
然后用莫队处理一下就好了。
注意：
1.在添加和删除时，对前缀和的记录的值，不应该影响原来的答案。
2.l在移动的时候，注意移动后l-1的值要存在或消失。
代码：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn=2e6;
typedef long long ll;
int a[maxn];//异或的前缀和
int pos[maxn];
ll sum[maxn];
ll res[maxn];
ll ans;
int n,m,k;
struct node
{
int l,r,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];
}
void add(int x)
{
ans+=sum[a[x]^k];//加的另一端点前缀和的次数
sum[a[x]]++;//加的x端点前缀和出现次数
}
void del(int x)
{
sum[a[x]]--;//与add的顺序相反才能把这点对答案的贡献移除，对结果不影响
ans-=sum[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++)
{
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;
sum[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--;
res[q[i].id]=ans;
}
for(int i=1;i<=m;i++)
cout<<res[i]<<endl;
return 0;
}