codeforces 617E 莫队算法

又是玄之又玄的bug改到深夜。

先附上代码和题目吧

题目：

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 pairli andri
and asks you to count the number of pairs of integersi andj, such thatl ≤ i ≤ j ≤ r and the xor of the numbersai, ai + 1, ..., aj
is equal tok.

Input

The first line of the input contains integers n,m andk (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 andri (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.

Example

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 andj 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

大致题意是：给你一段区间，让你求L，R区间有多少个子序列的异或值为k。莫队算法的思路http://blog.csdn.net/hzj1054689699/article/details/51866615和http://blog.csdn.net/bossup/article/details/39236275已经讲的非常详细了，我肯定没dalao讲的好啊，就不说了，主要提醒的一点是注意排序要分块，我到现在也不知道为什么不分块就会超时，仍然没有参透。哪天参透了再说吧，哈哈哈，分块大法好，关键时刻把命保

代码如下：

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const int maxn = 1<<20;
int n,m,k,nn;
struct node{
int l,r,id;
};
node Q[maxn];
LL ans[maxn],ANS;
LL flag[maxn];
int sum[maxn];
int cmp(node a,node b){
if((a.l/nn) == (b.l/nn))return a.r<b.r;
return (a.l/nn)<(b.l/nn);
}
void del(int rt){
flag[sum[rt]]--;
ANS-=flag[sum[rt]^k];
}
void add(int rt){
ANS+=flag[sum[rt]^k];
flag[sum[rt]]++;
}
int main(){
scanf("%d%d%d",&n,&m,&k);
nn=sqrt(n);
memset(flag,0,sizeof(flag));
sum[0]=0;
for(int i=1;i<=n;i++){
scanf("%d",&sum[i]);
sum[i]=sum[i]^sum[i-1];
}
//    for(int i=1;i<=n;i++){
//        printf("%d ",sum[i]);
//    }
//    printf("\n");
for(int i=1;i<=m;i++){
scanf("%d%d",&Q[i].l,&Q[i].r);
// Q[i].l--;
Q[i].id=i;
}
sort(Q+1,Q+m+1,cmp);
memset(ans,0,sizeof(ans));
ANS=0;
int L=1,R=0;
flag[0]=1;
for(int i=1;i<=m;i++){
//Q[i].l--;
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]);
}
}