XOR
Consider an array A with n elements. Each of its element is A[i] (1 ≤ i ≤ n). Then gives two integers
Q, K, and Q queries follow. Each query, give you L, R, you can get Z by the following rules.
To get Z, at first you need to choose some elements from A[L] to A[R], we call them A[i1], A[i2],
. . . , A[it], Then you can get number Z = K or (A[i1], A[i2], . . . , A[it]).
Please calculate the maximum Z for each query .
Input
Several test cases.
First line an integer T (1 ≤ T ≤ 10). Indicates the number of test cases.
Then T test cases follows. Each test case begins with three integer N, Q, K (1 ≤ N ≤ 10000,1 ≤
Q ≤ 100000, 0 ≤ K ≤ 100000). The next line has N integers indicate A[1] to A[N] (0 ≤ A[i] ≤ 108).
Then Q lines, each line two integer L, R (1 ≤ L ≤ R ≤ N).
Output
For each query, print the answer in a single line.
Sample Input
1
5 3 0
1 2 3 4 5
1 3
2 4
3 5
Sample Output
3
7
7
题意：t组数据，然后输入n，k，q，接着给出一个1个长度为n的数组,q个询问，对于每个询问，询问在下标为[l,r]的数中，选取一部分数，使得其异或值再OR上k后最大，输出这个最大值。
思路：根据题意，选取一部分值得到异或最大值，可以想到线性基，但是最后要OR上k，所以要消除k对其影响，我们就把每个数转化成二进制，然后将k为1的位置对于每个数其位置就变为0，这样就可以消除其k的影响了，最后在OR上k变回来，即为正确答案。比如数a[i]=6(110) ,k = 4(100),则a[i]应该变为(010)=2这样来消除k的影响。由于多次区间询问，所以用线段树维护一下即可。
AC代码：
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