D. Mishka and Interesting sum
memory limit per test: 256 megabytes
input: standard input
output: standard output
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 a**l, a**l + 1, …, a**r) 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, …, x**k, 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.
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, …, a**n (1 ≤ a**i ≤ 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.
Print m non-negative integers — the answers for the queries in the order they appear in the input.
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 .
给你一个数组，若干个询问。询问区间 [l, r] 上出个次数为偶数的数的异或和。
对于区间所有数字的异或和，我们可以先预处理出一个前缀异或和 $sum$，区间$[l,r]$ 上所有数字的异或和即为
sum[r] ^ sum[l-1]
区间上distinct数字的异或和我们用树状数组来维护。用一个 map 来维护某个数上次出现的位置，离线处理答案，走到 $a[i]$ 时，如果 $map[ a[i] ]$ 不为0，（说明 $a[i] $ 之前出现过）， 那么从上次的位置开始把 $a[i]$ 异或掉，并在此位置开始异或上 $a[i]$，更新map。只把出现多次的值放在最后出现的位置（这个思想很常见）。