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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,?...,?a

n 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:

1. Two integers l and r
   (1?≤?l?≤?r?≤?n) are specified — bounds of query segment.
2. Integers, presented in array segment [l,??r] (in sequence of integers al,?al?+?1,?...,?ar) even number of times, are written down.
3. 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 x
   1,?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 Copy

3
3 7 8
1
1 3

output Copy

0

input Copy

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

output Copy

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 .

題解：

考慮異或的性質，那麽，對於一個區間出現偶數次數的異或和，等價於於求該區間的異或和?該區間所有不同數字的異或和（因為一個數異或奇數次等於原數，異或偶數次為0，你把該區間異或和異或上所有不同數字異或和，相當於所有數字出現次數加了1次後再求區間異或和）

而求區間異或和，我們可以像前綴和那樣，先預處理一個前綴異或和，對於每個詢問，剩下的問題就是如何搞該區間所有不同數字的異或和

在線不好做，所以可以先離線詢問，按區間右端點排序

離散化後，考慮使用一個last數組記錄每個數字最後的出現位置。開一個樹狀數組，再從左到右掃一遍，對於一個數a[i]，若last[a[i]]==0，則直接加入，反之，則證明之前a[i]出現過，再異或一次（即取消該last[a[i]]），然後遇到一個詢問的右端點就處理該詢問的答案

代碼如下：

#include<bits/stdc++.h>
#define MAXN 1000005
using namespace std;
struct qz{
    int l,r,ans,num;
}q[MAXN];
int n,m,pt=1;
int arr[MAXN],sxor[MAXN],node[MAXN],last[MAXN],tp[MAXN];
bool cmp(qz a,qz b)
{
    return a.r<b.r;
}
bool cmp1(qz a,qz b)
{
    return a.num<b.num;
}
int lowbit(int a)
{
    return a&(-a);
}
void add(int pos,int val)
{
    for(;pos<=n;pos+=lowbit(pos))
        node[pos]^=val;
}
int query(int l,int r)
{
    int ret=0;
    l--;
    for(;l!=0;l-=lowbit(l))
        ret^=node[l];
    for(;r!=0;r-=lowbit(r))
        ret^=node[r];    
    return ret;    
}
int main()
{
    memset(last,0,sizeof(last));
    scanf("%d",&n);
    sxor[0]=0;
    for(int i=1;i<=n;i++)
        {
            scanf("%d",&arr[i]);
            tp[i]=arr[i];
            sxor[i]=sxor[i-1]^arr[i];
        }
    sort(tp+1,tp+1+n);    
    scanf("%d",&m);    
    for(int i=1;i<=m;i++)
        {
             scanf("%d%d",&q[i].l,&q[i].r);
             q[i].num=i;
        }
    sort(q+1,q+1+m,cmp);
    for(int i=1;i<=m;i++)
        {
            while(pt<=q[i].r)
                  {
                        int r=lower_bound(tp+1,tp+1+n,arr[pt])-tp;
                      if(last[r]!=0)
                         add(last[r],arr[pt]);
                      last[r]=pt;
                      add(pt,arr[pt]);              
                        pt++;
                  }     
            q[i].ans=query(q[i].l,q[i].r)^sxor[q[i].l-1]^sxor[q[i].r];      
        } 
    sort(q+1,q+1+m,cmp1);
    for(int i=1;i<=m;i++)
        printf("%d\n",q[i].ans);
    return 0;  
}

View Code

CF703D Mishka and Interesting sum