Given a sequence of $n$ numbers $a_1,a_2,?,a_n$ 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 $a_1,a_2,?,a_n(1\le a_i\le 10^9)$

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.

樣例輸入
1
5
1 2 3 4 5
5
1 3
1 5
1 4
4 5
3 5
樣例輸出
2
4
0
1
4

解題思路：
emmmm，我這道題是通過打表尋找規律來解答的。。。
打表的代碼

#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]++;}}}}}//輸出的1代表最后的答案這項存在，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");//例如假設區間為[l,r],第一行代表當區間長度為1時，其答案就是a[l]//當區間長度為2時,其答案為a[l]^a[r]//當區間長度為3時，其答案為a[l+1]//當區間長度為4時，其答案為0}return 0; }

通過表格我們可以發現，
當$(r?l+1)\%4=0$時，答案為$0$；
當$(r?l+1)\%4=1$時，答案為$a[l]\oplus a[l+4]\oplus a[l+8]?a[r]$
當$(r?l+1)\%4=2$時，答案為$a[l]\oplus a[l+1]\oplus a[l+4]\oplus a[l+5]?a[r?1]\oplus a[r]$
當$(r?l+1)\%4=3$時，答案為$a[l+1]\oplus a[l+5]\oplus a[l+9]?a[r?1]$
因為有$10^6$組查詢，所以還需要求一個間隔為$4$的前綴異或和，這樣就可以實現$O(1)$查詢
代碼：

#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; }

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