Given a sequence of $n$ numbers $a_1,a_2,\cdots ,a_n$ and three functions.

Define a function $f(l,r)$ which returns $\oplus a[x](l\le x\le r)$. The \oplus⊕ represents exclusive OR.

Define a function $g(l,r)$ which returns $\oplus f(x,y)(l\le x\le y\le r)\oplus f(x,y)$.

Define a function $w(l,r)$ which returns $\oplus g(x,y)(l\le x\le y\le r)\oplus g(x,y)$.

You are also given a number of xor-queries. A xor-query is a pair $(i,j)(1\le i\le j\le n)$. For each xor-query $(i,j)$, you have to answer the result of function $w(l,r)$.

Input
Line $1$: $t(1\le t\le 20)$.

For each test case:

Line $1$: $n(1\le n\le 100000)$.

Line $2$: $n$ numbers $a_1,a_2,\cdots ,a_n(1\le a_i\le 10^9)$

Line $3$: $q(1\le q\le 100000)$, the number of xor-queries.

In the next q lines, each line contains $2$ numbers $i,j$ representing a xor-query $(1\le i\le j\le n)$.

It is guaranteed that sum of $n$ and $q\le 10^6$

Output
For each xor-query $(i,j)$, print the result of function $w(i,j)$ in a single line.

样例输入
1
5
1 2 3 4 5
5
1 3
1 5
1 4
4 5
3 5
样例输出
2
4
0
1
4

解题思路：
emmmm，我这道题是通过打表寻找规律来解答的。。。
打表的代码

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
int arr[100];
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endiffreopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);for(int l1=1,r1=1;r1<=50;r1++) {memset(arr,0,sizeof arr);for(int l2=l1;l2<=r1;l2++) {for(int r2=l2;r2<=r1;r2++) {for(int l3=l2;l3<=r2;l3++) {for(int r3=l3;r3<=r2;r3++) {for(int k=l3;k<=r3;k++) {arr[k]++;}}}}}//输出的1代表最后的答案这项存在，0代表不存在for(int i=1;i<=r1;i++) {if(arr[i]%2==0) printf("0 ");else printf("1 ");}printf("\n");//例如假设区间为[l,r],第一行代表当区间长度为1时，其答案就是a[l]//当区间长度为2时,其答案为a[l]^a[r]//当区间长度为3时，其答案为a[l+1]//当区间长度为4时，其答案为0}return 0;
}

通过表格我们可以发现，
当$(r-l+1)\%4=0$时，答案为$0$；
当$(r-l+1)\%4=1$时，答案为$a[l]\oplus a[l+4]\oplus a[l+8]\cdots a[r]$
当$(r-l+1)\%4=2$时，答案为$a[l]\oplus a[l+1]\oplus a[l+4]\oplus a[l+5]\cdots a[r-1]\oplus a[r]$
当$(r-l+1)\%4=3$时，答案为$a[l+1]\oplus a[l+5]\oplus a[l+9]\cdots a[r-1]$
因为有$10^6$组查询，所以还需要求一个间隔为$4$的前缀异或和，这样就可以实现$O(1)$查询
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
int arr[100100];
int sum[4][100100];
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int t;scanf("%d",&t);while(t--) {int n;scanf("%d",&n);rep(i,1,n) {scanf("%d",&arr[i]);sum[0][i]=sum[1][i]=sum[2][i]=sum[3][i]=0;}sum[1][1]=arr[1];sum[2][2]=arr[2];sum[3][3]=arr[3];sum[0][4]=arr[4];rep(i,5,n) {sum[1][i]=sum[1][i-4]^arr[i];sum[2][i+1]=sum[2][i+1-4]^arr[i+1];sum[3][i+2]=sum[3][i+2-4]^arr[i+2];sum[0][i+3]=sum[0][i+3-4]^arr[i+3];}int q;scanf("%d",&q);rep(i,1,q) {int l,r;scanf("%d%d",&l,&r);int len=r-l+1,ans;if(len%4==0) printf("0\n");else if(len%4==1) {ans=sum[r%4][r]^sum[l%4][l]^arr[l];printf("%d\n",ans);}else if(len%4==2) {ans=sum[r%4][r]^sum[(l+1)%4][l+1]^arr[l+1];ans=ans^(sum[(r-1)%4][r-1]^sum[l%4][l]^arr[l]);printf("%d\n",ans);}else {ans=sum[(r-1)%4][r-1]^sum[(l+1)%4][l+1]^arr[l+1];printf("%d\n",ans);}}}return 0;
}