【codeforces 703 D】【离线询问 树状数组 前驱思想 前缀异或和】D. Mishka and Interesting sum【 区间内出现次数偶数的数的异或和】

传送门：D. Mishka and Interesting sum

描述：

D. Mishka and Interesting sum

time limit per test
3.5 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Little Mishka enjoys programming. Since her birthday has just passed, her friends decided to present her with array of non-negative integers a1, a2, ..., an of n elements!

Mishka loved the array and she instantly decided to determine its beauty value, but she is too little and can't process large arrays. Right because of that she invited you to visit her and asked you to process m queries.

Each query is processed in the following way:
Two integers l and r (1 ≤ l ≤ r ≤ n)
are specified — bounds of query segment.
Integers, presented in array segment [l,  r] (in
sequence of integers al, al + 1, ..., ar) even
number of times, are written down.
XOR-sum of written down integers is calculated, and this value is the answer for a query. Formally, if integers written down in point 2 are x1, x2, ..., xk,
then Mishka wants to know the value 

,
where 

 —
operator of exclusive bitwise OR.

Since only the little bears know the definition of array beauty, all you are to do is to answer each of queries presented.

Input

The first line of the input contains single integer n (1 ≤ n ≤ 1 000 000) —
the number of elements in the array.

The second line of the input contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109) —
array elements.

The third line of the input contains single integer m (1 ≤ m ≤ 1 000 000) —
the number of queries.

Each of the next m lines describes corresponding query by a pair of integers l and r (1 ≤ l ≤ r ≤ n) —
the bounds of query segment.

Output

Print m non-negative integers — the answers for the queries in the order they appear in the input.

Examples

input

3
3 7 8
1
1 3

output

0

input

7
1 2 1 3 3 2 3
5
4 7
4 5
1 3
1 7
1 5

output

03
1
3
2

Note

In the second sample:

There is no integers in the segment of the first query, presented even number of times in the segment — the answer is 0.

In the second query there is only integer 3 is presented even number of times — the answer is 3.

In the third query only integer 1 is written down — the answer is 1.

In the fourth query all array elements are considered. Only 1 and 2 are
presented there even number of times. The answer is 

.

In the fifth query 1 and 3 are
written down. The answer is 

.

题意：

有一个长度为n（1e6）的数组a[]，权值范围为[1,1e9]

有m（1e6）个询问，对于每个询问[l,r]

我们要使得在区间[l,r]范围内的，

出现次数为偶数次的权值做异或和 

思路：

题目没有修改操作，那么可以考虑离线的做法，考虑到异或有一个性质a^a=0，即出现次数为偶数的话那么异或起来为0，假设有一个序列为1,2,1,3,2的话，如果我查询区间[1,5]，那么可以留意到把这个区间所有的数1^2^1^3^2算出来，然后再异或这个区间的不同的数字1^2^3，那么最后结果就是1^2，恰好就是出现偶数次的数字的异或和，那么就可以预处理出所有的前缀异或和。

那么如何快速计算一个区间内不同的数的异或和呢？离线处理,结构体存储每个查询区间的左右边界，按照右边界排序，从左向右遍历序列 树状数组维护不断的将数添加到树状数组，若当前位置的数存在前驱，则删除前驱 (删除就是再进行一次异或a^a=0)，最后的目的就是保证每个数只保留最后一个进行异或，对于共右边界的查询区间 一次遍历得到答案，然后继续遍历。

PS:这题的弱化版就是求区间中不同的数的异或和，就和HDU3333的求求区间中不同的数的和差不多，题解见Here

复杂度：

O(mlongn)


代码：

#include <bits/stdc++.h>
#define pr(x) cout << #x << "= " << x << " " ；
#define pl(x) cout << #x << "= " << x << endl;
#define ll __int64
using namespace std;

template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
const int N=1e6+10;
struct node{
int l,r,id;
bool operator < (const node& tmp) const {
return r<tmp.r;
}
}q
;
int ans
,a
;
map<int, int>vis;
int pre
;
int bit
,sum
;
int n,m;

void update(int i,int x){
while(i<=n){
bit[i]^=x;
i+=i&-i;
}
}

int query(int i){
int s=0;
while(i>0){
s^=bit[i];
i-=i&-i;
}
return s;
}

int main(){
read(n);
for(int i=1; i<=n; i++){
read(a[i]);
sum[i]=sum[i-1]^a[i];
pre[i]=vis[a[i]];//前驱思想
vis[a[i]]=i;
}
read(m);
for(int i=1; i<=m; i++){
read(q[i].l);read(q[i].r);
q[i].id=i;
}
sort(q+1, q+m+1);
for(int i=1,r=1; i<=m; i++){
while(r<=q[i].r){
if(pre[r])
update(pre[r], a[r]);//删除前驱
update(r, a[r]);//添加最右边的数
r++;
}
ans[q[i].id]=query(q[i].r)^query(q[i].l-1)^sum[q[i].r]^sum[q[i].l-1];
}
for(int i=1; i<=m; i++)printf("%d\n", ans[i]);
return 0;
}