Mishka and Interesting sum CodeForces - 703D(离线处理）

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.

Example
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
0
3
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 .

一个比较厉害的思维，边做便学。
最后这个题就是转化为如何求一个区间内有多少个不同的数的异或值（关键是这个）。看代码；

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include <time.h>
#include<fstream>
#define maxn 1000003
using namespace std;
struct seg
{
    int l,r,index;
    bool operator <(const seg& rhs)const{return r<rhs.r;}
};//n个询问的结构，
seg p[maxn];
int a[maxn],c[maxn],sum[maxn],pre[maxn],res[maxn];
int n;
int lowbit(int x)//下面三个函数是异或版的树状数组，当然线段树也是可以的
{
    return x&(-x);
}
void add(int x,int v)
{
    for(int i=x;i<=n;i+=lowbit(i))
        c[i]^=v;
}
int getSum(int x)
{
    int ans=0;
    for(int i=x;i>=1;i-=lowbit(i))
        ans^=c[i];
    return ans;
}
int main()
{

     map<int,int>mp;//这个用来映射一个数最新出现的坐标
     memset(c,0,sizeof(c));
     scanf("%d",&n);
     sum[0]=0;
     for(int i=1;i<=n;i++)//数组的输入
     {
         scanf("%d",&a[i]);
         sum[i]=sum[i-1]^a[i];//同时求各个点的前缀异或值，和前缀和的写法是一样的
         pre[i]=mp[a[i]];//这个是一个链状结构，在这个结构中，所对应的a数组的值都是一样的，也就是说map的作用就是这个
         mp[a[i]]=i;//更新最新位置
     }
     int m;
     scanf("%d",&m);
     for(int i=1;i<=m;i++)
     {
         scanf("%d%d",&p[i].l,&p[i].r);
         p[i].index=i;
     }//输入m个询问
     sort(p+1,p+1+m);//以r为标准从小到大排序
     int j=1;
     for(int i=1;i<=m;i++)
     {
        for(;j<=p[i].r;j++)//以每次查询的r为界限，把所有的值尽可能的往右更新，因为查询已经排过序了，所以不会出现我在某个区间上的一个值已经更新更大的坐标的这种情况
        {
            if(pre[j])add(pre[j],a[j]);//如果前面有节点，就把这个点置为0,就是只要在异或一次
            add(j,a[j]);//因为初始化每个点的值都是0，所有只要异或一次，
        }
        res[p[i].index]=sum[p[i].r]^sum[p[i].l-1]^getSum(p[i].r)^getSum(p[i].l-1);，求出这个区间上的不同的异或值
     }
     for(int i=1;i<=m;i++)
        printf("%d\n",res[i]);//把存好的答案输出即可
    return 0;
}